The mass–spring oscillator and the pendulum
Why does every swing, pendulum and body bobbing on a spring oscillate at its own pace, and what determines that pace?
Prior knowledge
- The period $T$ is the duration of one cycle, the frequency is $f = \tfrac{1}{T}$ and the amplitude is the greatest displacement from the equilibrium positionchapter Describing oscillation: period, frequency and amplitude
- A harmonic force $F = -kx$ points towards the equilibrium position, and the elastic potential energy of a spring is $\tfrac{1}{2}kx^2$chapter Spring force and spring energy
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 7.
Think about it
In a playground there are two swings side by side with chains of the same length. A small child swings on one and her father on the other. On the edge of the sandpit there is a toy in which a wooden bird bobs up and down on a spring, and the child pulls it a little at one moment and a lot at another. Each of these oscillates back and forth at its own pace.
Which quantities determine how long one oscillation back and forth takes, and which do not affect it at all?
Optional Recall: the harmonic force
Predict
Make your predictions before investigating 2 questions
Prediction 1: a larger oscillation
Prediction 2: a heavier pendulum bob
Investigate
Choose an oscillator, adjust its quantities and start it. Use the graph to see how long one cycle takes.
Simulation
How to use
Choose Oscillator: a mass on a spring or a pendulum. Use the sliders to set the mass and the spring constant or the length, and the amplitude or the starting angle, and press Start. The body starts from rest at an extreme position. The graph plots the position $x$ or the angle $\alpha$ as a function of time, and the label $T$ shows the interval between the two most recent extreme positions on the same side. Above the graph you see the stopwatch time $t$ and the number of complete cycles $n$. The bar under the canvas shows how the energy is divided between kinetic energy $E_\text{k}$ and potential energy $E_\text{p}$. For the spring, the lengths of the arrows are relative to the greatest value of each quantity during the motion; for the pendulum, the arrows are drawn to the same scale in all settings, so in a small swing they are short. Speed changes only the pace of the animation, not the period. When you change a value, the body returns to its starting position.
Assumptions of the model
The surface of the mass on a spring is frictionless, and the only force along the motion is the harmonic force of the spring, $F = -kx$. The pendulum is a simple pendulum: a point mass on a massless string without air resistance, and its motion is calculated from the exact equation also at large angles, so the small-angle approximation has not been made. There is no damping, so the energy is conserved. The period, the frequency and the energies are values calculated by the model, not measurements.
How the calculation works
The motion is calculated in small time steps: at each step the acceleration is obtained from the force ($a = -\tfrac{k}{m}x$ for the spring and the tangential acceleration $a_\text{t} = -g\sin\alpha$ for the pendulum), and the speed and the position are updated from it. The period $T$ is the time measured from the motion, from the starting moment until the body has again stopped at the same extreme position; it is not calculated from the formula for the period. The frequency is $f = \tfrac{1}{T}$. The energy is $E = \tfrac{1}{2}kA^2$ for the spring and $E = mgl(1 - \cos\alpha_0)$ for the pendulum, when the zero level of the potential energy is at the lowest point of the pendulum; $g = 9.81\,\text{m/s}^2$.
Experiment 1: only the amplitude changes
Test prediction 1. The mass of the body and the spring constant stay constant; change only the amplitude. Record the period, the greatest speed and the energy with at least the amplitudes $0.05\,\text{m}$ and $0.10\,\text{m}$, and if you like also $0.20\,\text{m}$. Start the oscillator from time to time and see how long a cycle is in the graph.
Initial state: oscillator: mass on a spring, mass $m$ = $0{.}5\,\text{kg}$, spring constant $k$ = $20\,\text{N/m}$, amplitude $A$ = $0{.}05\,\text{m}$. Kept constant: oscillator, mass $m$, spring constant $k$. Varied: amplitude $A$.
| amplitude $A$ | $T$ $(\text{s})$ | $v_\text{max}$ $(\text{m/s})$ | $E$ $(\text{J})$ |
|---|
Experiment 2: only the mass changes
The spring constant and the amplitude stay constant; change only the mass of the body. Record the period with at least the masses $0.50\,\text{kg}$ and $2.00\,\text{kg}$, and if you like also $0.20\,\text{kg}$ and $0.80\,\text{kg}$.
Initial state: oscillator: mass on a spring, mass $m$ = $0{.}5\,\text{kg}$, spring constant $k$ = $20\,\text{N/m}$, amplitude $A$ = $0{.}1\,\text{m}$. Kept constant: oscillator, spring constant $k$, amplitude $A$. Varied: mass $m$.
| mass $m$ | $T$ $(\text{s})$ |
|---|
Experiment 3: the mass of the pendulum changes
Test prediction 2. The length of the pendulum is $1.00\,\text{m}$ and the starting angle is $10^\circ$; change only the mass of the bob. Record the period and the energy with at least the masses $0.10\,\text{kg}$ and $1.00\,\text{kg}$.
Initial state: oscillator: pendulum, mass $m$ = $0{.}1\,\text{kg}$, length $l$ = $1\,\text{m}$, starting angle $\alpha_0$ = $10^\circ$. Kept constant: oscillator, length $l$, starting angle $\alpha_0$. Varied: mass $m$.
| mass $m$ | $T$ $(\text{s})$ | $E$ $(\text{J})$ |
|---|
Experiment 4: the length of the pendulum changes
The mass of the bob is $0.50\,\text{kg}$ and the starting angle is $10^\circ$; change only the length of the pendulum. Record the period with at least the lengths $0.25\,\text{m}$ and $1.00\,\text{m}$. By what factor does the period change when the length is quadrupled?
Initial state: oscillator: pendulum, mass $m$ = $0{.}5\,\text{kg}$, length $l$ = $0{.}25\,\text{m}$, starting angle $\alpha_0$ = $10^\circ$. Kept constant: oscillator, mass $m$, starting angle $\alpha_0$. Varied: length $l$.
| length $l$ | $T$ $(\text{s})$ |
|---|
Optional Extension: large swings
The length of the pendulum is $1.00\,\text{m}$; change only the starting angle. Record the period with, for example, the angles $5^\circ$, $20^\circ$, $40^\circ$ and $60^\circ$, and compare it with the small-angle value $2\pi\sqrt{\tfrac{l}{g}} = 2.006\,\text{s}$. At what angle does the difference start to show?
Initial state: oscillator: pendulum, mass $m$ = $0{.}5\,\text{kg}$, length $l$ = $1\,\text{m}$, starting angle $\alpha_0$ = $5^\circ$. Kept constant: oscillator, mass $m$, length $l$. Varied: starting angle $\alpha_0$.
| starting angle $\alpha_0$ | $T$ $(\text{s})$ |
|---|
What does your table tell you about the period of a pendulum at large starting angles? Why does the period increase at large angles? When can the formula $T = 2\pi\sqrt{\tfrac{l}{g}}$ be used?
Model answer
At small angles the period is almost the same: $2.007\,\text{s}$ at $5^\circ$ and $2.010\,\text{s}$ at $10^\circ$, less than $0.2\,\%$ above the value $2.006\,\text{s}$ given by the formula. At large angles the period increases: $2.021\,\text{s}$ at $20^\circ$, $2.069\,\text{s}$ at $40^\circ$ and $2.153\,\text{s}$ at $60^\circ$, about $7\,\%$ above. The restoring force $mg\sin\alpha$ remains smaller at large angles than in the harmonic model, so the pendulum returns more slowly. The formula can be used when the swings are small, below about $10^\circ$. The values are calculated by the model.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Conclusion: what does the period depend on?
Write, on the basis of your tables: which quantities changed the period of the mass on a spring and which did not? What about the pendulum? By what factor did the period change when the mass or the length was quadrupled? Why does the mass matter for the spring but not for the pendulum?
Model answer
The period of the mass on a spring stayed at $0.993\,\text{s}$ when the amplitude increased from $0.05\,\text{m}$ to $0.10\,\text{m}$; the greatest speed doubled and the energy quadrupled. When the mass quadrupled from $0.50\,\text{kg}$ to $2.00\,\text{kg}$, the period doubled to $1.99\,\text{s}$. The period of the pendulum was $2.01\,\text{s}$ for all the masses, but when the length quadrupled from $0.25\,\text{m}$ to $1.00\,\text{m}$, the period doubled from $1.00\,\text{s}$ to $2.01\,\text{s}$. So the period is proportional to the square root of the mass or of the length. For the spring the mass matters because the force of the spring does not grow with the mass; for the pendulum the restoring force is a part of the weight and grows with the mass. The values are calculated by the model, without friction and air resistance.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
In the chapter Spring force and spring energy the force of a spring was $F = -kx$: it is directly proportional to the displacement $x$ from the equilibrium position and always points towards the equilibrium position. When a body is pulled aside and released, the force accelerates it towards the equilibrium position. At the equilibrium position the force is zero, but the body has speed, so it carries on to the other side. There the force slows it down, stops it at the extreme position and turns it back. The motion repeats itself in the same way: the body oscillates. When the oscillation is caused by a harmonic force, the motion is called simple harmonic motion.
The figure shows the rhythm of the motion. At the extreme positions $x = A$ and $x = -A$ the speed is zero, but the force and the acceleration are greatest. At the equilibrium position the situation is the opposite: the force and the acceleration are zero and the speed is greatest. The acceleration always points towards the equilibrium position, in the same direction as the force.
Position as a function of time
When the position of a simple harmonic oscillator is plotted as a function of time, the result is a sine-shaped curve. The concepts of the chapter Describing oscillation: period, frequency and amplitude can be seen in it directly. The amplitude $A$ is the greatest displacement from the equilibrium position. The period $T$ is the time in which the body makes one complete oscillation back and forth and returns to the same point moving in the same direction, for example the interval between two successive peaks. The frequency $f = \tfrac{1}{T}$ tells us the number of oscillations per second.
Why the amplitude does not change the period
A simple harmonic oscillator has a remarkable property. If the amplitude is doubled, the distance travelled in one cycle doubles. At the same time the force at every corresponding point also doubles, because $F$ is proportional to the displacement. A doubled force gives a doubled acceleration, and in the same time a doubled acceleration makes the speed grow twice as much. So the longer distance is covered at a greater speed, and the time taken is the same. The period of a simple harmonic oscillator does not depend on the amplitude. It depends only on how large the mass in motion is and how stiff the spring that restores it is.
Period of a simple harmonic oscillator
Mass on a spring
When a body oscillates on a spring and the only force along the motion is the harmonic force of the spring, the period is
$$T = 2\pi\sqrt{\frac{m}{k}}$$
where $T$ is the period ($\text{s}$), $m$ the mass of the oscillating body ($\text{kg}$) and $k$ the spring constant of the spring ($\text{N/m}$). A greater mass accelerates more slowly under the same force, so the period grows as the mass grows. A stiffer spring gives a greater force, so the period becomes shorter. The square root means that the mass must be quadrupled to double the period. The formula holds for a spring hanging vertically too: the weight only shifts the equilibrium position, and the displacement from it obeys the same law.
The formula can be justified with circular motion. When a body moves uniformly around a circle of radius $A$, its shadow on a wall in the direction of a diameter moves back and forth. The acceleration of the shadow is a component of the centripetal acceleration $a_\text{n} = \tfrac{4\pi^2 A}{T^2}$, and it is $-\tfrac{4\pi^2}{T^2}x$, that is, proportional to the displacement. For a mass on a spring Newton's second law gives $a = -\tfrac{k}{m}x$. When the coefficients are set equal, $\tfrac{4\pi^2}{T^2} = \tfrac{k}{m}$, and the period can be solved from this.
Simple pendulum
In a simple pendulum a point mass hangs on a light string and swings without air resistance. At the extreme position the speed is zero, and the tension $\overline{S}$ of the string cancels the component of the weight along the string. The symbol $S$ is used here for the tension, because $T$ is the period. The motion is driven by the component of the weight along the path, $G_\text{t} = mg\sin\alpha$, which points towards the equilibrium position.
For small angles $\sin\alpha$ is approximately the angle in radians, and the arc length from the equilibrium position is $s = l\alpha$. The restoring force is then $\tfrac{mg}{l}s$, which is proportional to the displacement, and the pendulum is a simple harmonic oscillator with "spring constant" $\tfrac{mg}{l}$. When this is substituted into the formula for the spring, the mass cancels. For small swings the period of a simple pendulum is
$$T = 2\pi\sqrt{\frac{l}{g}}$$
where $T$ is the period ($\text{s}$), $l$ the length of the pendulum, that is, the distance from the point of suspension of the string to the centre of mass of the body ($\text{m}$), and $g$ the acceleration of free fall ($9.81\,\text{m/s}^2$). The period does not depend on the mass of the body, nor, for small swings, on the amplitude.
The formula is an approximation that holds for small swing angles. As the starting angle grows, $\sin\alpha$ remains smaller than the angle in radians, the restoring force remains smaller than the harmonic one and the period increases slightly. The table gives the true period $T$ relative to the value $T_0$ given by the formula:
| Starting angle | $T : T_0$ |
|---|---|
| $10^\circ$ | $1.002$ |
| $20^\circ$ | $1.008$ |
| $40^\circ$ | $1.031$ |
| $60^\circ$ | $1.073$ |
For swings of less than $10^\circ$ the difference is less than $0.2\,\%$, which is smaller than the uncertainty of timing by hand.
Energy in oscillation
When friction and air resistance are neglected, the mechanical energy of the oscillator is conserved. The energy of a mass on a spring is the sum of the kinetic energy and the elastic potential energy of the spring. At the extreme position the speed is zero and all the energy is potential energy, so
$$\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2$$
where $m$ is the mass ($\text{kg}$), $v$ the speed ($\text{m/s}$), $k$ the spring constant ($\text{N/m}$), $x$ the displacement from the equilibrium position ($\text{m}$) and $A$ the amplitude ($\text{m}$). The right-hand side is the energy of the oscillation, $E = \tfrac{1}{2}kA^2$. At the equilibrium position $x = 0$, so all the energy is kinetic energy and the speed is greatest. In a pendulum the energy changes in the same way between gravitational potential energy and kinetic energy.
Common misconception
Many people think that a heavy pendulum bob swings faster, because a greater weight acts on it. A greater force does act on a heavier bob, but it is also harder to accelerate: the acceleration $a = \tfrac{F}{m}$ stays the same, as in free fall. That is why the period of a pendulum does not depend on the mass. For a mass on a spring, however, the mass does matter, because the force of the spring does not grow with the mass.
Another mistake is to think that a bigger swing takes longer. For a simple harmonic oscillator the longer distance is covered at a greater speed, so the period stays the same; for a pendulum this holds for small swings. A third mistake is to confuse the cycle with half a cycle: one cycle is a motion back and forth that ends at the same extreme position where it started.
Summary
- A harmonic force $F = -kx$ makes a body perform simple harmonic motion: position as a function of time is a sine-shaped curve with amplitude $A$ and period $T$.
- The period of a mass on a spring is $T = 2\pi\sqrt{\tfrac{m}{k}}$: it increases as the mass increases and decreases as the spring gets stiffer, but it does not depend on the amplitude.
- The period of a simple pendulum with small swings is $T = 2\pi\sqrt{\tfrac{l}{g}}$: it depends on the length and the acceleration of free fall, not on the mass. At large angles the period is slightly longer.
- The frequency is $f = \tfrac{1}{T}$. The period is measured from several cycles, so that the uncertainty of timing is divided among many cycles.
- The energy of the oscillation $E = \tfrac{1}{2}kA^2$ is conserved when friction and air resistance are neglected. Damping reduces the energy and the amplitude.
Optional Return to your predictions
Compare your predictions with the values of the model in the experiments and with the formulas $T = 2\pi\sqrt{\tfrac{m}{k}}$ and $T = 2\pi\sqrt{\tfrac{l}{g}}$. Then return to the question at the beginning: do the child and her father swing at the same pace if they swing on chains of equal length with small swings? What determines the pace at which the wooden bird bobs?
Would you change your prediction?
Optional Damping
Damping
In real oscillators air resistance and friction change some of the mechanical energy into internal energy. The energy $E = \tfrac{1}{2}kA^2$ then decreases, and the amplitude decreases from cycle to cycle. This is called damping. When the damping is small, the period stays almost the same as without damping, so a pendulum clock that keeps time by the period does not run fast or slow as the amplitude changes. In a pendulum clock a weight or a spring supplies energy to replace the damping, so the amplitude stays almost constant. The shock absorbers of a car, on the other hand, are made to damp the oscillation quickly, so that the car does not keep rocking after every pothole.
Apply
Worked example
Problem
A body of mass $0.400\,\text{kg}$ is attached to a spring and oscillates on a frictionless horizontal surface. A student measures with a stopwatch that ten complete oscillations take $8.00\,\text{s}$. What is the spring constant of the spring?
Step 1/7
Given and find. Mass $m = 0.400\,\text{kg}$, time of ten cycles $t = 8.00\,\text{s}$. We want the spring constant $k$.
Step 2/7
Principle. The only force along the motion acting on the body is the harmonic force of the spring, so the motion is simple harmonic motion and the period is $T = 2\pi\sqrt{\tfrac{m}{k}}$. The period is obtained by dividing the time of ten cycles by ten.
Step 3/7
Formula solved. Square the formula for the period and solve $k$:
$$\begin{aligned} T^2 &= \frac{4\pi^2 m}{k} \\ k &= \frac{4\pi^2 m}{T^2} \end{aligned}$$
Step 4/7
Unit conversions. None are needed: the mass is in kilograms and the time in seconds, and $\tfrac{\text{kg}}{\text{s}^2} = \tfrac{\text{N}}{\text{m}}$.
Step 5/7
Substitution. The period is
so
$$\begin{aligned} k &= \frac{4\pi^2 m}{T^2} \\ &= \frac{4\pi^2 \cdot 0.400\,\text{kg}}{(0.800\,\text{s})^2} \\ &= 24.67\,\text{N/m} \end{aligned}$$
Step 6/7
Answer. The spring constant is about $24.7\,\text{N/m}$, to three significant figures as in the mass $0.400\,\text{kg}$ and the time $8.00\,\text{s}$.
Step 7/7
Reasonableness. Check the other way round: if the spring constant were exactly $25\,\text{N/m}$, the period would be $2\pi\sqrt{0.400\,\text{kg} : 25\,\text{N/m}} \approx 0.795\,\text{s}$, very close to the measured $0.800\,\text{s}$. A spring of about $25\,\text{N/m}$ stretches $1\,\text{cm}$ when pulled with a force of $0.25\,\text{N}$, which suits a light school spring. The frequency is $f = \tfrac{1}{T} = 1.25\,\text{Hz}$, so there is a little over one oscillation per second, and they are easy to count.
Experimental work: determining the period of a pendulum
Made-up example data
A student hung a weight on a thin string so that the length of the pendulum from the point of suspension to the centre of mass of the weight was $1.000\,\text{m}$, and displaced it by about $5^\circ$. With a stopwatch she measured the time of ten complete cycles five times. The watch was started and stopped when the weight passed the equilibrium position moving in the same direction, because at that point the weight moves fastest and the moment of passing can be seen most accurately.
| Measurement | $10T$ (s) |
|---|---|
| 1 | $20.12$ |
| 2 | $19.94$ |
| 3 | $20.21$ |
| 4 | $19.88$ |
| 5 | $20.05$ |
The uncertainty of timing by hand is about $0.2\,\text{s}$, even though the watch shows hundredths. When ten cycles are measured, the same uncertainty is divided among ten cycles.
Calculate the period $T$ of the pendulum from the mean of the measurements. Answer in seconds.
Hint 1: a guiding question
The table gives the time of ten cycles. What must you calculate first, and what do you divide it by?
Hint 2: what is the connection
Calculate the mean of the five measurements and divide it by ten.
Hint 3: an intermediate step
The mean is $\tfrac{100.20\,\text{s}}{5} = 20.04\,\text{s}$.
Show the solution
Given
length of the pendulum $l = 1.000\,\text{m}$, times of ten cycles $20.12\,\text{s}$, $19.94\,\text{s}$, $20.21\,\text{s}$, $19.88\,\text{s}$ and $20.05\,\text{s}$, uncertainty of timing by hand about $0.2\,\text{s}$
Find
the period $T$ in $\text{s}$
Principle
The mean of repeated measurements reduces the effect of random error. The time of ten cycles is ten times the period, and because the uncertainty of timing concerns only the starting and stopping moments, measuring ten cycles reduces the uncertainty of the period to one tenth.
Formula
$T = \dfrac{\overline{t}_{10}}{10}$, $\Delta T = \dfrac{\Delta t}{10}$
Conversions
The times are in seconds, so no conversions are needed.
Substitution
Answer
$T = 2.004\,\text{s}$
Or with its uncertainty $T = (2.00 \pm 0.02)\,\text{s}$. The uncertainty is hundredths of a second, so the result is given to the accuracy of hundredths.
Reasonableness
The small-swing formula gives a period of $2.006\,\text{s}$ for a pendulum one metre long, which is within the uncertainty interval of the measurement. The spread of the measurements is of the same order as the estimated uncertainty: half of the range is $\tfrac{20.21\,\text{s} - 19.88\,\text{s}}{2} \approx 0.17\,\text{s}$.
Why ten cycles?
Optional The acceleration of free fall from a pendulum
Calculate the acceleration of free fall $g$ from the period $T = 2.004\,\text{s}$ and the length $l = 1.000\,\text{m}$ of the previous measurement. The swings were small. Answer in $\text{m/s}^2$.
Hint 1: a guiding question
Which formula for the period do you use, and which quantity in it must now be solved?
Hint 2: what is the connection
Square $T = 2\pi\sqrt{\tfrac{l}{g}}$ and solve $g$.
Hint 3: an intermediate step
Show the solution
Given
length of the pendulum $l = 1.000\,\text{m}$, period $T = (2.004 \pm 0.02)\,\text{s}$, small swings
Find
the acceleration of free fall $g$ in $\text{m/s}^2$
Principle
For small swings the period of a pendulum depends only on the length and the acceleration of free fall, so when the length and the period have been measured, the acceleration of free fall can be solved. A starting angle of about $5^\circ$ is so small that the error of the small-angle formula is below $0.1\,\%$.
Formula
$T = 2\pi\sqrt{\dfrac{l}{g}}$, so $T^2 = \dfrac{4\pi^2 l}{g}$ and $g = \dfrac{4\pi^2 l}{T^2}$
Conversions
The length is in metres and the time in seconds, so no conversions are needed.
Substitution
Answer
$g \approx 9.8\,\text{m/s}^2$
The uncertainty of the period is about $1\,\%$, and because $T$ is squared in the formula, the uncertainty of $g$ is about $2\,\%$: with $T = 2.024\,\text{s}$ and $T = 1.984\,\text{s}$ we get $9.64\,\text{m/s}^2$ and $10.03\,\text{m/s}^2$. The result is therefore given to two significant figures, $g = (9.8 \pm 0.2)\,\text{m/s}^2$.
Reasonableness
The table value $9.81\,\text{m/s}^2$ is within the uncertainty interval. A more accurate result is obtained by measuring the period with several lengths and determining the acceleration of free fall from the slope of the $(l, T^2)$ graph; that is done in the investigation Determining the acceleration of free fall with a pendulum.
Optional Find the error
Apply to a new situation
On a space station scales do not work, because everything is weightless. The mass of an astronaut is therefore measured with a chair that oscillates on a spring. The mass of the chair itself is $10.0\,\text{kg}$ and the spring constant of the spring is $1200\,\text{N/m}$. When the astronaut sits in the chair, ten cycles take $16.6\,\text{s}$.
What is the mass of the astronaut? Answer in kilograms.
Hint 1: a guiding question
Why does the period of a spring work in weightlessness too, even though weight does not? Which quantity in the formula for the period is unknown?
Hint 2: what is the connection
First calculate $T$ from the time of ten cycles, and solve the total oscillating mass from the formula $T = 2\pi\sqrt{\tfrac{m}{k}}$: $m = \tfrac{kT^2}{4\pi^2}$.
Hint 3: an intermediate step
The total mass is
What must still be subtracted from it?
Show the solution
Given
mass of the chair $m_\text{c} = 10.0\,\text{kg}$, spring constant $k = 1200\,\text{N/m}$, time of ten cycles $t = 16.6\,\text{s}$
Find
the mass of the astronaut $m_\text{a}$ in $\text{kg}$
Principle
The period of a mass on a spring depends on the mass, that is, the inertia, and not on the weight, so the measurement works in weightlessness. The chair and the astronaut oscillate together on the spring, and the only force along the motion is the harmonic force of the spring.
Formula
so
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
Answer
$73.8\,\text{kg}$
Three significant figures as in the time $16.6\,\text{s}$.
Reasonableness
The mass of an adult is typically $60$–$90\,\text{kg}$, so the result is believable. Check: with the total mass $83.76\,\text{kg}$ the formula gives a period of $1.66\,\text{s}$. A heavier astronaut would make the chair oscillate more slowly.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- The spring of a mass–spring system is replaced with a spring whose spring constant is four times as large. The mass stays the same. How does the period change?
Answer
The period halves. In the formula $T = 2\pi\sqrt{\tfrac{m}{k}}$ the spring constant is in the denominator under the square root, and $\sqrt{4} = 2$: a stiffer spring pulls the body with a greater force, so it oscillates faster.
- Where is the acceleration of a body in simple harmonic motion greatest, and where is the speed greatest?
Answer
The acceleration is greatest at the extreme positions, where the displacement and therefore the harmonic force $F = -kx$ are greatest. The speed is greatest at the equilibrium position, where the force and the acceleration are zero and all the energy is kinetic energy.
1. Concept
2. From a graph to the spring constant
The graph shows the position of a $0.25\,\text{kg}$ body oscillating without friction on a spring. Determine the period from the graph and calculate the spring constant of the spring. Answer in $\text{N/m}$.
Hint 1: a guiding question
Where in the graph is the period read? Find two successive peaks.
Hint 2: what is the connection
The peaks are at the times $0\,\text{s}$ and $0.40\,\text{s}$. Solve $k$ from the formula $T = 2\pi\sqrt{\tfrac{m}{k}}$.
Hint 3: an intermediate step
Show the solution
Given
mass $m = 0.25\,\text{kg}$, frictionless oscillation; the period $T$ from the graph
Find
the spring constant $k$ in $\text{N/m}$
Principle
The only force along the motion is the harmonic force of the spring, so the period is $2\pi\sqrt{\tfrac{m}{k}}$. When the period is read from the graph, the spring constant can be solved.
Reading the graph
The peaks of the curve, $x = 3.0\,\text{cm}$, are at the times $0\,\text{s}$, $0.40\,\text{s}$ and $0.80\,\text{s}$. The interval between two successive peaks is one cycle, so $T = 0.40\,\text{s}$. The interval of two cycles divided by two, $0.80\,\text{s} : 2 = 0.40\,\text{s}$, confirms the reading. The amplitude $3.0\,\text{cm}$ is not needed, because the period does not depend on it.
Formula
$T = 2\pi\sqrt{\dfrac{m}{k}}$, so $T^2 = \dfrac{4\pi^2 m}{k}$ and $k = \dfrac{4\pi^2 m}{T^2}$
Conversions
The mass is in kilograms and the time in seconds, so no conversions are needed.
Substitution
Answer
$62\,\text{N/m}$
Two significant figures as in the mass $0.25\,\text{kg}$ and the period $0.40\,\text{s}$.
Reasonableness
The frequency is $\tfrac{1}{0.40\,\text{s}} = 2.5\,\text{Hz}$, so a light body on a fairly stiff spring oscillates quickly. Check: with the spring constant $62\,\text{N/m}$ the formula gives a period of about $0.40\,\text{s}$.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
Optional 3. Explain with energy
A body at the end of a spring oscillates without friction. Using conservation of energy, explain where the speed of the body is greatest, why doubling the amplitude doubles the greatest speed, and what this tells us about the period.
Model answer
In frictionless oscillation the mechanical energy is conserved: $\tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}kA^2$. The elastic potential energy of the spring is zero at the equilibrium position $x = 0$, so there all the energy is kinetic energy and the speed is greatest. At the extreme positions the speed is zero. For the greatest speed we get $\tfrac{1}{2}mv_\text{max}^2 = \tfrac{1}{2}kA^2$, that is, $v_\text{max} = A\sqrt{\tfrac{k}{m}}$, so it is directly proportional to the amplitude. Doubling the amplitude quadruples the energy, and because the kinetic energy is proportional to the square of the speed, the speed doubles. That is why double the distance is covered in the same period.
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- Damping
- The decrease of the amplitude as energy passes to the surroundings, for example through friction or drag.
- Spring force
- The restoring force produced by the deformation of a spring.
- Simple harmonic motion
- Oscillation in which the displacement varies sinusoidally with time.
- Harmonic oscillator
- An oscillator whose period does not depend on the amplitude, because the restoring force is directly proportional to the displacement.
- Pendulum
- A suspended object that swings to both sides of its equilibrium position under its weight.
- Simple pendulum
- A point-like object swinging on a massless string; for small angles:$$T = 2\pi\sqrt{\dfrac{l}{g}}$$
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 9-1Core
A body of mass $0.80\,\text{kg}$ oscillates on a spring on a frictionless horizontal surface. The spring constant of the spring is $50\,\text{N/m}$.
What is the period of the oscillation?
Answer
$0.79\,\text{s}$
Show the solution
Given
mass $m = 0.80\,\text{kg}$, spring constant $k = 50\,\text{N/m}$, frictionless surface
Find
the period $T$ in $\text{s}$
Principle
The surface is frictionless, so the only force along the motion acting on the body is the harmonic force of the spring, $F = -kx$. The motion is then simple harmonic motion, and its period depends only on the mass and the spring constant.
Formula
$T = 2\pi\sqrt{\dfrac{m}{k}}$
Conversions
The mass is in kilograms and the spring constant in $\text{N/m}$, so no conversions are needed.
Substitution
$T$$= 2\pi\sqrt{\dfrac{0.80\,\text{kg}}{50\,\text{N/m}}}$$= 2\pi\sqrt{0.016\,\text{s}^2}$$= 2\pi \cdot 0.1265\,\text{s}$$= 0.7948\,\text{s}$Answer
$0.79\,\text{s}$
Two significant figures as in the mass $0.80\,\text{kg}$ and the spring constant $50\,\text{N/m}$.
Reasonableness
Unit:
$\tfrac{\text{kg}}{\text{N/m}}$$= \tfrac{\text{kg}\cdot\text{m}}{\text{kg}\cdot\text{m/s}^2}$$= \text{s}^2$so the square root gives seconds. A period of less than a second is usual for a school spring and a weight of less than a kilogram: the oscillations can be counted by eye.
- Exercise 9-2Core
A pendulum of length $0.600\,\text{m}$ hangs in the corner of a bookcase as a decoration. The bob of the pendulum swings with small swings, and air resistance is neglected.
What is the period of the pendulum?
Answer
$1.55\,\text{s}$
Show the solution
Given
length of the pendulum $l = 0.600\,\text{m}$, small swings, air resistance neglected, $g = 9.81\,\text{m/s}^2$
Find
the period $T$ in $\text{s}$
Principle
The pendulum is a simple pendulum: a small bob on a string without air resistance. For small swings the component of the weight along the path is proportional to the displacement, so the motion is simple harmonic and the period depends only on the length and the acceleration of free fall.
Formula
$T = 2\pi\sqrt{\dfrac{l}{g}}$
Conversions
The length is in metres, so no conversions are needed.
Substitution
$T$$= 2\pi\sqrt{\dfrac{0.600\,\text{m}}{9.81\,\text{m/s}^2}}$$= 2\pi\sqrt{0.06116\,\text{s}^2}$$= 2\pi \cdot 0.2473\,\text{s}$$= 1.554\,\text{s}$Answer
$1.55\,\text{s}$
Three significant figures as in the length $0.600\,\text{m}$ and the acceleration of free fall $9.81\,\text{m/s}^2$.
Reasonableness
The period of a pendulum about a metre long is about $2\,\text{s}$. The length is $0.6$ times as large, so the period is $\sqrt{0.6} \approx 0.77$ times as large, about $1.55\,\text{s}$. The result fits.
- Exercise 9-3
In a toy a figure of mass $0.20\,\text{kg}$ oscillates on a spring. The spring constant of the spring is $80\,\text{N/m}$. What is the frequency of the oscillation?
Answer
$3.2\,\text{Hz}$
Show the solution
Given
mass $m = 0.20\,\text{kg}$, spring constant $k = 80\,\text{N/m}$
Find
the frequency $f$ in $\text{Hz}$
Principle
The figure oscillates under the harmonic force of the spring, so the period is obtained from the formula for a mass on a spring. The frequency is the reciprocal of the period: it tells us how many cycles occur in a second.
Formula
$T = 2\pi\sqrt{\dfrac{m}{k}}$ and $f = \dfrac{1}{T}$, so
$f$$= \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$f$$= \dfrac{1}{2\pi}\sqrt{\dfrac{80\,\text{N/m}}{0.20\,\text{kg}}}$$= \dfrac{1}{2\pi}\sqrt{400\,\tfrac{1}{\text{s}^2}}$$= \dfrac{20\,\tfrac{1}{\text{s}}}{2\pi}$$= 3.183\,\text{Hz}$Answer
$3.2\,\text{Hz}$
Two significant figures as in the mass $0.20\,\text{kg}$ and the spring constant $80\,\text{N/m}$.
Reasonableness
The period is $\tfrac{1}{3.2\,\text{Hz}} \approx 0.31\,\text{s}$, so the figure bobs about three times a second, which suits a light toy on a stiff spring.
- Exercise 9-4
In a school oscillation experiment the spring constant of a spring is $120\,\text{N/m}$. The body at the end of the spring oscillates without friction on a horizontal track, and the amplitude of the oscillation is $0.050\,\text{m}$.
What is the energy of the oscillation?
Answer
$0.15\,\text{J}$
Show the solution
Given
spring constant $k = 120\,\text{N/m}$, amplitude $A = 0.050\,\text{m}$, frictionless oscillation
Find
the energy of the oscillation $E$ in $\text{J}$
Principle
In frictionless oscillation the mechanical energy is conserved. At the extreme position the speed is zero, so all the energy is the elastic potential energy of the spring at the displacement $A$.
Formula
$E = \dfrac{1}{2}kA^2$
Conversions
The amplitude is in metres, so no conversions are needed.
Substitution
$E$$= \dfrac{1}{2} \cdot 120\,\text{N/m} \cdot (0.050\,\text{m})^2$$= 60\,\text{N/m} \cdot 0.0025\,\text{m}^2$$= 0.15\,\text{J}$Answer
$0.15\,\text{J}$
Two significant figures as in the amplitude $0.050\,\text{m}$.
Reasonableness
The unit
$\tfrac{\text{N}}{\text{m}} \cdot \text{m}^2$$= \text{N}\cdot\text{m}$$= \text{J}$The force of the spring grows from zero to
$kA$$= 120\,\text{N/m} \cdot 0.050\,\text{m}$$= 6.0\,\text{N}$so the mean force is $3.0\,\text{N}$ and the work over the distance $0.050\,\text{m}$ is $3.0\,\text{N} \cdot 0.050\,\text{m} = 0.15\,\text{J}$. The same value confirms that the energy of a small oscillation is a fraction of a joule.
- Exercise 9-5
The amplitude of a body oscillating on a spring without friction is $3.0\,\text{cm}$. How far, in metres, does the body travel during one cycle?
Answer
$0.12\,\text{m}$
Show the solution
Given
amplitude $A = 3.0\,\text{cm}$
Find
the distance $s$ travelled during one cycle in $\text{m}$
Principle
During one cycle the body starts from an extreme position, passes through the equilibrium position to the other extreme position and returns. An extreme position is an amplitude away from the equilibrium position, so a cycle consists of four parts, each as long as the amplitude.
Formula
$s = 4A$
Conversions
$3.0\,\text{cm}$$= 3.0 \cdot 0.01\,\text{m}$$= 0.030\,\text{m}$Substitution
$s$$= 4 \cdot 0.030\,\text{m}$$= 0.12\,\text{m}$Answer
$0.12\,\text{m}$
Two significant figures as in the amplitude $3.0\,\text{cm}$.
Reasonableness
The distance between the extreme positions is $2A = 6.0\,\text{cm}$, and during a cycle it is covered twice, first in one direction and then in the other: $2 \cdot 6.0\,\text{cm} = 12\,\text{cm}$.
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 9-6Core
A pendulum in a science centre is to make one full swing, that is, one cycle, in exactly $1.50\,\text{s}$. The swings are small, and air resistance is neglected.
How long must the pendulum be?
Answer
$0.559\,\text{m}$
Show the solution
Given
period $T = 1.50\,\text{s}$, small swings, air resistance neglected, $g = 9.81\,\text{m/s}^2$
Find
the length of the pendulum $l$ in $\text{m}$
Principle
For small swings a pendulum is a simple harmonic oscillator whose period depends only on the length and the acceleration of free fall. The length is solved from the formula for the period.
Formula
$T = 2\pi\sqrt{\dfrac{l}{g}}$, so $T^2 = 4\pi^2\dfrac{l}{g}$ and $l = \dfrac{gT^2}{4\pi^2}$
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$l$$= \dfrac{9.81\,\text{m/s}^2 \cdot (1.50\,\text{s})^2}{4\pi^2}$$= \dfrac{22.07\,\text{m}}{39.48}$$= 0.5591\,\text{m}$Answer
$0.559\,\text{m}$
Three significant figures as in the period $1.50\,\text{s}$.
Reasonableness
Check: with the length $0.559\,\text{m}$ the formula gives a period of $1.50\,\text{s}$. Also, the period of a pendulum about a metre long is about $2\,\text{s}$, so a pendulum with a period of less than $2\,\text{s}$ is shorter than a metre.
- Exercise 9-7Core
A weight of mass $0.30\,\text{kg}$ oscillates on a spring. The period is to be doubled with the same spring by adding mass to the weight.
How much mass must be added?
Answer
$0.90\,\text{kg}$
Show the solution
Given
mass of the weight $m_1 = 0.30\,\text{kg}$, the spring stays the same, the period doubles: $T_2 = 2T_1$
Find
the mass to be added $\Delta m$ in $\text{kg}$
Principle
The period of a mass on a spring is proportional to the square root of the mass when the spring constant stays the same. Doubling the period therefore requires four times the mass.
Formula
$\dfrac{T_2}{T_1}$$= \sqrt{\dfrac{m_2}{m_1}}$so
$m_2$$= \left(\dfrac{T_2}{T_1}\right)^2 m_1$$= 4m_1$$\Delta m = m_2 - m_1 = 3m_1$
Conversions
The mass is in kilograms, so no conversions are needed.
Substitution
$\Delta m$$= 3 \cdot 0.30\,\text{kg}$$= 0.90\,\text{kg}$Answer
$0.90\,\text{kg}$
Two significant figures as in the mass $0.30\,\text{kg}$.
Reasonableness
The total mass is $1.20\,\text{kg}$, four times the original, and $\sqrt{4} = 2$. A common mistake is to double the mass, but then the period would grow by a factor of only $\sqrt{2} \approx 1.4$.
- Exercise 9-8
A cart of mass $0.40\,\text{kg}$ on a frictionless air track is attached to a spring with a spring constant of $25\,\text{N/m}$. The cart is pulled $8.0\,\text{cm}$ from the equilibrium position and released from rest.
What is the greatest speed of the cart?
Answer
$0.63\,\text{m/s}$
Show the solution
Given
mass $m = 0.40\,\text{kg}$, spring constant $k = 25\,\text{N/m}$, amplitude $A = 8.0\,\text{cm}$, frictionless track, released from rest
Find
the greatest speed $v_\text{max}$ in $\text{m/s}$
Principle
There is no friction on the track, so the mechanical energy of the oscillator is conserved. At the extreme position all the energy is the elastic potential energy of the spring, at the equilibrium position all of it is kinetic energy, and there the speed is greatest.
Formula
$\dfrac{1}{2}mv_\text{max}^2$$= \dfrac{1}{2}kA^2$so $v_\text{max} = A\sqrt{\dfrac{k}{m}}$
Conversions
$8.0\,\text{cm}$$= 8.0 \cdot 0.01\,\text{m}$$= 0.080\,\text{m}$Substitution
$v_\text{max}$$= 0.080\,\text{m} \cdot \sqrt{\dfrac{25\,\text{N/m}}{0.40\,\text{kg}}}$$= 0.080\,\text{m} \cdot 7.906\,\tfrac{1}{\text{s}}$$= 0.6325\,\text{m/s}$Answer
$0.63\,\text{m/s}$
Two significant figures as in the data.
Reasonableness
Energy check:
$\tfrac{1}{2} \cdot 25\,\text{N/m} \cdot (0.080\,\text{m})^2$$= 0.080\,\text{J}$$\tfrac{1}{2} \cdot 0.40\,\text{kg} \cdot (0.6325\,\text{m/s})^2$$= 0.080\,\text{J}$At its fastest the cart moves a little over half a metre per second, which is usual on an air track.
- Exercise 9-9
At the surface of the Earth the period of a pendulum is $2.00\,\text{s}$. The same pendulum is taken to Mars, where the acceleration of free fall is $3.71\,\text{m/s}^2$. The swings are small.
What is the period of the pendulum on Mars?
Answer
$3.25\,\text{s}$
Show the solution
Given
period on the Earth $T_\text{E} = 2.00\,\text{s}$, $g_\text{E} = 9.81\,\text{m/s}^2$, on Mars $g_\text{Mars} = 3.71\,\text{m/s}^2$, small swings
Find
the period on Mars $T_\text{Mars}$ in $\text{s}$
Principle
For small swings the period of a pendulum is $2\pi\sqrt{\tfrac{l}{g}}$. The length of the pendulum stays the same, so the period is inversely proportional to the square root of the acceleration of free fall. The length does not need to be calculated.
Formula
$\dfrac{T_\text{Mars}}{T_\text{E}}$$= \sqrt{\dfrac{g_\text{E}}{g_\text{Mars}}}$so
$T_\text{Mars}$$= T_\text{E}\sqrt{\dfrac{g_\text{E}}{g_\text{Mars}}}$Conversions
The unit of both accelerations of free fall is $\text{m/s}^2$, and the unit cancels, so no conversions are needed.
Substitution
$T_\text{Mars}$$= 2.00\,\text{s} \cdot \sqrt{\dfrac{9.81\,\text{m/s}^2}{3.71\,\text{m/s}^2}}$$= 2.00\,\text{s} \cdot \sqrt{2.644}$$= 2.00\,\text{s} \cdot 1.626$$= 3.252\,\text{s}$Answer
$3.25\,\text{s}$
Three significant figures as in the period $2.00\,\text{s}$ and the acceleration of free fall $3.71\,\text{m/s}^2$.
Reasonableness
On Mars the restoring force is smaller, so the pendulum swings more slowly and the period increases. The acceleration of free fall is about a third, and $\sqrt{3} \approx 1.7$, so the period is about $1.6$ to $1.7$ times as large.
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 9-10
A body of mass $0.50\,\text{kg}$ oscillates without friction on a spring. The spring constant of the spring is $200\,\text{N/m}$ and the amplitude of the oscillation is $0.100\,\text{m}$.
What is the speed of the body when it is $0.0500\,\text{m}$ from the equilibrium position?
Answer
$1.7\,\text{m/s}$
Show the solution
Given
mass $m = 0.50\,\text{kg}$, spring constant $k = 200\,\text{N/m}$, amplitude $A = 0.100\,\text{m}$, displacement $x = 0.0500\,\text{m}$, frictionless oscillation
Find
the speed $v$ at the displacement $x$ in $\text{m/s}$
Principle
In frictionless oscillation the mechanical energy is conserved. At every point the sum of the kinetic energy and the elastic potential energy of the spring equals the energy at the extreme position, where the speed is zero. At the displacement $x$ some of the energy is still in the spring, so the speed is smaller than at the equilibrium position.
Formula
$\dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2$$= \dfrac{1}{2}kA^2$so $v^2 = \dfrac{k}{m}(A^2 - x^2)$ and $v = \sqrt{\dfrac{k}{m}(A^2 - x^2)}$
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
First
$\dfrac{k}{m}$$= \dfrac{200\,\text{N/m}}{0.50\,\text{kg}}$$= 400\,\tfrac{1}{\text{s}^2}$$A^2 - x^2$$= (0.100\,\text{m})^2 - (0.0500\,\text{m})^2$$= 0.00750\,\text{m}^2$so
$v$$= \sqrt{400\,\tfrac{1}{\text{s}^2} \cdot 0.00750\,\text{m}^2}$$= \sqrt{3.00\,\tfrac{\text{m}^2}{\text{s}^2}}$$= 1.732\,\text{m/s}$Answer
$1.7\,\text{m/s}$
Two significant figures as in the mass $0.50\,\text{kg}$.
Reasonableness
At the equilibrium position the speed would be
$A\sqrt{\tfrac{k}{m}}$$= 0.100\,\text{m} \cdot 20\,\tfrac{1}{\text{s}}$$= 2.0\,\text{m/s}$At half the amplitude the spring holds only a quarter of the energy, because $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$, so three quarters of it is kinetic energy and the speed is $\sqrt{0.75} \approx 0.87$ times the greatest, that is, $1.7\,\text{m/s}$. So the speed is not half of the greatest.
- Exercise 9-11
When a passenger of mass $80\,\text{kg}$ sits in a car of mass $1200\,\text{kg}$, the body of the car sinks $3.0\,\text{cm}$ lower on its suspension. Model the whole car with the passenger as one body on one spring and neglect the shock absorbers.
What is the period of the vertical oscillation of the car body when the car drives into a pothole with the passenger on board?
Answer
$1.4\,\text{s}$
Show the solution
Given
mass of the passenger $m_\text{p} = 80\,\text{kg}$, mass of the car $1200\,\text{kg}$, sinking $\Delta x = 3.0\,\text{cm}$, $g = 9.81\,\text{m/s}^2$; model: one body on one spring, no damping
Find
the period $T$ of the oscillation of the car body in $\text{s}$
Principle
Two ideas. First the spring constant: when the passenger has sat down and the car is at rest again, the extra force of the spring equals the weight of the passenger, so $k\Delta x = m_\text{p}g$. Then the period: in vertical oscillation the weight only shifts the equilibrium position, so the period is $2\pi\sqrt{\tfrac{m}{k}}$, where $m$ is the whole oscillating mass, the car and the passenger.
Formula
$k = \dfrac{m_\text{p}g}{\Delta x}$ and $T = 2\pi\sqrt{\dfrac{m}{k}}$, so
$T$$= 2\pi\sqrt{\dfrac{m\,\Delta x}{m_\text{p}g}}$Conversions
$\Delta x$$= 3.0\,\text{cm}$$= 3.0 \cdot 0.01\,\text{m}$$= 0.030\,\text{m}$the oscillating mass
$m$$= 1200\,\text{kg} + 80\,\text{kg}$$= 1280\,\text{kg}$Substitution
$k$$= \dfrac{80\,\text{kg} \cdot 9.81\,\text{m/s}^2}{0.030\,\text{m}}$$= 26\,160\,\text{N/m}$$T$$= 2\pi\sqrt{\dfrac{1280\,\text{kg}}{26\,160\,\text{N/m}}}$$= 2\pi\sqrt{0.04893\,\text{s}^2}$$= 2\pi \cdot 0.2212\,\text{s}$$= 1.390\,\text{s}$Answer
$1.4\,\text{s}$
Two significant figures as in the sinking $3.0\,\text{cm}$.
Reasonableness
Cars rock on their suspension with periods of about a second, so the result is of the right order of magnitude. In reality the shock absorbers damp the oscillation to one or two swings, and each wheel has its own spring, so the model gives only the order of magnitude.
- Exercise 9-12
The pendulum of a pendulum clock is $0.994\,\text{m}$ long, and in spring the clock runs exactly right. In the heat of summer the rod of the pendulum lengthens by $0.50\,\text{mm}$ because of thermal expansion. The clock counts swings and assumes that every cycle is as long as in spring. The swings are small.
By how many seconds per day does the clock run slow?
Answer
$22\,\text{s}$
Show the solution
Given
in spring the length of the pendulum $l = 0.994\,\text{m}$, in summer it is longer by $\Delta l = 0.50\,\text{mm}$, small swings, a day $t = 86\,400\,\text{s}$
Find
how much the clock runs slow in a day, $\Delta t$ in $\text{s}$
Principle
For small swings the period is $2\pi\sqrt{\tfrac{l}{g}}$, so a longer pendulum swings more slowly. The clock counts cycles and adds the period $T$ it had in spring for each of them, although the real cycle lasts $T'$. In a day it manages to make only $\tfrac{t}{T'}$ of the longer cycles $T'$, so it shows the time $\tfrac{t}{T'}T$ and falls behind. The acceleration of free fall cancels in the ratio of the periods.
Formula
$\dfrac{T'}{T}$$= \sqrt{\dfrac{l + \Delta l}{l}}$$\Delta t = t - \dfrac{t}{T'}T$, so
$\Delta t$$= t\left(1 - \dfrac{T}{T'}\right)$$= t\left(1 - \sqrt{\dfrac{l}{l + \Delta l}}\right)$Conversions
$\Delta l$$= 0.50\,\text{mm}$$= 0.50 \cdot 0.001\,\text{m}$$= 0.00050\,\text{m}$a day
$t$$= 24 \cdot 3600\,\text{s}$$= 86\,400\,\text{s}$Substitution
$\Delta t$$= 86\,400\,\text{s} \cdot \left(1 - \sqrt{\dfrac{0.994\,\text{m}}{0.9945\,\text{m}}}\right)$$= 86\,400\,\text{s} \cdot (1 - 0.999\,749)$$= 86\,400\,\text{s} \cdot 0.000\,251$$= 21.7\,\text{s}$The intermediate result must not be rounded: even rounding $1 - 0.9997 = 0.0003$ would give $26\,\text{s}$.
Answer
$22\,\text{s}$
Two significant figures as in the lengthening $0.50\,\text{mm}$.
Reasonableness
The length increases by the relative amount $\tfrac{0.50\,\text{mm}}{994\,\text{mm}} \approx 0.050\,\%$, and because of the square root the period increases by about half of that, $0.025\,\%$. Of a day that is $0.00025 \cdot 86\,400\,\text{s} \approx 22\,\text{s}$. That is why accurate pendulum clocks used pendulums that compensate for changes in temperature.
