Simple Harmonic Motion

Lessons

Senior 5 · Physics

Simple Harmonic Motion

4 lessons

Lesson 2 of 4

Kinematics and Energy of Simple Harmonic Motion

25 min

Where is it fastest?

Hang a mass on a spring above a motion sensor, pull it down and let go. The data logger plots its displacement, and from that its velocity and acceleration, against time. Three curves appear, all sinusoidal, all with the same period, but none of them in step with each other. The mass is fastest where its displacement is zero, and its acceleration is greatest where it stops. In this lesson you derive these curves from $x = A\cos(\omega t + \phi)$ , find a formula for the speed at any displacement, and follow the energy as it flows back and forth between kinetic and potential.

Velocity and acceleration as functions of time

Take a body released from rest at $x = + A$ at $t = 0$ , so $\phi = 0$ . Differentiating once and then again:

x = A\cos\omega t \qquad v = \frac{dx}{dt} = -A\omega\sin\omega t \qquad a = \frac{dv}{dt} = -A\omega^2\cos\omega t = -\omega^2 x

Read the graphs column by column:

At $t = 0$ and $t = \frac{T}{2}$ the body is at an extreme: $v = 0$ and $|a| = A\omega^2$ , the largest value.
At $t = \frac{T}{4}$ and $t = \frac{3T}{4}$ it passes through equilibrium: $a = 0$ and $|v| = A\omega$ , the largest value.
Writing $v = A\omega\cos\left(\omega t + \frac{\pi}{2}\right)$ shows that the velocity leads the displacement by $\frac{\pi}{2}$ (a quarter cycle).
Writing $a = A\omega^2\cos(\omega t + \pi)$ shows that the acceleration is in antiphase with the displacement: a phase difference of $π \pi$ .

v_{\max} = A\omega \qquad a_{\max} = A\omega^2

Speed at a given displacement

Often you know where the body is, not when. Eliminate $t$ using $\sin^2\omega t + \cos^2\omega t = 1$ : from $\cos\omega t = \frac{x}{A}$ and $\sin\omega t = -\frac{v}{A\omega}$ ,

\frac{x^2}{A^2} + \frac{v^2}{A^2\omega^2} = 1 \quad\Longrightarrow\quad v = \pm\,\omega\sqrt{A^2 - x^2}

The sign is $+$ when the body moves in the positive direction. The formula holds whatever the phase constant, because $\phi$ cancels when you eliminate the time.

Worked example 1: speed and acceleration at a point

A mass on a spring oscillates with amplitude 5.0 cm and frequency 2.0 Hz. Find (a) its greatest speed, (b) its greatest acceleration, (c) its speed and acceleration when it is 3.0 cm from equilibrium.

Solution

$\omega = 2\pi f = 4\pi = 12.6\,\mathrm{rad\,s^{-1}}$ and $A = 0.050\,\mathrm{m}$ .

(a) $v_{\max} = A\omega = 0.050 \times 12.57 = 0.63\,\mathrm{m\,s^{-1}}$ .

(b) $a_{\max} = A\omega^2 = 0.050 \times 12.57^2 = 7.9\,\mathrm{m\,s^{-2}}$ .

(c) $v = \omega\sqrt{A^2 - x^2} = 12.57\sqrt{0.050^2 - 0.030^2} = 12.57 \times 0.040 = 0.50\,\mathrm{m\,s^{-1}}$ . $a = -\omega^2 x = -157.9 \times 0.030 = -4.7\,\mathrm{m\,s^{-2}}$ : its size is $4.7\,\mathrm{m\,s^{-2}}$ , directed towards equilibrium.

Worked example 2: SHM is not uniform

Show that a body in SHM takes $\frac{T}{12}$ to go from equilibrium to $x = \frac{A}{2}$ , but $\frac{T}{6}$ to go from $\frac{A}{2}$ on to $A$ .

Solution

Start the clock at equilibrium, moving outwards: $x = A\sin\omega t$ . Then $x = \frac{A}{2}$ when $\sin\omega t = \frac{1}{2}$ , so $\omega t = \frac{\pi}{6}$ and $t = \frac{\pi}{6} \times \frac{T}{2\pi} = \frac{T}{12}$ . The body reaches $x = A$ at $\omega t = \frac{\pi}{2}$ , that is $t = \frac{T}{4}$ . So the outer half takes $\frac{T}{4} - \frac{T}{12} = \frac{T}{6}$ : twice as long as the inner half, because the body slows down as it moves out.

Energy in simple harmonic motion

The kinetic energy at displacement $x$ follows from the speed formula:

E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}m\omega^2\left(A^2 - x^2\right)

The potential energy is the work done against the restoring force $F = k x$ (with $k = m\omega^2$ ) in moving the body from equilibrium to $x$ . The force grows steadily from 0, so

E_p = \int_0^x kx'\,dx' = \tfrac{1}{2}kx^2 = \tfrac{1}{2}m\omega^2 x^2

Adding the two, the $x^{2}$ terms cancel:

E = E_k + E_p = \tfrac{1}{2}m\omega^2 A^2 = \tfrac{1}{2}kA^2

The total energy is constant, and it is proportional to the square of the amplitude. Doubling the amplitude stores four times the energy.

Against time, $E_p = E\cos^2\omega t$ and $E_k = E\sin^2\omega t$ . Since $\cos^2\omega t = \frac{1}{2}(1 + \cos 2\omega t)$ , each energy oscillates at twice the frequency of the motion: the potential energy peaks at both extremes, and the kinetic energy peaks each time the body passes through equilibrium, in either direction.

Energy changes in real oscillators

The same exchange happens in every oscillator; only the form of the potential energy changes. In a pendulum it is gravitational potential energy, highest at the ends of the swing. In a horizontal spring it is elastic energy. In a vertical spring both change, but if you measure the displacement from the equilibrium position (where the spring already supports the weight), the gravitational and elastic terms combine into exactly $\frac{1}{2}ky^2$ , so all the results above still hold. In a real system some energy is also converted to heat and sound by friction and air resistance in every cycle, so the amplitude slowly falls: that is the subject of Unit 2.

Definition

The total energy of a simple harmonic oscillator is the sum of its kinetic and potential energies, $E = \frac{1}{2}m\omega^2A^2 = \frac{1}{2}kA^2$ . In the absence of friction it is conserved: it stays constant while energy flows between the kinetic and potential stores twice per cycle. Example: a 0.20 kg mass on an $80\,\mathrm{N\,m^{-1}}$ spring with amplitude 0.10 m has $E = \frac{1}{2} \times 80 \times 0.10^2 = 0.40\,\mathrm{J}$ .

Worked example 3: sharing the energy

For the 0.20 kg mass on the $80\,\mathrm{N\,m^{-1}}$ spring with amplitude 0.10 m, find (a) the greatest speed from energy, (b) the kinetic and potential energies at $x = 0.050\,\mathrm{m}$ , (c) the displacement at which the two energies are equal.

Solution

(a) At equilibrium all 0.40 J is kinetic: $v_{\max} = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 0.40}{0.20}} = 2.0\,\mathrm{m\,s^{-1}}$ . Check: $\omega = \sqrt{\frac{k}{m}} = \sqrt{400} = 20\,\mathrm{rad\,s^{-1}}$ , and $A\omega = 0.10 \times 20 = 2.0\,\mathrm{m\,s^{-1}}$ .

(b) $E_p = \frac{1}{2} \times 80 \times 0.050^2 = 0.10\,\mathrm{J}$ , so $E_k = 0.40 - 0.10 = 0.30\,\mathrm{J}$ . Halfway out, three quarters of the energy is still kinetic.

Experiment 1.1: is energy conserved in a swinging pendulum?

Aim: to compare the gravitational potential energy lost by a pendulum bob with the kinetic energy it gains at the bottom of its swing.

Apparatus: a heavy bob of known mass and diameter on a string about 1 m long, a clamp stand with a G-clamp, a light gate and timer (or data logger), a metre rule, vernier calipers.

Method:

Measure the diameter $d$ of the bob with the calipers and its mass $m$ on a balance.
Set the light gate so that the bob, hanging at rest, sits in the beam.
Pull the bob aside until its centre is a height $h$ above the lowest position (measure with the metre rule held vertically), and release it without pushing.
Record the time $\Delta t$ for which the bob blocks the beam. Its speed at the bottom is $v = \frac{d}{\Delta t}$ .
Compare $\frac{1}{2}mv^2$ with $m g h$ . Repeat for four or five heights.

Illustrative results: a 50 g bob of diameter 2.00 cm released from $h = 5.0\,\mathrm{cm}$ blocks the beam for 20.4 ms. Then $v = \frac{0.0200}{0.0204} = 0.980\,\mathrm{m\,s^{-1}}$ , $E_k = \frac{1}{2} \times 0.050 \times 0.980^2 = 0.0240\,\mathrm{J}$ and $mgh = 0.050 \times 9.8 \times 0.050 = 0.0245\,\mathrm{J}$ . The kinetic energy is about 2% less than the potential energy lost: some energy was transferred to the air, and the measurement of $h$ is the largest uncertainty.

Warning

Safety. Clamp the stand to the bench, or weight its base, so that a swinging bob cannot topple it. Keep faces out of the plane of the swing, and use a bob no heavier than a few hundred grams.

Warning

Common mistakes. (1) Putting $v_{\max}$ at the extremes: the body is momentarily at rest there; the speed is greatest at equilibrium. (2) Writing $E = \frac{1}{2}kx^2$ for the total energy: that is only the potential energy; the total uses the amplitude, $\frac{1}{2}kA^2$ . (3) Saying the energy oscillates with the same frequency as the body: $E_{k}$ and $E_{p}$ each go through two cycles per oscillation. (4) Forgetting the square root's sign: $v = \pm\omega\sqrt{A^2 - x^2}$ gives the speed; the direction comes from the motion.

Tip

Each time you differentiate, multiply by one more $\omega$ and move a quarter cycle ahead: $x$ has amplitude $A$ , $v$ has amplitude $A\omega$ , $a$ has amplitude $A\omega^2$ .

Energy graphs for simple harmonic motion (Khan Academy). Watch how the kinetic and potential energy curves are built from the displacement, and why the total stays flat.

✦

Practice

A body performs SHM with amplitude 8.0 cm and period 0.50 s. Find its greatest speed and greatest acceleration.
For the body in question 1, find the speed when it is 5.0 cm from equilibrium.
Taking $x = A\cos\omega t$ , show that the velocity leads the displacement by $\frac{\pi}{2}$ .
A 0.40 kg mass on a spring of stiffness $160\,\mathrm{N\,m^{-1}}$ oscillates with amplitude 5.0 cm. Find its total energy, its greatest speed, and its kinetic energy at $x = 3.0\,\mathrm{cm}$ .
At what displacement, as a fraction of the amplitude, is the kinetic energy three times the potential energy?
A body oscillates at 1.5 Hz. At what frequency does its kinetic energy vary? Explain.
A pendulum bob of diameter 2.50 cm blocks a light gate for 31.2 ms at the bottom of its swing. It was released 4.0 cm above the gate. Use energy to predict its speed, and compare with the measured speed.
Mutesi says: "If I double the amplitude of a mass on a spring, its greatest speed doubles, so its energy doubles too." Evaluate her statement.

Show answersHide answers

$\omega = \frac{2\pi}{0.50} = 12.6\,\mathrm{rad\,s^{-1}}$ ; $v_{\max} = 0.080 \times 12.57 = 1.0\,\mathrm{m\,s^{-1}}$ ; $a_{\max} = 0.080 \times 12.57^2 = 12.6\,\mathrm{m\,s^{-2}}$ .
$v = 12.57\sqrt{0.080^2 - 0.050^2} = 12.57 \times 0.0624 = 0.79\,\mathrm{m\,s^{-1}}$ .
$v = -A\omega\sin\omega t = A\omega\cos\left(\omega t + \frac{\pi}{2}\right)$ : its phase is $\frac{\pi}{2}$ larger than that of $x$ .
$E = \frac{1}{2} \times 160 \times 0.050^2 = 0.20\,\mathrm{J}$ ; $v_{\max} = \sqrt{\frac{2 \times 0.20}{0.40}} = 1.0\,\mathrm{m\,s^{-1}}$ ; $E_k = \frac{1}{2} \times 160 \times (0.050^2 - 0.030^2) = 0.128\,\mathrm{J}$ .
$E_{k} = 3 E_{p}$ means $E = 4 E_{p}$ , so $\frac{1}{2}kA^2 = 4 \times \frac{1}{2}kx^2$ and $x = \frac{A}{2}$ .
3.0 Hz. $E_k = E\sin^2\omega t = \frac{E}{2}(1 - \cos 2\omega t)$ , which varies at $2\omega$ : the kinetic energy peaks each time the body passes equilibrium, twice per cycle.
Predicted: $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.040} = 0.885\,\mathrm{m\,s^{-1}}$ . Measured: $\frac{0.0250}{0.0312} = 0.801\,\mathrm{m\,s^{-1}}$ , about 10% lower. That is more than air resistance alone explains, so the height was probably over-measured or the gate was not at the lowest point.
The first part is right: $v_{\max} = A\omega$ doubles. The second is wrong: $E = \frac{1}{2}kA^2$ (or $\frac{1}{2}mv_{\max}^2$ ) depends on the square, so the energy becomes four times as large.

Key points

$v = -A\omega\sin(\omega t + \phi)$ , $a = -A\omega^2\cos(\omega t + \phi)$ ; $v$ leads $x$ by $\frac{\pi}{2}$ , $a$ is in antiphase with $x$ .
$v = \pm\omega\sqrt{A^2 - x^2}$ ; $v_{\max} = A\omega$ at equilibrium, $a_{\max} = A\omega^2$ at the extremes.
$E_k = \frac{1}{2}m\omega^2(A^2 - x^2)$ , $E_p = \frac{1}{2}m\omega^2x^2$ , total $E = \frac{1}{2}m\omega^2A^2$ , constant without friction.
Each energy varies at twice the frequency of the motion.

Sources and credits

15.1 Simple Harmonic Motion and 15.2 Energy in Simple Harmonic Motion, from University Physics Volume 1 by William Moebs, Samuel J. Ling and Jeff Sanny, OpenStax, licensed under CC BY 4.0. Adapted by Ganzaa.
Video: Energy graphs for simple harmonic motion, Khan Academy, embedded from YouTube.

The experiment, the four diagrams, the worked examples and the practice exercises are written by Ganzaa.

1Where is it fastest?

2Velocity and acceleration as functions of time

3Speed at a given displacement

4Energy in simple harmonic motion

Energy changes in real oscillators

5Experiment 1.1: is energy conserved in a swinging pendulum?

Where is it fastest?

Velocity and acceleration as functions of time

Speed at a given displacement

Energy in simple harmonic motion

Experiment 1.1: is energy conserved in a swinging pendulum?