IB Physics Quiz: Understand Simple Harmonic Motion

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Understand Simple Harmonic MotionQuestion 1 of 20

The motion of a simple pendulum is only approximately simple harmonic. This approximation is valid under what condition?

The mass of the pendulum bob is sufficiently small.

The length of the string is significantly greater than the bob's radius.

The gravitational field is uniform over the swing.

The angle of displacement from the vertical is small.

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IB Physics Quiz

IB Physics Quiz: Understand Simple Harmonic Motion

Practice Understand Simple Harmonic Motion in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Understand Simple Harmonic Motion, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The motion of a simple pendulum is only approximately simple harmonic. This approximation is valid under what condition?

The mass of the pendulum bob is sufficiently small.
The length of the string is significantly greater than the bob's radius.
The gravitational field is uniform over the swing.
The angle of displacement from the vertical is small. (correct answer)

Explanation: The restoring force for a pendulum is

F = -mg\sin\theta

. For the motion to be simple harmonic, the restoring force must be proportional to the displacement,

F \propto -x

. The arc length displacement is

x = l\theta

. The condition for SHM becomes

-mg\sin\theta \propto -l\theta

. This requires the approximation

\sin\theta \approx \theta

(for

\theta

in radians), which is only valid for small angles of displacement.

Question 2

A mass on a spring undergoes SHM. The total mechanical energy of the system is $E$ . What is the kinetic energy of the mass when its displacement is half of the amplitude?

$\frac{1}{4}E$
$\frac{1}{2}E$
$\frac{3}{4}E$ (correct answer)
$\frac{\sqrt{3}}{2}E$

Explanation: The total energy in SHM is constant and can be expressed as the maximum potential energy:

E = \frac{1}{2}kA^2

, where

k

is the spring constant and

A

is the amplitude. At any displacement

x

, the potential energy is

E_p = \frac{1}{2}kx^2

. When

x = A/2

, the potential energy is

E_p = \frac{1}{2}k(\frac{A}{2})^2 = \frac{1}{4}(\frac{1}{2}kA^2) = \frac{1}{4}E

. By conservation of energy,

E = E_k + E_p

, so the kinetic energy

E_k = E - E_p = E - \frac{1}{4}E = \frac{3}{4}E

Question 3

A mass $m$ attached to a spring oscillates with a certain amplitude. The maximum kinetic energy is $E_k$ . If the mass is doubled to (2m) while the spring and amplitude remain unchanged, what is the new maximum kinetic energy?

$E_k / 2$
$E_k / \sqrt{2}$
$E_k$ (correct answer)
$2E_k$

Explanation: The maximum kinetic energy in SHM is equal to the total mechanical energy of the system. The total energy of a mass-spring system is given by

E_{total} = \frac{1}{2}kA^2

, where

k

is the spring constant and

A

is the amplitude. Since both

k

and

A

remain unchanged, the total energy of the system is constant. Therefore, the maximum kinetic energy, which occurs when all the potential energy has been converted to kinetic energy, remains the same,

E_k

Question 4

A mass $m$ oscillates on a spring with constant $k$ and period $T$ . The spring is replaced by two new springs, each with spring constant $k$ , connected in series. What is the new period of oscillation for mass $m$ ?

$T/\sqrt{2}$
$T/2$
$\sqrt{2}T$ (correct answer)
$2T$

Explanation: For springs in series, the reciprocal of the effective spring constant

k_{eff}

is the sum of the reciprocals of the individual constants:

\frac{1}{k_{eff}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}

, so

k_{eff} = k/2

. The original period is

T = 2\pi\sqrt{m/k}

. The new period

T'

T' = 2\pi\sqrt{m/k_{eff}} = 2\pi\sqrt{m/(k/2)} = 2\pi\sqrt{2m/k} = \sqrt{2} (2\pi\sqrt{m/k}) = \sqrt{2}T

Question 5

A mass $M$ on a spring of constant $k$ has period $T$ . The mass is replaced by (4M) and the spring constant is changed to $k/4$ . What is the new period?

$T$
$2T$
$4T$ (correct answer)
$16T$

Explanation: The period of a mass-spring system is

T = 2\pi\sqrt{M/k}

. The new mass is

M' = 4M

and the new spring constant is

k' = k/4

. The new period

T'

T' = 2\pi\sqrt{M'/k'} = 2\pi\sqrt{(4M)/(k/4)} = 2\pi\sqrt{16M/k} = \sqrt{16} \times (2\pi\sqrt{M/k}) = 4T

Question 6

A mass $m$ oscillates on a horizontal frictionless surface attached to a spring of constant $k$ . The same mass and spring are then hung vertically and allowed to oscillate. How does the period of oscillation $T$ change?

It remains the same because gravity only shifts the equilibrium position. (correct answer)
It decreases because gravity opposes the motion upwards.
It increases because gravity assists the motion downwards.
It changes in a way that depends on the mass and spring constant.

Explanation: The period of a mass-spring system is given by

T = 2\pi\sqrt{m/k}

. This depends only on the mass and the spring constant. When the system is hung vertically, the force of gravity

mg

acts on the mass. This constant downward force shifts the equilibrium position to a point where the upward spring force balances the weight (

kx_{eq} = mg

). The oscillations then occur about this new equilibrium point, but the restoring force for a displacement

y

from this new equilibrium is still

F_{net} = k(x_{eq}-y) - mg = kx_{eq} - ky - mg = -ky

. Since the net restoring force is still proportional to the displacement from equilibrium with the same effective spring constant, the period remains unchanged.

Question 7

For an object in simple harmonic motion, which pair of quantities are exactly $\pi$ radians (180°) out of phase?

Velocity and acceleration
Displacement and acceleration (correct answer)
Displacement and velocity
Force and acceleration

Explanation: In SHM, acceleration is given by

a = -\omega^2 x

. The negative sign indicates that the acceleration vector is always in the opposite direction to the displacement vector

x

. Two vectors pointing in opposite directions are

\pi

radians (180°) out of phase. Velocity is

\pi/2

radians out of phase with displacement, and force is in phase (0 radians) with acceleration according to Newton's second law (

F=ma

Question 8

A mass attached to a spring oscillates with simple harmonic motion. At the moment when the displacement is $x = \frac{A}{2}$ (where $A$ is the amplitude), the ratio of kinetic energy to potential energy is:

$\frac{1}{3}$
$\frac{3}{1}$ (correct answer)
$\frac{1}{4}$
$\frac{4}{1}$

Explanation: For SHM,

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

and

PE = \frac{1}{2}m\omega^2x^2

. At

x = A/2

KE = \frac{1}{2}m\omega^2(A^2 - A^2/4) = \frac{3}{8}m\omega^2A^2

and

PE = \frac{1}{2}m\omega^2(A/2)^2 = \frac{1}{8}m\omega^2A^2

. Therefore

\frac{KE}{PE} = \frac{3/8}{1/8} = 3

. Choice A reverses the ratio. Choice C uses

(A/2)^2 = A^2/4

incorrectly. Choice D squares the correct ratio.

Question 9

A mass $m$ is attached to two identical springs of constant $k$ , arranged so that one spring connects the mass to a fixed wall on the left, and the other connects the mass to a fixed wall on the right. If the mass is displaced from its equilibrium position and released, the period of oscillation will be:

$2\pi\sqrt{\frac{m}{k}}$ , the same as a single spring system
$2\pi\sqrt{\frac{m}{2k}}$ , because the effective spring constant is $2k$ (correct answer)
$\pi\sqrt{\frac{m}{k}}$ , because both springs contribute to the restoring force
$4\pi\sqrt{\frac{m}{k}}$ , because the springs are in series configuration

Explanation: When displaced by distance

x

from equilibrium, both springs exert restoring forces toward equilibrium. The left spring exerts force

kx

rightward, the right spring exerts force

kx

leftward, giving total restoring force

F = -2kx

. This means effective spring constant is

k_{eff} = 2k

, so

T = 2\pi\sqrt{\frac{m}{2k}}

. Choice A ignores the second spring. Choice C has incorrect coefficient. Choice D confuses this parallel arrangement with series configuration.

Question 10

A particle executes SHM according to $x = A\cos(\omega t + \phi)$ . If the particle passes through the equilibrium position at $t = 0$ with positive velocity, and reaches its first maximum displacement at $t = \frac{T}{4}$ , the phase constant $\phi$ is:

$0$
$\frac{\pi}{2}$
$\pi$
$\frac{3\pi}{2}$ (correct answer)

Explanation: At

t = 0

x(0) = A\cos(\phi) = 0

, so

\phi = \frac{\pi}{2}

\frac{3\pi}{2}

. The velocity is

v = -A\omega\sin(\omega t + \phi)

, so

v(0) = -A\omega\sin(\phi) > 0

, which requires

\sin(\phi) < 0

. This means

\phi = \frac{3\pi}{2}

. We can verify: at

t = \frac{T}{4}

x = A\cos(\frac{\pi}{2} + \frac{3\pi}{2}) = A\cos(2\pi) = A

(maximum). Choice A gives

x(0) = A \neq 0

. Choice B gives

v(0) < 0

. Choice C gives

x(0) = -A \neq 0

Question 11

A horizontal mass-spring system undergoes SHM with period $T$ . If the system is rotated 90° so the spring-mass oscillates vertically, assuming the spring can support the weight, the period of small oscillations about the new equilibrium position will be:

less than $T$ because gravity provides additional restoring force throughout the motion
greater than $T$ because the spring must first overcome gravitational potential energy changes
equal to $T$ because the equilibrium position shifts but the restoring force constant remains unchanged (correct answer)
equal to $T$ only if the amplitude is much smaller than the gravitational extension of the spring

Explanation: When vertical, the spring extends by

\Delta x_0 = \frac{mg}{k}

to reach equilibrium. For small displacements

y

from this new equilibrium, the net force is

F = -ky

(spring force)

+ mg - mg = -ky

. The

mg

terms cancel, leaving the same restoring force law, so

T = 2\pi\sqrt{\frac{m}{k}}

unchanged. Choice A incorrectly assumes gravity adds to restoring force. Choice B misunderstands energy considerations. Choice D suggests amplitude-dependence that doesn't exist in SHM.

Question 12

A mass-spring system oscillates with amplitude $A$ and angular frequency $\omega$ . At what displacement from equilibrium does the magnitude of acceleration equal half the maximum acceleration?

$\frac{A}{4}$
$\frac{A}{2}$ (correct answer)
$\frac{A}{\sqrt{2}}$
$\frac{A\sqrt{3}}{2}$

Explanation: In SHM, acceleration

a = -\omega^2x

, so

|a| = \omega^2|x|

. Maximum acceleration occurs at

x = \pm A

|a_{max}| = \omega^2A

. For

|a| = \frac{1}{2}|a_{max}|

\omega^2|x| = \frac{1}{2}\omega^2A

, which gives

|x| = \frac{A}{2}

. Choice A confuses this with energy relationships. Choice C applies the relationship for velocity at half-maximum. Choice D uses trigonometric ratios inappropriately.

Question 13

A simple pendulum of length $L$ oscillates with small amplitude in simple harmonic motion. If the pendulum is moved to a location where the gravitational field strength is reduced to $\frac{g}{4}$ , and the length is simultaneously increased to $4L$ , the new period will be:

equal to the original period
twice the original period
four times the original period (correct answer)
half the original period

Explanation: The period of a simple pendulum is

T = 2\pi\sqrt{\frac{L}{g}}

. Originally:

T_1 = 2\pi\sqrt{\frac{L}{g}}

. After changes:

T_2 = 2\pi\sqrt{\frac{4L}{g/4}} = 2\pi\sqrt{\frac{16L}{g}} = 4 \times 2\pi\sqrt{\frac{L}{g}} = 4T_1

. Choice A assumes the effects cancel. Choice B only accounts for one change. Choice D inverts the relationship.

Question 14

The total energy of a system in simple harmonic motion is doubled, while the mass and spring constant are kept the same. By what factor does the maximum speed of the oscillating mass change?

$\sqrt{2}$ (correct answer)
2
4
It does not change.

Explanation: The total energy

E

of the system is equal to the maximum kinetic energy,

E = \frac{1}{2}mv_{max}^2

. Therefore,

v_{max} = \sqrt{2E/m}

. This shows that the maximum speed is proportional to the square root of the total energy (

v_{max} \propto \sqrt{E}

). If the energy

E

is doubled to

2E

, the new maximum speed

v'_{max}

will be

v'_{max} \propto \sqrt{2E}

, which means it increases by a factor of

\sqrt{2}

Question 15

An object executes simple harmonic motion with amplitude $A$ and angular frequency $\omega$ . What is the speed of the object when its potential energy is equal to its kinetic energy?

$\omega A / 2$
$\omega A / \sqrt{2}$ (correct answer)
$\omega A$
$\omega A / 4$

Explanation: Total energy is

E = E_k + E_p

. When

E_k = E_p

, we have

E = 2E_k

. The total energy is also equal to the maximum kinetic energy,

E = \frac{1}{2}mv_{max}^2 = \frac{1}{2}m(\omega A)^2

. The kinetic energy at any speed

v

E_k = \frac{1}{2}mv^2

. Substituting into

E = 2E_k

gives

\frac{1}{2}m(\omega A)^2 = 2(\frac{1}{2}mv^2) = mv^2

. Simplifying gives

\frac{1}{2}(\omega A)^2 = v^2

, so

v = \frac{\omega A}{\sqrt{2}}

Question 16

Which statement correctly describes the relationship between the net force $F$ and displacement $x$ for an object undergoing simple harmonic motion?

$F$ is constant and directed towards equilibrium.
$F$ is proportional to $x$ and in the same direction.
$F$ is proportional to $x^2$ and directed towards equilibrium.
$F$ is proportional to $x$ and in the opposite direction. (correct answer)

Explanation: From Newton's second law,

F = ma

. The defining equation for SHM is

a = -\omega^2 x

. Substituting for

a

gives

F = m(-\omega^2 x) = -(m\omega^2)x

. Since

m

and

\omega

are constants, this shows that the net force

F

is directly proportional to the displacement

x

and the negative sign indicates it is always in the opposite direction to the displacement (i.e., it is a restoring force).

Question 17

An object oscillates in simple harmonic motion. What are the average velocity and average speed of the object over one complete period of motion?

Both are zero.
Average velocity is zero; average speed is non-zero. (correct answer)
Average velocity is non-zero; average speed is zero.
Both are non-zero.

Explanation: Average velocity is defined as total displacement divided by total time. Over one complete period, the object returns to its starting position, so its total displacement is zero. Therefore, its average velocity is zero. Average speed is total distance travelled divided by total time. Over one period, the object moves from one amplitude to the other and back again, covering a non-zero distance. Therefore, its average speed is non-zero.

Question 18

A simple pendulum has a length of 0.25 m. It is displaced and released, oscillating with an amplitude of 0.05 m. Assuming $g = 10$ m s⁻², what is the maximum speed of the pendulum bob?

0.32 m s⁻¹ (correct answer)
0.22 m s⁻¹
0.10 m s⁻¹
0.50 m s⁻¹

Explanation: First, calculate the angular frequency

\omega

for the pendulum using

\omega = \sqrt{g/l}

. Given

l=0.25

m and

g=10

m s⁻²,

\omega = \sqrt{10/0.25} = \sqrt{40} \approx 6.32

rad s⁻¹. The maximum speed in SHM is

v_{max} = \omega A

, where

A

is the amplitude. Given

A = 0.05

v_{max} = (6.32 \text{ rad s⁻¹}) \times (0.05 \text{ m}) \approx 0.316

m s⁻¹. The closest answer is 0.32 m s⁻¹.

Question 19

An object in SHM is at its position of maximum positive displacement. It then moves to the equilibrium position. How do its kinetic energy (KE) and potential energy (PE) change during this interval?

KE increases from zero to maximum; PE decreases from maximum to zero. (correct answer)
KE decreases from maximum to zero; PE increases from zero to maximum.
KE remains constant; PE decreases from maximum to zero.
KE increases from zero to maximum; PE remains constant.

Explanation: At the maximum positive displacement (the amplitude), the object is momentarily at rest, so its kinetic energy is zero. The displacement is maximum, so the potential energy stored in the system (e.g., a spring) is at its maximum. As the object moves towards the equilibrium position, its speed increases, and its displacement decreases. At the equilibrium position (

x=0

), the speed is maximum, so kinetic energy is maximum. The displacement is zero, so potential energy is zero. Thus, during this interval, KE increases from zero to its maximum value, and PE decreases from its maximum value to zero.

Question 20

A system undergoing simple harmonic motion is subject to light damping. How does this damping affect the amplitude and the period of the oscillations?

Amplitude decreases; period decreases significantly.
Amplitude decreases; period remains approximately constant. (correct answer)
Amplitude remains constant; period increases.
Amplitude increases; period remains approximately constant.

Explanation: Damping is a process where energy is removed from an oscillating system, usually due to friction or air resistance. This loss of energy causes the amplitude of the oscillations to decrease over time. For light damping, the period of oscillation increases very slightly, but it is often considered to be approximately constant, especially when compared to the significant decrease in amplitude. The primary and most noticeable effect of light damping is the decay of amplitude.