IB Physics Quiz: Apply Simple Harmonic Motion

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Apply Simple Harmonic MotionQuestion 1 of 20

A mass $m$ attached to a spring with spring constant $k$ oscillates with period $T$ . What is the period of oscillation if the mass is changed to (2m) and the spring constant is changed to $k/2$ ?

T

T\sqrt{2}

2T

4T

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IB Physics Quiz

IB Physics Quiz: Apply Simple Harmonic Motion

Practice Apply 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.

What this quiz covers

This quiz focuses on Apply 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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Question 1

A mass $m$ attached to a spring with spring constant $k$ oscillates with period $T$ . What is the period of oscillation if the mass is changed to (2m) and the spring constant is changed to $k/2$ ?

$T$
$T\sqrt{2}$
$2T$ (correct answer)
$4T$

Explanation: The period of a mass-spring system is

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

. The new mass is

m' = 2m

and the new spring constant is

k' = k/2

. The new period is

T' = 2\pi\sqrt{m'/k'} = 2\pi\sqrt{2m/(k/2)} = 2\pi\sqrt{4m/k} = 2 \cdot (2\pi\sqrt{m/k}) = 2T

Question 2

Which statement correctly describes the relationship between the acceleration $a$ and displacement $x$ of an object in simple harmonic motion?

Acceleration is constant and in the same direction as displacement.
Acceleration is proportional to displacement and is in the same direction.
Acceleration is constant and in the opposite direction to displacement.
Acceleration is proportional to displacement and is in the opposite direction. (correct answer)

Explanation: The defining equation of simple harmonic motion is

a = -\omega^2x

. This equation shows that acceleration

a

is directly proportional to the displacement

x

(since

\omega^2

is a positive constant). The negative sign indicates that the acceleration vector always points in the opposite direction to the displacement vector.

Question 3

A simple pendulum is suspended from the ceiling of an elevator. The elevator is accelerating upwards at a constant rate. How does the period of the pendulum's oscillation compare to its period when the elevator is at rest?

The period is greater.
The period is less. (correct answer)
The period is the same.
The period becomes zero.

Explanation: When the elevator accelerates upwards, the apparent weight of the pendulum bob increases. This is equivalent to an increase in the effective gravitational field strength,

g_{eff} = g + a

. Since the period of a pendulum is

T = 2\pi\sqrt{l/g_{eff}}

, an increase in

g_{eff}

results in a decrease in the period

T

Question 4

The total energy of a mass-spring system in simple harmonic motion is doubled. By what factor does the amplitude of the oscillation change?

$1/\sqrt{2}$
$\sqrt{2}$ (correct answer)
2
4

Explanation: The total energy

E

of a mass-spring system in SHM is proportional to the square of the amplitude

A

, given by

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

. If the energy is doubled to

2E

, then

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

. Taking the ratio of the new energy to the old energy gives

2 = (A'/A)^2

. Therefore, the new amplitude

A'

A\sqrt{2}

. The factor of change is

\sqrt{2}

Question 5

A particle undergoes simple harmonic motion. When its displacement is at its maximum positive value, what can be said about its velocity and acceleration?

Velocity is zero and acceleration is zero.
Velocity is zero and acceleration is maximum negative. (correct answer)
Velocity is maximum positive and acceleration is zero.
Velocity is maximum negative and acceleration is maximum positive.

Explanation: At the maximum positive displacement (the amplitude), the particle momentarily stops before changing direction, so its velocity is zero. According to the defining equation of SHM,

a = -\omega^2x

, acceleration is proportional to the negative of displacement. Therefore, when displacement

x

is maximum positive, acceleration

a

is maximum negative.

Question 6

Two simple pendulums, P and Q, are on the same planet. The length of pendulum P is four times the length of pendulum Q. The mass of the bob of P is half the mass of the bob of Q. What is the ratio of the period of P to the period of Q, $T_P / T_Q$ ?

1/2
$\sqrt{2}$
2 (correct answer)
4

Explanation: The period of a simple pendulum is given by

T = 2\pi\sqrt{l/g}

. It is independent of the mass of the bob. The ratio of the periods is

T_P / T_Q = (2\pi\sqrt{l_P/g}) / (2\pi\sqrt{l_Q/g}) = \sqrt{l_P/l_Q}

. Given that

l_P = 4l_Q

, the ratio becomes

\sqrt{4l_Q/l_Q} = \sqrt{4} = 2

Question 7

A mass attached to a vertical spring is at its lowest point in its oscillation. Which statement is correct about the net force and acceleration at this instant?

Both the net force and acceleration are at their maximum upward value. (correct answer)
The net force is downwards and the acceleration is upwards.
Both the net force and acceleration are zero.
The net force is zero but the acceleration is at its maximum downward value.

Explanation: At the lowest point of oscillation, the displacement from the equilibrium position is maximum downwards. According to the SHM definition

a = -\omega^2 x

, if

x

is maximum negative, acceleration

a

is maximum positive (upwards). By Newton's second law

F_{net} = ma

, the net force must also be maximum and in the upward direction. This is because the upward spring force is at its maximum and is greater than the downward force of gravity.

Question 8

A simple harmonic oscillator has maximum kinetic energy $K_{max}$ and maximum potential energy $U_{max}$ . What is the kinetic energy of the oscillator when its displacement is half of its amplitude?

$3K_{max} / 4$ (correct answer)
$K_{max} / 2$
$K_{max} / 4$
$K_{max}$

Explanation: The total energy is

E_T = K_{max} = U_{max} = \frac{1}{2}kA^2

. The potential energy at displacement

x

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

. At

x = A/2

, the potential energy is

U = \frac{1}{2}k(A/2)^2 = \frac{1}{4}(\frac{1}{2}kA^2) = U_{max}/4

. By conservation of energy,

K + U = E_T

. Therefore,

K = E_T - U = U_{max} - U_{max}/4 = 3U_{max}/4

. Since

U_{max} = K_{max}

, the kinetic energy is

3K_{max}/4

Question 9

A block attached to an ideal spring oscillates horizontally on a frictionless surface with period $T$ . The same system is then suspended vertically and set into oscillation. What is the new period of oscillation?

Less than $T$ , as gravity assists the motion downwards.
Greater than $T$ , as gravity opposes the motion upwards.
Exactly $T$ . (correct answer)
Dependent on the amplitude of the vertical oscillation.

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

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

. This depends only on the mass

m

and the spring constant

k

. When the system is vertical, gravity exerts a constant downward force, which shifts the equilibrium position but does not change the restoring force for displacements from this new equilibrium. Therefore, the period of oscillation remains unchanged.

Question 10

A pendulum oscillates with simple harmonic motion. When the pendulum bob is at 60% of its maximum displacement from equilibrium, what is the ratio of its kinetic energy to its total mechanical energy?

0.36
0.64 (correct answer)
0.60
0.80

Explanation: In SHM, total energy E = ½kA² where A is amplitude. At displacement x = 0.6A, potential energy U = ½kx² = ½k(0.6A)² = 0.36(½kA²) = 0.36E. Since energy is conserved, kinetic energy K = E - U = E - 0.36E = 0.64E. Therefore K/E = 0.64. Choice A gives the potential energy ratio. Choice C incorrectly uses the displacement ratio directly. Choice D uses an incorrect energy relationship.

Question 11

A mass-spring system has total energy $E = 0.50$ J and amplitude $A = 0.10$ m. At what position will the particle's speed equal 75% of its maximum speed?

$\pm 0.066$ m from equilibrium (correct answer)
$\pm 0.075$ m from equilibrium
$\pm 0.085$ m from equilibrium
$\pm 0.043$ m from equilibrium

Explanation: Maximum speed occurs at equilibrium: v_max = ωA. When v = 0.75v_max, we use energy conservation: ½mv² + ½kx² = ½mω²A². Substituting v = 0.75ωA: ½m(0.75ωA)² + ½kx² = ½mω²A². This gives ½m(0.5625ω²A²) + ½kx² = ½mω²A². Since k = mω², we get: 0.5625(½mω²A²) + ½mω²x² = ½mω²A². Dividing by ½mω²: 0.5625A² + x² = A². Therefore x² = A²(1 - 0.5625) = 0.4375A². So |x| = A√0.4375 = 0.10√0.4375 ≈ 0.066 m. Choice B uses √0.5. Choice C uses an incorrect energy fraction. Choice D uses 0.75² incorrectly.

Question 12

Two pendulums A and B have the same length but different masses ( $m_A = 2m_B$ ). Both are displaced by the same small angle and released simultaneously. After pendulum A completes exactly 10 oscillations, how many oscillations has pendulum B completed?

Exactly 10 oscillations, since period is independent of mass (correct answer)
Exactly 7.1 oscillations, due to the mass difference
Exactly 14.1 oscillations, since lighter objects oscillate faster
Between 9 and 11 oscillations, depending on the amplitude

Explanation: For a simple pendulum, the period T = 2π√(L/g) depends only on length L and gravitational acceleration g, not on mass m. Since both pendulums have identical length, they have identical periods and frequencies. Therefore, they complete oscillations at exactly the same rate. After any given time interval, both will have completed the same number of oscillations. Choice B incorrectly assumes mass affects period. Choice C shows the common misconception that lighter objects oscillate faster. Choice D incorrectly suggests amplitude dependence for small oscillations.

Question 13

Two identical masses undergo SHM with the same amplitude but different periods. Mass 1 has period $T_1 = 2.0$ s and mass 2 has period $T_2 = 4.0$ s. When both masses have the same displacement from equilibrium, what is the ratio of their kinetic energies $KE_1/KE_2$ ?

The ratio depends on the specific displacement value
The ratio equals 2.0 at all displacement values
The ratio equals 4.0 at all displacement values (correct answer)
The ratio equals 1.0 at all displacement values

Explanation: When you encounter SHM problems comparing two systems with different periods, focus on how energy varies with position and the fundamental relationships between period, frequency, and energy. For SHM, the kinetic energy at any displacement

x

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

, where

\omega = \frac{2\pi}{T}

is the angular frequency. Since both masses are identical and have the same amplitude

A

, the ratio of their kinetic energies at the same displacement becomes:

\frac{KE_1}{KE_2} = \frac{\omega_1^2}{\omega_2^2} = \frac{(2\pi/T_1)^2}{(2\pi/T_2)^2} = \frac{T_2^2}{T_1^2}

Substituting the given periods:

\frac{KE_1}{KE_2} = \frac{(4.0)^2}{(2.0)^2} = \frac{16}{4} = 4.0

This ratio is independent of displacement because the

(A^2 - x^2)

term cancels out when taking the ratio. Option A is incorrect because the displacement terms cancel, making the ratio constant. Option B gives the wrong numerical value—this would result from incorrectly using

T_2/T_1

instead of the squared ratio. Option D would occur if you mistakenly thought identical masses and amplitudes meant equal kinetic energies, ignoring the period difference entirely. Remember: In SHM energy comparisons, the key factor is

\omega^2

, which scales as

1/T^2

. When periods differ, the system with the shorter period has dramatically higher kinetic energy due to this squared relationship.

Question 14

A particle undergoes SHM with angular frequency $\omega = 5.0$ rad/s. At $t = 0$ , the particle is at equilibrium moving in the positive direction with speed $v_0 = 2.0$ m/s. What is the phase constant $\phi$ in the equation $x(t) = A\cos(\omega t + \phi)$ ?

$\phi = +\pi/2$ radians, to account for the initial velocity direction
$\phi = 0$ radians, since the particle starts at equilibrium
$\phi = -\pi/2$ radians, since cosine starts at maximum (correct answer)
$\phi = \pi$ radians, since the motion begins at the center

Explanation: When analyzing simple harmonic motion problems, you need to match the given initial conditions to the correct form of the position equation. The key is understanding how position, velocity, and phase constants relate at

t = 0

. Given

x(t) = A\cos(\omega t + \phi)

, the velocity is

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

. At

t = 0

, you have initial conditions:

x(0) = 0

(at equilibrium) and

v(0) = +2.0

m/s (positive direction). For the position:

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

. This means

\cos(\phi) = 0

, which occurs when

\phi = \pm\pi/2

. For the velocity:

v(0) = -A\omega\sin(\phi) = +2.0

m/s. Since

\omega = 5.0

rad/s is positive and the velocity is positive, you need

-A\sin(\phi) > 0

. This requires

\sin(\phi) < 0

. When

\phi = -\pi/2

\sin(-\pi/2) = -1 < 0

, which gives the correct positive velocity. Therefore,

\phi = -\pi/2

radians. Option A (

\phi = +\pi/2

) gives

\sin(+\pi/2) = +1

, resulting in negative initial velocity. Option B (

\phi = 0

) gives

\cos(0) = 1

, meaning the particle starts at maximum displacement, not equilibrium. Option D (

\phi = \pi

) gives

\cos(\pi) = -1

, placing the particle at minimum displacement initially. Study tip: Always check both position and velocity initial conditions. The phase constant must satisfy both simultaneously—don't just focus on where the particle starts.

Question 15

A mass attached to a vertical spring oscillates with SHM. The spring constant is $k = 200$ N/m and the mass is $m = 0.50$ kg. If the amplitude of oscillation about the equilibrium position is $A = 0.075$ m, what is the maximum acceleration experienced by the mass?

$6.0$ m/s² occurring at the equilibrium position
$12$ m/s² occurring at maximum displacement from equilibrium
$20$ m/s² occurring at the turning points of the motion
$30$ m/s² occurring when the spring force is maximum (correct answer)

Explanation: In SHM, acceleration a = -ω²x where ω = √(k/m) = √(200/0.50) = √400 = 20 rad/s. Maximum acceleration occurs at maximum displacement: |a|_max = ω²A = (20)² × 0.075 = 400 × 0.075 = 30 m/s². The maximum acceleration occurs at maximum displacement (turning points) when the spring force is maximum. Choice A incorrectly states maximum acceleration occurs at equilibrium (where a = 0). Choice B has the wrong magnitude. Choice C has wrong magnitude but mentions the correct location.

Question 16

An object in simple harmonic motion has a period of 4.0 s. What is the minimum time it takes for the object to travel from its maximum displacement to the equilibrium position?

1.0 s (correct answer)
2.0 s
4.0 s
8.0 s

Explanation: One full period (T) is the time taken to complete one full oscillation. This can be divided into four equal parts: from maximum displacement to equilibrium, from equilibrium to minimum displacement, from minimum displacement back to equilibrium, and from equilibrium back to maximum displacement. Each of these segments takes a time of T/4. Therefore, the time taken is

4.0 \text{ s} / 4 = 1.0 \text{ s}

Question 17

An object oscillates with simple harmonic motion of amplitude $x_0$ and angular frequency $\omega$ . What is the magnitude of its acceleration when it is at a displacement of $x = x_0/2$ ?

$\omega^2 x_0$
$\omega^2 x_0 / 2$ (correct answer)
$\omega^2 x_0 / 4$
$\omega x_0 / 2$

Explanation: The defining equation for SHM is

a = -\omega^2 x

. The magnitude of the acceleration is

|a| = \omega^2 |x|

. When the displacement is

x = x_0/2

, the magnitude of the acceleration is

a = \omega^2 (x_0/2) = \omega^2 x_0 / 2

Question 18

An object on a spring is oscillating in a system where light damping is present. Which statement correctly describes the effect of damping on the motion over several oscillations?

The amplitude decreases while the frequency increases.
The amplitude is constant but the period increases.
Both the amplitude and the frequency decrease to zero.
The amplitude decreases while the period remains approximately constant. (correct answer)

Explanation: Light damping causes a gradual loss of mechanical energy from the oscillating system, resulting in an exponential decrease in the amplitude of oscillation. For light damping, the effect on the period (and thus frequency) is negligible, so it is considered to remain approximately constant. The motion eventually stops, but the frequency doesn't decrease to zero during the oscillation.

Question 19

A mass of 200 g is attached to a spring with a spring constant of 50 N m⁻¹. The mass is displaced and released. What is the frequency of the resulting oscillation, to two significant figures?

0.080 Hz
2.5 Hz (correct answer)
16 Hz
40 Hz

Explanation: First, convert the mass to SI units:

m = 200 \text{ g} = 0.200 \text{ kg}

. The angular frequency

\omega

is given by

\omega = \sqrt{k/m} = \sqrt{50 \text{ N m}^{-1} / 0.200 \text{ kg}} = \sqrt{250} \approx 15.8 \text{ rad s}^{-1}

. The frequency

f

is related to

\omega

f = \omega / (2\pi)

. So,

f = 15.8 / (2\pi) \approx 2.516 \text{ Hz}

. To two significant figures, this is 2.5 Hz.

Question 20

A simple pendulum has a period $T$ on Earth. It is taken to a planet with twice Earth's mass and twice Earth's radius. What is the new period of the pendulum on this planet?

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

Explanation: The period of a simple pendulum is given by

T = 2\pi\sqrt{l/g}

. The gravitational field strength is

g = GM/R^2

. On the new planet,

M' = 2M

and

R' = 2R

. The new gravity is

g' = G(2M)/(2R)^2 = G(2M)/(4R^2) = \frac{1}{2}(GM/R^2) = g/2

. The new period is

T' = 2\pi\sqrt{l/g'} = 2\pi\sqrt{l/(g/2)} = 2\pi\sqrt{2l/g} = \sqrt{2} \cdot (2\pi\sqrt{l/g}) = T\sqrt{2}