IB Physics Quiz: Apply Wave Model

20 questions · exam conditions

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Apply Wave ModelQuestion 1 of 20

Two sound waves, X and Y, travel through the same body of air at the same temperature. The wavelength of wave X is twice the wavelength of wave Y. What is the ratio of the speed of wave X to the speed of wave Y ( $v_X / v_Y$ )?

1/2

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IB Physics Quiz

IB Physics Quiz: Apply Wave Model

Practice Apply Wave Model 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 Wave Model, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

How to use this quiz

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.

All questions

Question 1

1/2
1 (correct answer)
2
4

Explanation: The speed of a mechanical wave, such as sound, depends only on the properties of the medium through which it travels (in this case, the air's temperature, pressure, and density). Since both waves travel through the same medium, their speeds must be identical. Therefore, the ratio

v_X / v_Y

is 1.

Question 2

A transverse wave travels from left to right along a horizontal string. At a particular instant, a point P on the string is at the crest of the wave (maximum positive displacement). What is the instantaneous velocity of the particle at point P?

Zero (correct answer)
Directed to the right
Directed upwards
Directed downwards

Explanation: The particles of the string oscillate in simple harmonic motion. At the point of maximum displacement (the crest), the particle momentarily stops before changing its direction to move downwards. Therefore, its instantaneous velocity is zero. The velocity of the wave is to the right, but this is different from the velocity of the individual particles of the medium.

Question 3

A continuous transverse wave is travelling along a string. Two points on the string, P and Q, are separated by a distance equal to three-quarters of the wavelength ( $3\lambda/4$ ). What is the phase difference between the oscillations of points P and Q?

$\pi/2$ radians
$\pi$ radians
$3\pi/2$ radians (correct answer)
$2\pi$ radians

Explanation: A full wavelength (

\lambda

) corresponds to a phase difference of

2\pi

radians. The phase difference is directly proportional to the separation distance. Therefore, for a separation of

3\lambda/4

, the phase difference

\Delta\phi

(3\lambda/4) / \lambda \times 2\pi = (3/4) \times 2\pi = 3\pi/2

radians.

Question 4

Which property is the same for all electromagnetic waves travelling in a vacuum?

Frequency
Wavelength
Photon energy
Speed (correct answer)

Explanation: All electromagnetic waves, from radio waves to gamma rays, travel at the same speed in a vacuum, which is the speed of light,

c \approx 3.00 \times 10^8

m s⁻¹. Frequency, wavelength, and photon energy (which is proportional to frequency,

E=hf

) vary across the electromagnetic spectrum.

Question 5

A sound wave is generated in air and then enters a large body of water. The speed of sound is greater in water than in air. Which property of the sound wave must remain constant as it crosses the boundary from air to water?

Frequency (correct answer)
Wavelength
Speed
Amplitude

Explanation: The frequency of a wave is determined by its source and does not change when the wave passes from one medium to another. The wave speed changes with the medium, as stated in the question. Since

v = f\lambda

, a change in speed with constant frequency implies that the wavelength must also change. The amplitude will change as some of the wave's energy is reflected at the boundary and some is transmitted.

Question 6

The intensity of a sound wave is proportional to the square of its displacement amplitude. A sound wave X has an intensity $I$ . A second sound wave Y, travelling in the same medium, has a displacement amplitude that is one-third the amplitude of wave X. What is the intensity of wave Y?

$9I$
$I/3$
$3I$
$I/9$ (correct answer)

Explanation: Let the amplitude of wave X be

A_X

and its intensity be

I_X = I

. We are given that

I \propto A^2

. The amplitude of wave Y is

A_Y = A_X/3

. The intensity of wave Y,

I_Y

, will be proportional to

(A_Y)^2

. Thus, the ratio of intensities is

I_Y / I_X = (A_Y)^2 / (A_X)^2 = (A_X/3)^2 / (A_X)^2 = (A_X^2/9) / A_X^2 = 1/9

. Therefore,

I_Y = I_X/9 = I/9

Question 7

A point on a transverse wave oscillates with a period $T$ . In a time interval equal to one-quarter of the period, the wave itself travels a distance $d$ . What is the wavelength of the wave?

$d/4$
$d$
$2d$
$4d$ (correct answer)

Explanation: The speed of the wave is

v = \text{distance} / \text{time}

. The time interval is

t = T/4

and the distance is

d

, so

v = d / (T/4) = 4d/T

. The wavelength

\lambda

is the distance the wave travels in one full period, so

\lambda = v \times T

. Substituting the expression for

v

, we get

\lambda = (4d/T) \times T = 4d

Question 8

A monochromatic light wave travels from air (refractive index ≈ 1.0) into a block of glass (refractive index ≈ 1.5). Which row correctly describes the changes, if any, to the frequency, wavelength, and speed of the light wave?

The frequency is constant, the wavelength decreases, and the speed decreases. (correct answer)
The frequency decreases, the wavelength is constant, and the speed decreases.
The frequency is constant, the wavelength increases, and the speed increases.
The frequency increases, the wavelength decreases, and the speed is constant.

Explanation: The frequency of a wave is determined by its source and does not change when the wave enters a new medium. The speed of light decreases in a denser medium (higher refractive index). According to the wave equation

v = f\lambda

, if

v

decreases and

f

remains constant, the wavelength

\lambda

must also decrease.

Question 9

A wave of frequency $f$ travels along a stretched string where its speed is $v$ . The tension in the string is then adjusted so that the wave speed becomes $2v$ . If the source continues to generate waves with the same frequency $f$ , what is the new wavelength?

$\lambda / 2$
$\lambda$
$2\lambda$ (correct answer)
$4\lambda$

Explanation: The relationship between wave speed (v), frequency (f), and wavelength (

\lambda

) is

v = f\lambda

. Initially,

\lambda = v/f

. The frequency

f

is determined by the source and remains constant. The new speed is

v' = 2v

. The new wavelength is

\lambda' = v'/f = (2v)/f = 2(v/f) = 2\lambda

Question 10

A radio station broadcasts at a frequency of 102.5 MHz. The waves travel from the transmitter to a receiver 60.0 km away. Approximately how many full wavelengths fit between the transmitter and the receiver? (Speed of light $c = 3.00 \times 10^8$ m s⁻¹)

2.93
5.13 \times 10^3
2.05 \times 10^4 (correct answer)
2.05 \times 10^7

Explanation: First, calculate the wavelength using

\lambda = v/f

. The frequency is

f = 102.5 \times 10^6

Hz and the speed is

v = c = 3.00 \times 10^8

m s⁻¹. So,

\lambda = (3.00 \times 10^8) / (102.5 \times 10^6) \approx 2.927

m. Next, calculate the number of wavelengths (N) in the distance (d = 60.0 km = 60.0

\times

10^3

m).

N = d / \lambda = (60.0 \times 10^3) / 2.927 \approx 20499

. This is approximately

2.05 \times 10^4

Question 11

An observer sees a flash of lightning and hears the corresponding thunder 4.5 s later. Assuming the speed of sound in air is 340 m s⁻¹, what is the best estimate for the distance to the lightning strike?

76 m
340 m
1.5 km (correct answer)
1.4 \times 10^9 m

Explanation: The speed of light is extremely high (

3.0 \times 10^8

m s⁻¹), so the time taken for the light to reach the observer is negligible. The distance can be calculated using the time it takes for the sound to travel. Using the formula distance = speed × time, we get

d = 340 \text{ m s⁻¹} \times 4.5 \text{ s} = 1530 \text{ m}

, which is approximately 1.5 km.

Question 12

A source produces a wave of period $T$ that travels through a medium with speed $v$ . What is the shortest distance between two points on the wave that are oscillating with a phase difference of $\pi/2$ radians?

$vT/8$
$vT/4$ (correct answer)
$vT/2$
$vT$

Explanation: One full wavelength,

\lambda

, corresponds to a phase difference of

2\pi

radians. The wavelength can be expressed as

\lambda = vT

. A phase difference of

\pi/2

(\pi/2) / (2\pi) = 1/4

of a full cycle. Therefore, the distance between these two points is

1/4

of a wavelength. The distance is

\lambda/4 = (vT)/4

Question 13

The intensity $I$ of a mechanical wave is proportional to the square of its amplitude $A$ and the square of its frequency $f$ , such that $I \propto A^2 f^2$ .

The amplitude of a wave is doubled, and its frequency is halved. What is the effect on the intensity of the wave?

The intensity is quartered.
The intensity is halved.
The intensity remains unchanged. (correct answer)
The intensity is doubled.

Explanation: Let the initial intensity be

I_{old} \propto A^2 f^2

. The new amplitude is

A_{new} = 2A

and the new frequency is

f_{new} = f/2

. The new intensity is

I_{new} \propto (A_{new})^2 (f_{new})^2 = (2A)^2 (f/2)^2 = (4A^2)(f^2/4) = A^2 f^2

. Since the expression for the new intensity is the same as for the old intensity, the intensity remains unchanged.

Question 14

A wave is generated in a Slinky spring. An observer notes that the coils of the spring move back and forth along the same axis as the wave propagates. The distance between consecutive regions of maximum coil density is 0.80 m. The wave travels 4.0 m in 2.0 s. What are the type and frequency of the wave?

Longitudinal, 2.5 Hz (correct answer)
Transverse, 2.5 Hz
Longitudinal, 1.6 Hz
Transverse, 1.6 Hz

Explanation: The coils move parallel to the wave propagation, which defines a longitudinal wave. The distance between consecutive regions of maximum density (compressions) is the wavelength, so

\lambda = 0.80

m. The speed of the wave is

v = \text{distance} / \text{time} = 4.0 \text{ m} / 2.0 \text{ s} = 2.0

m s⁻¹. The frequency is found using

f = v / \lambda = 2.0 / 0.80 = 2.5

Hz.

Question 15

Two coherent wave sources $S_1$ and $S_2$ separated by distance $d = 4.0 \text{ m}$ emit waves in phase with wavelength $\lambda = 1.2 \text{ m}$ . A detector moves along a line parallel to the line connecting the sources, at distance $L = 12 \text{ m}$ from the midpoint. At what distance from the perpendicular bisector will the detector first encounter a minimum in the interference pattern?

$1.8 \text{ m}$ (correct answer)
$2.4 \text{ m}$
$3.0 \text{ m}$
$3.6 \text{ m}$

Explanation: For destructive interference, the path difference must be

(m + \frac{1}{2})\lambda

where

m

is an integer. The path difference is

\Delta = \sqrt{L^2 + (x + d/2)^2} - \sqrt{L^2 + (x - d/2)^2}

. For small angles,

\Delta \approx \frac{xd}{L}

. For the first minimum (

m = 0

\frac{xd}{L} = \frac{\lambda}{2}

. Substituting:

x = \frac{\lambda L}{2d} = \frac{1.2 \times 12}{2 \times 4.0} = 1.8 \text{ m}

Question 16

A standing wave is established on a string of length $L = 0.60 \text{ m}$ fixed at both ends. The wave speed is $v = 240 \text{ m/s}$ . When the string vibrates in its third harmonic, what is the minimum time required for a particle initially at rest at an antinode to return to rest at the same position?

$1.25 \times 10^{-3} \text{ s}$
$2.50 \times 10^{-3} \text{ s}$ (correct answer)
$5.00 \times 10^{-3} \text{ s}$
$7.50 \times 10^{-3} \text{ s}$

Explanation: For the third harmonic,

f_3 = \frac{3v}{2L} = \frac{3 \times 240}{2 \times 0.60} = 600 \text{ Hz}

. The period is

T = \frac{1}{f_3} = \frac{1}{600} = 5.0 \times 10^{-3} \text{ s}

. A particle at an antinode oscillates with displacement

y(t) = A\cos(2\pi f_3 t)

. Starting from rest at

t = 0

means the particle is at maximum displacement. It returns to rest at the same position after half a period:

\frac{T}{2} = \frac{5.0 \times 10^{-3}}{2} = 2.50 \times 10^{-3} \text{ s}

Question 17

A wave $y_1 = 0.05\sin(10t - 2x)$ interferes with wave $y_2 = 0.05\sin(10t - 2x + \phi)$ , where $x$ is in meters, $t$ is in seconds, and $\phi$ is the phase difference. If the resultant wave has an amplitude of $0.07 \text{ m}$ at all positions, what is the value of $\phi$ ?

$\frac{\pi}{6}$
$\frac{\pi}{4}$
$\frac{\pi}{3}$ (correct answer)
$\frac{\pi}{2}$

Explanation: When two waves with equal amplitudes

A_1 = A_2 = 0.05 \text{ m}

interfere, the resultant amplitude is

A_{res} = 2A_1\cos(\phi/2)

. Given

A_{res} = 0.07 \text{ m}

0.07 = 2 \times 0.05 \times \cos(\phi/2)

. Solving:

\cos(\phi/2) = \frac{0.07}{0.10} = 0.7

. Therefore

\phi/2 = \arccos(0.7) \approx 0.524 \text{ rad} = \pi/6

, which gives

\phi = \pi/3

Question 18

A wave undergoes total internal reflection at the boundary between two media. The incident wave has amplitude $A_i$ and the reflected wave has amplitude $A_r$ . Due to the phase change upon reflection, a standing wave pattern forms in the incident medium. What is the amplitude of oscillation at a point located $\lambda/8$ from the boundary?

$A_i$
$A_i\sqrt{2}$ (correct answer)
$\frac{A_i}{\sqrt{2}}$
$\frac{A_i}{2}$

Explanation: For total internal reflection,

|A_r| = |A_i|

but with a phase change of

\pi

. The standing wave amplitude at distance

x

from the boundary is

A(x) = 2A_i|\sin(kx)|

where

k = 2\pi/\lambda

. At

x = \lambda/8

kx = 2\pi \cdot \frac{\lambda/8}{\lambda} = \frac{\pi}{4}

. Therefore

A(\lambda/8) = 2A_i\sin(\pi/4) = 2A_i \cdot \frac{1}{\sqrt{2}} = A_i\sqrt{2}

Question 19

A wave pulse traveling on a rope encounters a junction where three identical ropes meet. The incident pulse has amplitude $A$ and carries power $P$ . Assuming no energy is lost at the junction and the transmitted pulses maintain their shape, what is the amplitude of each transmitted pulse?

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

Explanation: When a wave pulse splits at a junction, you need to apply two fundamental principles: conservation of energy and the relationship between wave power and amplitude. Since the three ropes are identical, symmetry tells us each transmitted pulse will have the same amplitude. The power carried by a wave is proportional to the square of its amplitude:

P \propto A^2

. Since no energy is lost at the junction, the total power must be conserved. The incident pulse carries power

P

, and this power splits equally among the three transmitted pulses, so each carries power

\frac{P}{3}

. If each transmitted pulse has amplitude

A_t

, then

\frac{P}{3} \propto A_t^2

. Since the original power satisfies

P \propto A^2

, we can write:

\frac{P}{3} = \frac{A^2}{3} \times \text{(proportionality constant)}

Therefore:

A_t^2 = \frac{A^2}{3}

, which gives

A_t = \frac{A}{\sqrt{3}}

. Answer A (

\frac{A\sqrt{3}}{3}

) is actually equivalent to the correct answer when rationalized, but it's written in an unnecessarily complex form. Answer B (

\frac{A}{2}

) incorrectly assumes the amplitude scales linearly with the number of branches. Answer C (

\frac{A}{3}

) makes the same linear scaling error, dividing amplitude directly by three instead of considering that power depends on amplitude squared. Remember: when waves split symmetrically, power divides equally among branches, but since power depends on amplitude squared, you must take the square root when finding the new amplitude.

Question 20

Which phenomenon is characteristic of transverse waves but not of longitudinal waves?

Polarization (correct answer)
Refraction
Diffraction
Superposition

Explanation: Polarization is the restriction of the orientation of oscillations to a single plane. This is only possible for transverse waves, where the oscillations are perpendicular to the direction of energy transfer. Longitudinal waves, where oscillations are parallel to the direction of energy transfer, cannot be polarized. Refraction, diffraction, and superposition are phenomena exhibited by all types of waves.