IB Physics Quiz: Understand Wave Model
20 questions · exam conditions
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Understand Wave ModelQuestion 1 of 20

A sound wave is travelling through air. The wave has a period T. Which of the following best describes the motion of a single air molecule due to the passage of this sound wave?

It travels a distance equal to one wavelength during the time T.
It remains stationary, as only energy is propagated.
It oscillates about an equilibrium position with period T.
It moves with a constant velocity equal to the wave speed.
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IB Physics Quiz

IB Physics Quiz: Understand Wave Model

Practice Understand 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.

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This quiz focuses on Understand Wave Model, 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 sound wave is travelling through air. The wave has a period T. Which of the following best describes the motion of a single air molecule due to the passage of this sound wave?

  1. It travels a distance equal to one wavelength during the time T.
  2. It remains stationary, as only energy is propagated.
  3. It oscillates about an equilibrium position with period T. (correct answer)
  4. It moves with a constant velocity equal to the wave speed.
Explanation: A sound wave is a mechanical longitudinal wave. The individual molecules of the medium (air) do not travel along with the wave. Instead, they oscillate back and forth about their fixed equilibrium positions. The period of this oscillation for a single molecule is the same as the period of the wave itself, T. They do not have a net displacement and their velocity is not constant.

Question 2

An electromagnetic wave has a frequency of 1.5×1014 Hz1.5 \times 10^{14} \text{ Hz} in a vacuum. It enters a transparent medium where the speed of light is 2.0×108 m s−12.0 \times 10^8 \text{ m s}^{-1}. What is the wavelength of the wave inside this medium?

  1. 5.0×10−7 m5.0 \times 10^{-7} \text{ m}
  2. 1.3×10−6 m1.3 \times 10^{-6} \text{ m} (correct answer)
  3. 2.0×10−6 m2.0 \times 10^{-6} \text{ m}
  4. 3.0×10−6 m3.0 \times 10^{-6} \text{ m}
Explanation: When a wave enters a new medium, its frequency remains constant at 1.5×1014 Hz1.5 \times 10^{14} \text{ Hz}. Using the wave equation v=fλv = f\lambda, we can find the wavelength: λ=vf=2.0×1081.5×1014=1.33×10−6 m\lambda = \frac{v}{f} = \frac{2.0 \times 10^8}{1.5 \times 10^{14}} = 1.33 \times 10^{-6} \text{ m}, which rounds to 1.3×10−6 m1.3 \times 10^{-6} \text{ m}.

Question 3

A wave in a certain medium has a speed vv, frequency ff, and wavelength λ\lambda. The source of the wave is adjusted so that the frequency is halved to f/2f/2. The wave then enters a second medium where its speed is doubled to 2v2v. What is the wavelength of the wave in the second medium?

  1. λ\lambda
  2. 2λ2\lambda
  3. 4λ4\lambda (correct answer)
  4. λ/4\lambda/4
Explanation: Initially, λ=v/f\lambda = v/f. The new frequency is fnew=f/2f_{new} = f/2. The frequency remains f/2f/2 when the wave enters the new medium. The new speed is vnew=2vv_{new} = 2v. The new wavelength is λnew=vnew/fnew=(2v)/(f/2)=4(v/f)=4λ\lambda_{new} = v_{new} / f_{new} = (2v) / (f/2) = 4(v/f) = 4\lambda.

Question 4

A transverse wave pulse travels along a horizontal rope. At a specific instant, a point P on the rope has maximum positive vertical displacement. What is the vertical velocity of point P at this instant?

  1. Maximum and upwards.
  2. Maximum and downwards.
  3. Zero. (correct answer)
  4. Equal to the horizontal speed of the wave pulse.
Explanation: The motion of any point on the rope is an oscillation (simple harmonic motion) in the vertical direction. At the point of maximum displacement (a crest), the particle is momentarily at rest as it changes direction from moving upwards to moving downwards. Therefore, its instantaneous vertical velocity is zero. The horizontal speed is the speed of the wave, not the speed of the particle.

Question 5

A water wave is an example of a wave that is a combination of transverse and longitudinal motion. A small cork floating on the water is observed as the wave passes. What is the overall motion of the cork?

  1. It moves in a circular or elliptical path, returning to its starting point. (correct answer)
  2. It is carried along horizontally in the direction of the wave's propagation.
  3. It oscillates purely vertically, perpendicular to the wave's direction.
  4. It oscillates purely horizontally, parallel to the wave's direction.
Explanation: Surface water waves are complex; the particles of the medium (water) move both up and down (transverse component) and back and forth (longitudinal component). The combination of these two oscillations results in a circular or elliptical motion for a particle on the surface. Crucially, like other waves, there is no net transport of the medium, so the cork returns to its approximate original position after the wave passes.

Question 6

Two points, X and Y, are on a string carrying a sinusoidal transverse wave. The distance between X and Y is exactly half a wavelength. How does the vertical motion of particle Y compare to that of particle X?

  1. Y moves in phase with X, reaching its crest at the same time as X.
  2. Y is stationary while X oscillates at maximum amplitude.
  3. Y moves in antiphase with X, reaching its crest when X is at its trough. (correct answer)
  4. Y's motion is delayed by one-quarter of a period compared to X's motion.
Explanation: Points on a wave that are separated by half a wavelength (λ/2\lambda/2) are exactly out of phase, or in antiphase. This means that when one point is at its maximum positive displacement (a crest), the other is at its maximum negative displacement (a trough). Their motions are perfectly opposite at all times.

Question 7

A wave is generated in a medium with a period of 0.50 s and a wavelength of 2.0 m. The wave passes into a second medium where its wavelength changes to 1.5 m. What is the speed of the wave in the second medium?

  1. 1.5 m s−11.5 \text{ m s}^{-1}
  2. 3.0 m s−13.0 \text{ m s}^{-1} (correct answer)
  3. 4.0 m s−14.0 \text{ m s}^{-1}
  4. 6.0 m s−16.0 \text{ m s}^{-1}
Explanation: First, calculate the frequency of the wave from its period: f=1/T=1/0.50 s=2.0 Hzf = 1/T = 1 / 0.50 \text{ s} = 2.0 \text{ Hz}. The frequency of the wave remains constant when it enters the second medium. Now, use the wave equation with the new wavelength to find the new speed: vnew=f×λnew=2.0 Hz×1.5 m=3.0 m s−1v_{new} = f \times \lambda_{new} = 2.0 \text{ Hz} \times 1.5 \text{ m} = 3.0 \text{ m s}^{-1}.

Question 8

A wave source generates a wave of wavelength λ\lambda in a medium where its speed is vv. The wave propagates into a different region of the medium where, due to a change in the medium's properties, the wave speed is reduced to v/3v/3. The source frequency remains constant. What is the wavelength in the new region?

  1. 3λ3\lambda
  2. 9λ9\lambda
  3. λ/9\lambda/9
  4. λ/3\lambda/3 (correct answer)
Explanation: The relationship between wave speed, frequency, and wavelength is v=fλv = f\lambda. The frequency ff is determined by the source and remains constant as the wave moves between regions. In the first region, f=v/λf = v/\lambda. In the second region, the new speed is vnew=v/3v_{new} = v/3. The new wavelength is λnew=vnew/f=(v/3)/(v/λ)=(v/3)×(λ/v)=λ/3\lambda_{new} = v_{new}/f = (v/3) / (v/\lambda) = (v/3) \times (\lambda/v) = \lambda/3.

Question 9

Two waves, P and Q, are generated by different sources and travel through the same uniform body of water. The wavelength of P is three times the wavelength of Q (λP=3λQ\lambda_P = 3\lambda_Q). What is the ratio of the frequency of P to the frequency of Q (fP/fQf_P / f_Q)?

  1. 1/9
  2. 1/3 (correct answer)
  3. 3
  4. 9
Explanation: Since both waves travel through the same uniform medium, their speeds must be equal: vP=vQ=vv_P = v_Q = v. Using the wave equation v=fλv = f\lambda, we have f=v/λf = v/\lambda. Therefore, fP=v/λPf_P = v/\lambda_P and fQ=v/λQf_Q = v/\lambda_Q. The ratio is fP/fQ=(v/λP)/(v/λQ)=λQ/λPf_P / f_Q = (v/\lambda_P) / (v/\lambda_Q) = \lambda_Q / \lambda_P. Since λP=3λQ\lambda_P = 3\lambda_Q, the ratio is λQ/(3λQ)=1/3\lambda_Q / (3\lambda_Q) = 1/3.

Question 10

A continuous sinusoidal longitudinal wave travels from left to right through a uniform medium. Consider a single particle within the medium at the moment it has its maximum displacement to the right. What is the instantaneous velocity of this particle?

  1. Zero. (correct answer)
  2. Maximum and directed to the right.
  3. Maximum and directed to the left.
  4. Equal to the wave speed and directed to the right.
Explanation: The particles in the medium oscillate in simple harmonic motion about their equilibrium positions. Just like a pendulum or a mass on a spring, a particle's velocity is momentarily zero at the point of maximum displacement (the turning point) before it reverses direction. The wave speed is the speed of energy propagation, not the speed of the individual particles, which is variable.

Question 11

What is the ratio of the speed of a gamma ray to the speed of a radio wave in a vacuum?

  1. 10610^6
  2. 10−610^{-6}
  3. 1 (correct answer)
  4. The ratio depends on their respective frequencies.
Explanation: All electromagnetic waves, regardless of their frequency or wavelength (from radio waves to gamma rays), travel at the same speed in a vacuum. This speed is the speed of light, cc. Therefore, the ratio of their speeds is 1.

Question 12

A standing wave is established on a string of length L=1.2 mL = 1.2 \text{ m} fixed at both ends. The wave has frequency f=85 Hzf = 85 \text{ Hz} and corresponds to the fourth harmonic. If the tension in the string is increased by a factor of 1.44 while keeping the same harmonic number, what will be the new frequency?

  1. 93 Hz93 \text{ Hz}
  2. 102 Hz102 \text{ Hz} (correct answer)
  3. 118 Hz118 \text{ Hz}
  4. 122 Hz122 \text{ Hz}
Explanation: For a string fixed at both ends, the frequency of the nnth harmonic is fn=n2LTμf_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}, where TT is tension and μ\mu is linear mass density. The wave speed is v=T/μv = \sqrt{T/\mu}. Initially, f4=42LT1μ=85 Hzf_4 = \frac{4}{2L}\sqrt{\frac{T_1}{\mu}} = 85 \text{ Hz}. When tension increases to T2=1.44T1T_2 = 1.44T_1, the new frequency is f4′=42L1.44T1μ=42L1.44T1μ=1.2×85=102 Hzf_4' = \frac{4}{2L}\sqrt{\frac{1.44T_1}{\mu}} = \frac{4}{2L}\sqrt{1.44}\sqrt{\frac{T_1}{\mu}} = 1.2 \times 85 = 102 \text{ Hz}. Note that 1.44=1.2\sqrt{1.44} = 1.2.

Question 13

In a double-slit experiment, coherent light with wavelength λ=600 nm\lambda = 600 \text{ nm} illuminates two slits separated by d=1.5 mmd = 1.5 \text{ mm}. The interference pattern is observed on a screen L=2.5 mL = 2.5 \text{ m} away. If one of the slits is covered with a thin glass plate of thickness t=0.012 mmt = 0.012 \text{ mm} and refractive index n=1.5n = 1.5, by what distance will the central bright fringe be displaced?

  1. 0.8 mm0.8 \text{ mm}
  2. 1.0 mm1.0 \text{ mm} (correct answer)
  3. 1.6 mm1.6 \text{ mm}
  4. 2.0 mm2.0 \text{ mm}
Explanation: The glass plate introduces an additional optical path difference of (n−1)t=(1.5−1)×0.012=0.006 mm=6.0 μm(n-1)t = (1.5-1) \times 0.012 = 0.006 \text{ mm} = 6.0 \text{ μm}. This corresponds to a phase difference, but more importantly, it shifts the entire fringe pattern. The central bright fringe (where path difference was zero) now occurs where the path difference compensates for the glass plate. This happens when the geometric path difference equals −(n−1)t=−6.0 μm-(n-1)t = -6.0 \text{ μm}. Using the small angle approximation, a path difference of 6.0 μm6.0 \text{ μm} corresponds to a position shift of y=(n−1)tLd=6.0×10−6×2.51.5×10−3=0.010 m=1.0 mmy = \frac{(n-1)tL}{d} = \frac{6.0 \times 10^{-6} \times 2.5}{1.5 \times 10^{-3}} = 0.010 \text{ m} = 1.0 \text{ mm}.

Question 14

An unknown wave propagates through a region of space. Which observation provides conclusive evidence that the wave must be transverse?

  1. The wave's intensity decreases as the distance from the source increases.
  2. The wave can be polarized by passing it through a special filter. (correct answer)
  3. The wave undergoes reflection when it encounters a boundary.
  4. The wave's speed is observed to be constant within a uniform medium.
Explanation: Polarization is the phenomenon where the oscillations of a wave are restricted to a single plane. Only transverse waves, which have oscillations perpendicular to the direction of energy transfer, can be polarized. Longitudinal waves, whose oscillations are parallel to the direction of energy transfer, cannot be polarized. All waves (both transverse and longitudinal) exhibit intensity decrease with distance, reflection, and have a constant speed in a uniform medium.

Question 15

Which statement correctly describes the displacement of particles in a medium through which a sinusoidal mechanical wave is passing?

  1. All particles in the medium are permanently displaced in the direction of wave propagation.
  2. For a transverse wave, all particles reach their maximum displacement at the same instant in time.
  3. The particles move through the medium with a constant speed equal to the wave speed.
  4. The particles oscillate about fixed equilibrium positions with a net displacement of zero over one period. (correct answer)
Explanation: Waves transfer energy without a net transfer of matter. The particles of the medium oscillate about their equilibrium positions. Over a full cycle (one period), each particle returns to its starting point, so its net displacement is zero. Distractor A is incorrect because there is no net transport of matter. Distractor C confuses particle speed (which is variable) with wave speed. Distractor D is incorrect as different particles reach their maximum displacement at different times, which is what gives the wave its shape.

Question 16

Consider a point P on a string through which a continuous transverse wave is propagating. The wave speed is vv. Which statement correctly describes the motion of point P?

  1. Point P oscillates with a maximum speed that depends on the wave's amplitude and frequency. (correct answer)
  2. Point P moves along the string with a constant speed vv.
  3. Point P remains stationary as the wave's energy passes through it.
  4. Point P oscillates with a speed that is always equal to the wave speed vv.
Explanation: The wave speed vv is the speed at which the wave profile (and energy) propagates along the string. The point P itself does not travel along the string; it oscillates transversally (perpendicular to the string) about its equilibrium position. The speed of this oscillation is not constant and is not equal to vv. The maximum speed of the particle's oscillation depends on the wave's amplitude and frequency (specifically, vmax=Aω=A(2πf)v_{max} = A\omega = A(2\pi f)), which is a concept from SHM.

Question 17

A sound wave travels from a region of cool air into a region of warmer air. What happens to the frequency and wavelength of the sound?

  1. Frequency remains constant; wavelength increases. (correct answer)
  2. Frequency remains constant; wavelength decreases.
  3. Frequency increases; wavelength remains constant.
  4. Frequency decreases; wavelength increases.
Explanation: The frequency of the sound wave is determined by its source and does not change as it enters a new medium. The speed of sound is greater in warmer air than in cooler air. According to the wave equation v=fλv = f\lambda, if the speed vv increases and the frequency ff remains constant, the wavelength λ\lambda must increase.

Question 18

A light wave passes from a vacuum into a diamond. What is the ratio of the wave's frequency in the diamond to its frequency in the vacuum?

  1. Less than 1.
  2. Equal to the refractive index of diamond.
  3. Greater than 1.
  4. Equal to 1. (correct answer)
Explanation: A fundamental property of waves is that their frequency is determined by the source and does not change as the wave propagates from one medium to another. The speed and wavelength of the light will change upon entering the diamond, but the frequency remains constant. Therefore, the ratio of the frequency in the diamond to the frequency in the vacuum is exactly 1.

Question 19

Which statement provides the most fundamental distinction between mechanical waves and electromagnetic waves?

  1. Mechanical waves are always longitudinal, whereas electromagnetic waves are always transverse.
  2. Mechanical waves transfer energy by causing oscillations of particles, whereas electromagnetic waves require a medium to propagate.
  3. Mechanical waves require a medium for propagation, whereas electromagnetic waves can propagate through a vacuum. (correct answer)
  4. Mechanical waves travel at the speed of sound, whereas electromagnetic waves travel at the speed of light.
Explanation: The defining difference is their propagation requirements. Mechanical waves are disturbances in a medium and require the particles of that medium to transfer energy. Electromagnetic waves are oscillations of electric and magnetic fields and do not require a medium; they can travel through the vacuum of space. Distractor A is incorrect because mechanical waves can be transverse (e.g., on a string). Distractor B has the second clause incorrect. Distractor D is an oversimplification; the speed of both wave types depends on the medium they are traveling through.

Question 20

What property of a wave is determined solely by its source and is independent of the properties of the medium through which the wave travels?

  1. Speed
  2. Wavelength
  3. Amplitude
  4. Frequency (correct answer)
Explanation: The frequency of a wave is determined by the rate of oscillation of its source. When the wave passes into a new medium, this frequency remains constant. The wave's speed is determined by the properties of the medium. The wavelength depends on both speed and frequency (λ=v/f\lambda = v/f) and therefore changes with the medium. The amplitude can decrease due to damping or spreading, which also depends on the medium.