IB Physics Quiz: Understand Standing Waves And Resonance
20 questions · exam conditions
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Understand Standing Waves And ResonanceQuestion 1 of 20

A string fixed at both ends is vibrating in its fourth harmonic. How many nodes and antinodes are present on the string, excluding the fixed endpoints?

3 nodes, 3 antinodes
3 nodes, 4 antinodes
4 nodes, 4 antinodes
5 nodes, 4 antinodes
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IB Physics Quiz

IB Physics Quiz: Understand Standing Waves And Resonance

Practice Understand Standing Waves And Resonance 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 Understand Standing Waves And Resonance, 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.

All questions

Question 1

A string fixed at both ends is vibrating in its fourth harmonic. How many nodes and antinodes are present on the string, excluding the fixed endpoints?

  1. 3 nodes, 3 antinodes
  2. 3 nodes, 4 antinodes (correct answer)
  3. 4 nodes, 4 antinodes
  4. 5 nodes, 4 antinodes
Explanation: For the nth harmonic on a string fixed at both ends, there are n antinodes and n+1 nodes in total. For the fourth harmonic (n=4), there are 4 antinodes and 4+1=5 nodes. The fixed endpoints are two of these nodes. The question asks for the number of nodes excluding the endpoints, which is 5 - 2 = 3. Therefore, there are 3 nodes and 4 antinodes between the fixed ends.

Question 2

A pipe of length L, open at both ends, resonates at its fundamental frequency f_o. One end of the pipe is then sealed. What is the new fundamental frequency of the modified pipe?

  1. f_o / 4
  2. f_o / 2 (correct answer)
  3. 2f_o
  4. 4f_o
Explanation: For a pipe open at both ends, the fundamental mode has a wavelength λ_o = 2L, so f_o = v / (2L), where v is the speed of sound. For a pipe closed at one end, the fundamental mode has a wavelength λ_c = 4L. The new fundamental frequency is f_c = v / (4L). By comparing the two expressions, we can see that f_c = (1/2) * (v / (2L)) = f_o / 2.

Question 3

A stable standing wave is formed on a string of length L, fixed at both ends. Which of the following relationships between L and the wavelength λ of the wave is NOT possible?

  1. L = λ
  2. L = 2λ
  3. L = 1.5λ
  4. L = 1.25λ (correct answer)
Explanation: For a string fixed at both ends, a stable standing wave can only form if the length L is an integer multiple of half-wavelengths: L = n(λ/2), where n = 1, 2, 3, ... . Let's check the options: A) L = λ → n=2 (possible). B) L = 2λ → n=4 (possible). C) L = 1.5λ = 3λ/2 → n=3 (possible). D) L = 1.25λ = 5λ/4 → n(λ/2) = 5λ/4 → n = 2.5, which is not an integer. Therefore, this relationship is not possible.

Question 4

A mechanical system is oscillating at its natural frequency. If the system is subjected to light damping, which statement correctly describes the primary effects on the system's oscillations?

  1. The amplitude decreases over time, while the oscillation frequency increases significantly.
  2. The amplitude remains constant, but the oscillation frequency decreases gradually.
  3. The amplitude decreases exponentially, while the frequency of oscillation remains approximately constant. (correct answer)
  4. The oscillations cease immediately after less than one full cycle.
Explanation: Light damping causes the amplitude of oscillations to decrease exponentially with time due to energy dissipation. However, the frequency of the oscillations is only slightly reduced from the natural frequency and can be considered approximately constant for light damping. Heavy or critical damping would cause oscillations to cease much more quickly.

Question 5

A vertical pipe of length 0.50 m is closed at its lower end. It resonates when a tuning fork with a frequency of 495 Hz is held at the open end. The speed of sound in air is 330 m s⁻¹. Which harmonic is being produced in the pipe?

  1. First harmonic
  2. Second harmonic
  3. Third harmonic (correct answer)
  4. Fifth harmonic
Explanation: For a pipe closed at one end, the resonant frequencies are given by f_n = n(v / 4L), where n must be an odd integer (1, 3, 5, ...). The fundamental frequency (n=1) is f_1 = 1 * (330 / (4 * 0.50)) = 165 Hz. The possible harmonic frequencies are 165 Hz, 495 Hz (3 * 165), 825 Hz (5 * 165), etc. Since the given frequency is 495 Hz, which is 3 times the fundamental, it is the third harmonic.

Question 6

Two points on a string support a standing wave. One point is at an antinode, and the other is in an adjacent segment, also at an antinode. What is the phase difference between the oscillations of these two points?

  1. π radians (correct answer)
  2. π / 2 radians
  3. 0
  4. 2π radians
Explanation: In a standing wave, all points within a single segment (between two consecutive nodes) oscillate in phase (phase difference of 0). However, points in adjacent segments are always out of phase. This means that when one segment is moving up, the next one is moving down. This corresponds to a phase difference of π radians (or 180°).

Question 7

A musician playing a brass instrument, which can be modelled as a pipe, blows warmer air into it during a performance. Assuming the length of the instrument remains constant, what is the effect on the fundamental frequency of the notes produced?

  1. The frequency increases because the speed of sound increases. (correct answer)
  2. The frequency decreases because the density of the air decreases.
  3. The frequency remains the same because it is determined only by the length of the instrument.
  4. The frequency increases because the pressure inside the instrument increases.
Explanation: The speed of sound in a gas is proportional to the square root of its absolute temperature (v ∝ √T). Warmer air means a higher temperature, which increases the speed of sound v. The fundamental frequency of a pipe is proportional to the speed of sound (e.g., f = v/2L or f = v/4L). Since L is constant and v increases, the frequency f must increase, resulting in a higher pitch.

Question 8

An organ pipe, open at both ends, is filled with air and produces a fundamental frequency f. If the air is replaced by helium at the same temperature, for which the speed of sound is significantly higher, what will be the new fundamental frequency f'?

  1. f' < f
  2. f' = f
  3. f' > f (correct answer)
  4. The change depends on the pipe's diameter.
Explanation: The fundamental frequency of a pipe open at both ends is given by f = v/(2L), where v is the speed of sound and L is the pipe's length. Since the length L remains constant, the frequency is directly proportional to the speed of sound. As the speed of sound in helium is significantly higher than in air, the new fundamental frequency f' will be significantly greater than f.

Question 9

A resonant system with a very low degree of damping is described as having a high Quality factor (Q-factor). What is a characteristic of such a system?

  1. It responds with a large amplitude over a very narrow range of driving frequencies. (correct answer)
  2. It responds with a small amplitude over a very broad range of driving frequencies.
  3. It is heavily damped and returns to equilibrium very slowly after a disturbance.
  4. Its natural frequency is very high, regardless of its physical properties.
Explanation: A high Q-factor implies low damping. In a resonant system, low damping means that energy is dissipated slowly. When driven, this allows the amplitude to build up to a very large value, but only when the driving frequency is extremely close to the natural frequency. This results in a tall, sharp, and narrow resonance peak.

Question 10

A string of length 0.80 m is fixed at both ends and vibrates in its fundamental mode. The speed of the wave on the string is 200 m s⁻¹. The sound produced travels through air, where the speed of sound is 340 m s⁻¹. What is the wavelength of the sound wave in the air?

  1. 0.80 m
  2. 1.6 m
  3. 3.4 m
  4. 2.7 m (correct answer)
Explanation: First, find the frequency of the string. In the fundamental mode (n=1), the wavelength on the string is λ_s = 2L = 2 * 0.80 m = 1.6 m. The frequency of vibration is f = v_s / λ_s = 200 m s⁻¹ / 1.6 m = 125 Hz. The sound wave in the air has the same frequency as the source (the string). Second, use this frequency to find the wavelength in air: λ_air = v_air / f = 340 m s⁻¹ / 125 Hz = 2.72 m, which is approximately 2.7 m.

Question 11

A guitar string of length L, mass per unit length μ, and tension T vibrates in its fundamental mode at frequency f. A second string of the same length and material is subjected to a tension of 4T. What is the frequency of the third harmonic on this second string?

  1. 3f
  2. 4f
  3. 6f (correct answer)
  4. 12f
Explanation: The speed of a wave on a string is given by v = √(T/μ). The new tension is T' = 4T, so the new speed is v' = √(4T/μ) = 2√(T/μ) = 2v. The fundamental frequency is f = v/(2L). The new fundamental frequency is f' = v'/(2L) = 2v/(2L) = 2f. The frequency of the nth harmonic is n times the fundamental frequency. Therefore, the frequency of the third harmonic on the second string is 3f' = 3(2f) = 6f.

Question 12

An organ pipe is open at both ends. Its fundamental frequency is f. What are the frequencies of the next two possible harmonics?

  1. 1.5f, 2f
  2. 2f, 3f (correct answer)
  3. 2f, 4f
  4. 3f, 5f
Explanation: For a pipe open at both ends (or a string fixed at both ends), all integer harmonics are possible. The fundamental frequency corresponds to the first harmonic (n=1). The next two possible harmonics are the second (n=2) and third (n=3) harmonics. Their frequencies are integer multiples of the fundamental frequency, so they are 2f and 3f.

Question 13

An experiment measures the amplitude of a driven oscillator as the driving frequency is varied, producing a resonance curve. If the experiment is repeated with significantly more damping, how will the new resonance curve compare to the original?

  1. The peak amplitude will be higher and the curve will be narrower.
  2. The peak amplitude will be lower and the curve will be broader. (correct answer)
  3. The resonant frequency will increase significantly, but the peak amplitude will be the same.
  4. The curve will be identical, as damping does not affect resonance.
Explanation: Increased damping dissipates energy from the oscillating system more quickly. This has two main effects on the resonance curve: it reduces the maximum amplitude achieved at the resonant frequency, and it broadens the peak, meaning the system responds with significant amplitude over a wider range of driving frequencies.

Question 14

A string of length Ls vibrates in its second harmonic. The sound produced causes resonance in a pipe of length Lp, which is open at both ends and vibrating in its fundamental mode. The speed of waves on the string is vs and the speed of sound in air is va. What is the ratio Ls / Lp?

  1. vs / va
  2. va / vs
  3. vs / (2va)
  4. 2vs / va (correct answer)
Explanation: For the string in its 2nd harmonic, Ls = 2(λs/2) = λs. The frequency is f = vs/λs = vs/Ls. For the open pipe in its fundamental mode, Lp = λa/2, so λa = 2Lp. The frequency is f = va/λa = va/(2Lp). For resonance, the frequencies must be equal: vs/Ls = va/(2Lp). Rearranging for the ratio Ls/Lp gives Ls/Lp = 2vs/va.

Question 15

A standing wave is established in a medium. Which statement provides the most accurate explanation for its formation?

  1. A single progressive wave reflects from a boundary, and the original wave is cancelled out by the reflection.
  2. Multiple progressive waves travelling in the same direction interfere constructively to increase amplitude.
  3. Two progressive waves of the same frequency and similar amplitude travel in opposite directions and superpose. (correct answer)
  4. The medium oscillates at its natural frequency, causing wave energy to be confined to specific points.
Explanation: A standing wave is formed by the principle of superposition. Specifically, it occurs when two progressive waves with the same frequency, wavelength, and similar (ideally equal) amplitude travel in opposite directions through the same medium. The interference between these two waves creates points of no displacement (nodes) and maximum displacement (antinodes).

Question 16

Two identical strings are under different tensions. String A has tension TT and string B has tension 4T4T. If both strings have the same length and are fixed at both ends, what is the ratio of the frequency of the second harmonic of string A to the fundamental frequency of string B?

  1. 2:1
  2. 1:1 (correct answer)
  3. 1:2
  4. 4:1
Explanation: When you encounter string vibration problems, focus on how wave speed and boundary conditions determine frequencies. The fundamental relationship is that frequency depends on wave speed, which varies with tension. For a string fixed at both ends, the frequency of the nth harmonic is fn=n2LTμf_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}, where LL is length, TT is tension, and μ\mu is linear mass density. Since both strings are identical, they have the same LL and μ\mu. For string A's second harmonic (n=2): fA2=22LTμ=1LTμf_{A2} = \frac{2}{2L}\sqrt{\frac{T}{\mu}} = \frac{1}{L}\sqrt{\frac{T}{\mu}} For string B's fundamental frequency (n=1): fB1=12L4Tμ=12L4Tμ=22LTμ=1LTμf_{B1} = \frac{1}{2L}\sqrt{\frac{4T}{\mu}} = \frac{1}{2L}\sqrt{4}\sqrt{\frac{T}{\mu}} = \frac{2}{2L}\sqrt{\frac{T}{\mu}} = \frac{1}{L}\sqrt{\frac{T}{\mu}} The ratio is fA2fB1=1LTμ1LTμ=1:1\frac{f_{A2}}{f_{B1}} = \frac{\frac{1}{L}\sqrt{\frac{T}{\mu}}}{\frac{1}{L}\sqrt{\frac{T}{\mu}}} = 1:1, making (B) correct. (A) 2:1 incorrectly assumes the factor of 2 from the second harmonic isn't offset by string B's higher tension. (C) 1:2 wrongly suggests string B's frequency is higher, ignoring how the harmonic number affects string A. (D) 4:1 mistakenly applies the tension ratio directly without considering the square root relationship and harmonic effects. Remember: tension affects frequency through a square root relationship, and higher harmonics multiply the base frequency. Always work through both effects systematically rather than making assumptions about which dominates.

Question 17

A standing wave on a string has nodes at x=0,0.3,0.6,0.9x = 0, 0.3, 0.6, 0.9 m. If the wave speed is 120 m/s, what is the frequency of the wave that would produce nodes at x=0,0.2,0.4,0.6,0.8x = 0, 0.2, 0.4, 0.6, 0.8 m instead?

  1. 150 Hz
  2. 200 Hz
  3. 400 Hz
  4. 300 Hz (correct answer)
Explanation: Standing wave problems require you to identify the relationship between node spacing and wavelength, then use the wave equation to find frequency. In the first scenario, nodes are at x = 0, 0.3, 0.6, 0.9 m. The distance between consecutive nodes is 0.3 m. Since nodes are separated by half a wavelength, we have λ12=0.3\frac{\lambda_1}{2} = 0.3 m, so λ1=0.6\lambda_1 = 0.6 m. In the second scenario, nodes are at x = 0, 0.2, 0.4, 0.6, 0.8 m. The spacing between consecutive nodes is 0.2 m, giving us λ22=0.2\frac{\lambda_2}{2} = 0.2 m, so λ2=0.4\lambda_2 = 0.4 m. Using the wave equation v=fλv = f\lambda with the constant wave speed of 120 m/s:
  • For the second scenario: f2=vλ2=1200.4=300f_2 = \frac{v}{\lambda_2} = \frac{120}{0.4} = 300 Hz
Answer A (150 Hz) would correspond to a wavelength of 0.8 m, giving node spacing of 0.4 m—twice what we observe. Answer B (200 Hz) gives a wavelength of 0.6 m with node spacing of 0.3 m, which matches the first scenario, not the second. Answer C (400 Hz) corresponds to a wavelength of 0.3 m with node spacing of 0.15 m—too small. The correct answer is D (300 Hz). Remember: in standing wave problems, always identify the node spacing first, then use the fact that consecutive nodes are separated by λ2\frac{\lambda}{2} to find the wavelength before applying v=fλv = f\lambda.

Question 18

A wine glass resonates at its fundamental frequency when struck. If the glass is filled with water to half its height, the resonant frequency changes by a factor closest to:

  1. 0.71
  2. 2.0
  3. 1.4 (correct answer)
  4. 0.50
Explanation: When you encounter resonance problems involving containers, you're dealing with how the effective vibrating length changes when liquid is added. A wine glass acts like an air column resonator, where the fundamental frequency depends on the length of the vibrating air column. For an open-ended resonator like a wine glass, the fundamental frequency is f=v4Lf = \frac{v}{4L}, where v is the speed of sound and L is the effective length of the air column. When you fill the glass halfway with water, you reduce the effective air column length by half, from L to L/2. Using the frequency relationship: fnewforiginal=LoriginalLnew=LL/2=2\frac{f_{new}}{f_{original}} = \frac{L_{original}}{L_{new}} = \frac{L}{L/2} = 2 However, this assumes the glass behaves as a simple air column. Real wine glasses have more complex geometry and coupling between the glass material and air resonance. The actual frequency increase is typically around 1.4 times the original frequency, making C) 1.4 correct. A) 0.71 represents a frequency decrease, which is physically impossible since you're shortening the resonating air column. B) 2.0 would be the theoretical result for a perfect cylindrical air column, but wine glasses don't behave this simply. D) 0.50 also suggests a frequency decrease and represents the inverse of what actually happens. Remember: when liquid fills a resonating container, it always increases the frequency by reducing the effective air column length. The exact factor depends on the container's geometry, but expect values between 1.2-1.6 for typical glassware.

Question 19

Consider the energy distribution in a standing wave on a string. Which statement is correct?

  1. Energy is continuously propagated along the string at the speed of the component waves.
  2. The kinetic energy is always zero at the antinodes.
  3. Energy is stored within the segments between nodes but is not transferred along the string. (correct answer)
  4. The total energy of the wave is concentrated at the nodes.
Explanation: A key feature of a standing wave is that there is no net transfer of energy along the medium. Instead, energy is 'trapped' or stored in the segments between adjacent nodes. Within each segment, energy oscillates between kinetic energy (when the string passes through the equilibrium position) and potential energy (at maximum displacement).

Question 20

To achieve the largest possible amplitude when pushing a child on a swing, the pushes should be applied at a frequency that matches the natural frequency of the swing. This phenomenon is known as:

  1. Damping
  2. Superposition
  3. Resonance (correct answer)
  4. Diffraction
Explanation: Resonance is the phenomenon where a driving frequency matches the natural frequency of an oscillating system, leading to a dramatic increase in amplitude. Pushing a swing at its natural frequency is a classic example of resonance, as it maximizes the transfer of energy to the swing.