IB Physics Quiz: Apply Standing Waves And Resonance

20 questions · exam conditions

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Apply Standing Waves And ResonanceQuestion 1 of 20

A pipe of length L is open at both ends and resonates at its fundamental frequency $f_1$ . What are the frequencies of the next two higher resonant modes?

1.5

f_1

and 2.5

f_1

f_1

and 3

f_1

f_1

and 5

f_1

f_1

and 4

f_1

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IB Physics Quiz

IB Physics Quiz: Apply Standing Waves And Resonance

Practice Apply 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 Apply Standing Waves And Resonance, 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

A pipe of length L is open at both ends and resonates at its fundamental frequency $f_1$ . What are the frequencies of the next two higher resonant modes?

1.5 $f_1$ and 2.5 $f_1$
2 $f_1$ and 3 $f_1$ (correct answer)
3 $f_1$ and 5 $f_1$
2 $f_1$ and 4 $f_1$

Explanation: For a pipe open at both ends, all integer harmonics are present. The resonant frequencies are given by

f_n = n f_1

, where n = 1, 2, 3, ... The fundamental frequency corresponds to n=1. The next two higher resonant modes are the second harmonic (n=2) and the third harmonic (n=3). Their frequencies are therefore 2

f_1

and 3

f_1

Question 2

An object with a natural frequency $f_0$ is subjected to a periodic driving force of variable frequency $f$ . If the amount of damping in the system is increased, how does the resonance curve (amplitude vs. driving frequency) change?

The resonant frequency increases and the maximum amplitude decreases.
The resonant frequency decreases and the maximum amplitude increases.
The resonance peak becomes broader and the maximum amplitude decreases. (correct answer)
The resonance peak becomes sharper and the maximum amplitude increases.

Explanation: Damping is a process that removes energy from an oscillating system. Increasing damping has two main effects on the resonance curve: 1) It reduces the maximum amplitude achieved at the resonant frequency. 2) It broadens the resonance peak, meaning the system responds with a significant amplitude over a wider range of driving frequencies. A sharper peak and higher amplitude (option D) are characteristic of a system with less damping.

Question 3

A string fixed at both ends vibrates in its fundamental mode with frequency $f$ . The tension in the string is then increased by a factor of 4, while its length and mass per unit length remain unchanged. What is the new fundamental frequency?

$f/2$
$2f$ (correct answer)
$4f$
$16f$

Explanation: The fundamental frequency of a string is given by

f = v/(2L)

, where

v

is the wave speed and

L

is the length. The wave speed on the string is given by

v = \sqrt{T/\mu}

, where

T

is the tension and

\mu

is the mass per unit length. Combining these gives

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

. This shows that frequency is proportional to the square root of the tension (

f \propto \sqrt{T}

). If the tension

T

is increased by a factor of 4, the new frequency

f'

will be proportional to

\sqrt{4T} = 2\sqrt{T}

. Therefore, the new frequency is

2f

Question 4

A string vibrates in a standing wave pattern. Consider two points, P and Q, on the string. P is located between the first and second nodes, and Q is located between the second and third nodes. What is the phase difference between the oscillations of P and Q?

0
$\pi/4$
$\pi/2$
$\pi$ (correct answer)

Explanation: In a standing wave, all points within a single loop (i.e., between two adjacent nodes) oscillate in phase with each other (phase difference of 0). However, points in adjacent loops are in antiphase, meaning they oscillate with a phase difference of

\pi

radians (180°). Since P and Q are in adjacent loops, the phase difference between their motions is

\pi

Question 5

A tube closed at one end produces resonance for a sound source of 450 Hz. The next highest frequency at which resonance is produced is 750 Hz. What is the fundamental frequency of the tube?

75 Hz
150 Hz (correct answer)
225 Hz
300 Hz

Explanation: A tube closed at one end only supports odd harmonics (n=1, 3, 5, ...). The resonant frequencies are

f_n = n f_1

. The given frequencies, 450 Hz and 750 Hz, must be two consecutive odd harmonics. The difference in frequency between consecutive resonant modes is

f_{n+2} - f_n = (n+2)f_1 - nf_1 = 2f_1

. Therefore,

750 \text{ Hz} - 450 \text{ Hz} = 300 \text{ Hz} = 2f_1

. Solving for the fundamental frequency gives

f_1 = 300 / 2 = 150

Hz. We can check this: the 3rd harmonic would be

3 \times 150 = 450

Hz, and the 5th harmonic would be

5 \times 150 = 750

Hz, which matches the problem statement.

Question 6

A guitar string of length 0.50 m vibrates in its fundamental mode. The speed of waves on the string is 250 m s⁻¹. This vibration produces a sound wave that travels through the air at 340 m s⁻¹. What is the wavelength of the sound wave in the air?

0.74 m
1.00 m
1.36 m (correct answer)
1.70 m

Explanation: This is a two-step problem. First, find the frequency of the string's vibration. For the fundamental mode on a string fixed at both ends, the wavelength on the string is

\lambda_{\text{string}} = 2L = 2 \times 0.50 \text{ m} = 1.00 \text{ m}

. The frequency is then

f = v_{\text{string}} / \lambda_{\text{string}} = 250 \text{ m s⁻¹} / 1.00 \text{ m} = 250 \text{ Hz}

. This frequency is the same for the sound wave produced in the air. Second, use this frequency to find the wavelength in the air:

\lambda_{\text{air}} = v_{\text{air}} / f = 340 \text{ m s⁻¹} / 250 \text{ Hz} = 1.36 \text{ m}

Question 7

A tuning fork is held above a hollow tube open at the top and closed at the bottom by a movable piston. The first resonance is heard when the air column is 16.0 cm long. At what length of the air column will the second resonance be heard?

24.0 cm
32.0 cm
48.0 cm (correct answer)
64.0 cm

Explanation: This setup is a pipe closed at one end. Resonance occurs when the length of the air column is

L = n\lambda/4

for odd integers n=1, 3, 5, ... The first resonance (fundamental) occurs at

L_1 = \lambda/4

. The second resonance occurs at the next possible harmonic, which is n=3, so

L_3 = 3\lambda/4

. From the first resonance, we know

\lambda = 4L_1 = 4 \times 16.0 \text{ cm} = 64.0 \text{ cm}

. Therefore, the length for the second resonance is

L_3 = (3 \times 64.0 \text{ cm})/4 = 48.0 \text{ cm}

. Alternatively,

L_3 = 3L_1 = 3 \times 16.0 \text{ cm} = 48.0 \text{ cm}

Question 8

Consider three systems of length L, all resonating at their fundamental frequency:

I. A string fixed at both ends.

II. An air column in a pipe open at both ends.

III. An air column in a pipe closed at one end. Which list correctly ranks the wavelengths ( $\lambda$ ) of the fundamental modes from longest to shortest?

I = II > III
III > II > I
I > II > III
III > I = II (correct answer)

Explanation: We need to find the wavelength for the fundamental mode (n=1) in each case. I. String fixed at both ends:

L = \lambda_I/2 \implies \lambda_I = 2L

. II. Pipe open at both ends:

L = \lambda_{II}/2 \implies \lambda_{II} = 2L

. III. Pipe closed at one end:

L = \lambda_{III}/4 \implies \lambda_{III} = 4L

. Comparing the wavelengths, we see that

\lambda_{III} = 4L

is the longest, and

\lambda_I = \lambda_{II} = 2L

. Therefore, the correct ranking from longest to shortest is III > I = II.

Question 9

A mechanical system with low damping has a natural frequency of oscillation $f_0$ . The system is driven by an external force with a slowly increasing frequency $f$ . Which statement best describes the amplitude of the system's oscillation as $f$ increases from a value much less than $f_0$ to a value much greater than $f_0$ ?

The amplitude is initially large, decreases to a minimum at $f=f_0$ , and then increases again.
The amplitude increases approximately linearly with $f$ until it reaches a maximum at $f=f_0$ , then decreases.
The amplitude remains near zero until $f=f_0$ , at which point it becomes extremely large, then immediately returns to zero.
The amplitude is small for $f \ll f_0$ , rises to a sharp maximum at $f \approx f_0$ , and becomes small again for $f \gg f_0$ . (correct answer)

Explanation: This question describes the phenomenon of resonance. When a system is driven by a periodic force, its response amplitude depends on the driving frequency

f

. The amplitude is relatively small when

f

is far from the natural frequency

f_0

. As

f

approaches

f_0

, the amplitude increases dramatically, reaching a peak at or very near

f_0

. For a system with low damping, this peak is sharp and high. As

f

increases beyond

f_0

, the amplitude decreases again. Option C provides the most accurate physical description of this resonance curve.

Question 10

A pipe is open at one end and closed at the other. For standing waves of air in the pipe, which statement correctly describes the displacement of air molecules and the pressure variation at the two ends?

The closed end has a displacement node and the open end has a pressure node. (correct answer)
The closed end has a displacement node and the open end has a pressure antinode.
The closed end has a displacement antinode and the open end has a pressure node.
The closed end has a displacement antinode and the open end has a pressure antinode.

Explanation: At the closed end of the pipe, air molecules cannot move longitudinally, so this must be a point of zero displacement (a displacement node). At the open end, the air is free to move, resulting in maximum displacement (a displacement antinode). Pressure and displacement are out of phase by

\pi/2

. A displacement node corresponds to a pressure antinode (maximum pressure variation), and a displacement antinode corresponds to a pressure node (pressure is constant atmospheric pressure). Therefore, the closed end is a displacement node (and pressure antinode), and the open end is a displacement antinode (and pressure node). Option B correctly identifies two of these conditions.

Question 11

An organ pipe of length L, open at both ends, is sounding its fundamental frequency, $f_{open}$ . A cap is then placed over one of the ends, closing it. What is the new fundamental frequency, $f_{closed}$ , of the modified pipe?

$f_{open} / 2$ (correct answer)
$f_{open} / 4$
$f_{open}$
$2 f_{open}$

Explanation: For an open-open pipe of length L, the fundamental mode corresponds to half a wavelength fitting in the pipe, so

L = \lambda/2

. The frequency is

f_{open} = v/\lambda = v/(2L)

. When one end is closed, the pipe becomes a closed-open pipe. Its fundamental mode corresponds to a quarter of a wavelength fitting in the pipe, so

L = \lambda'/4

. The new fundamental frequency is

f_{closed} = v/\lambda' = v/(4L)

. By comparing the two expressions, we can see that

f_{closed} = \frac{v}{4L} = \frac{1}{2} \left( \frac{v}{2L} \right) = \frac{1}{2} f_{open}

Question 12

A pipe open at both ends resonates at its fundamental frequency in air (speed of sound ≈ 340 m s⁻¹). The pipe is then filled with helium, in which the speed of sound is approximately three times greater. What is the new fundamental frequency of the pipe?

It is unchanged.
It is approximately 1/3 of the original frequency.
It is approximately 3 times the original frequency. (correct answer)
It is approximately 9 times the original frequency.

Explanation: The fundamental frequency of a pipe open at both ends is given by

f = v/(2L)

, where

v

is the speed of sound in the gas inside the pipe and

L

is the length of the pipe. This relationship shows that the frequency is directly proportional to the speed of sound (

f \propto v

). If the air is replaced with helium, where the speed of sound is three times greater, the new fundamental frequency will also be three times greater than the original frequency.

Question 13

A string fixed at both ends has a fundamental frequency of 220 Hz when the tension is 400 N. If the tension is increased to 900 N while keeping all other parameters constant, what is the frequency of the third harmonic of the string?

495 Hz
990 Hz (correct answer)
330 Hz
660 Hz

Explanation: The fundamental frequency is proportional to the square root of tension:

f \propto \sqrt{T}

. When tension increases from 400 N to 900 N, the new fundamental frequency becomes

f' = 220 \times \sqrt{\frac{900}{400}} = 220 \times 1.5 = 330 Hz

. The third harmonic is three times the fundamental frequency:

3 \times 330 = 990 Hz

. Choice A incorrectly uses the square root relationship for the harmonic calculation. Choice C gives only the new fundamental frequency. Choice D incorrectly applies the tension ratio directly to the original frequency.

Question 14

In a resonance tube experiment, a tuning fork of unknown frequency is used. The first resonance occurs when the air column is 20.5 cm long, and the second resonance occurs when it is 62.0 cm long. If the speed of sound is 344 m/s, what is the frequency of the tuning fork?

415 Hz (correct answer)
398 Hz
425 Hz
385 Hz

Explanation: For consecutive resonances in a closed tube, the difference in lengths equals half a wavelength:

\frac{\lambda}{2} = 62.0 - 20.5 = 41.5 cm = 0.415 m

. Therefore,

\lambda = 0.83 m

. The frequency is

f = \frac{v}{\lambda} = \frac{344}{0.83} = 414.5 Hz ≈ 415 Hz

. Choice B incorrectly uses the first resonance length directly in calculations. Choice C uses an approximation that ignores end correction effects. Choice D uses an incorrect relationship between consecutive resonances.

Question 15

Two waves traveling in opposite directions on a string have equations $y_1 = 0.05\sin(4\pi t - 2\pi x)$ and $y_2 = 0.05\sin(4\pi t + 2\pi x)$ where distances are in meters and time in seconds. At what positions along the string do nodes occur?

$x = 0.25n$ m, where n = 0, 1, 2, ...
$x = 0.5n$ m, where n = 0, 1, 2, ...
$x = 0.125 + 0.5n$ m, where n = 0, 1, 2, ...
$x = 0.25 + 0.5n$ m, where n = 0, 1, 2, ... (correct answer)

Explanation: The superposition gives

y = y_1 + y_2 = 0.05[\sin(4\pi t - 2\pi x) + \sin(4\pi t + 2\pi x)] = 0.1\sin(4\pi t)\cos(2\pi x)

. Nodes occur where the amplitude is zero, i.e., where

\cos(2\pi x) = 0

. This happens when

2\pi x = \frac{\pi}{2} + n\pi

, giving

x = \frac{1}{4} + \frac{n}{2} = 0.25 + 0.5n

meters. Choice A gives antinode positions. Choice B gives positions spaced by half wavelength but starting at x = 0. Choice C incorrectly calculates the phase relationship.

Question 16

A pipe organ has pipes of different lengths to produce different notes. An open pipe produces a fundamental frequency of 264 Hz. To produce a note exactly one octave lower (132 Hz) using a closed pipe, what should be the ratio of the length of the closed pipe to the length of the open pipe?

1:1 (correct answer)
2:1
4:1
1:2

Explanation: For an open pipe:

f_{open} = \frac{v}{2L_{open}}

, so

264 = \frac{v}{2L_{open}}

. For a closed pipe:

f_{closed} = \frac{v}{4L_{closed}}

, so

132 = \frac{v}{4L_{closed}}

. From the first equation:

v = 528L_{open}

. From the second:

v = 528L_{closed}

. Therefore,

L_{closed} = L_{open}

, giving a ratio of 1:1. Choice B assumes the frequency ratio directly determines the length ratio. Choice C incorrectly applies the factor of 4 difference between open and closed pipe formulas. Choice D inverts the correct relationship.

Question 17

A guitar string of length 65 cm is plucked and vibrates in its fundamental mode with frequency 330 Hz. When the guitarist presses the string at the 5th fret (13 cm from the nut), reducing the vibrating length to 52 cm, and simultaneously increases the tension by 21%, what is the new fundamental frequency?

450 Hz (correct answer)
485 Hz
425 Hz
520 Hz

Explanation: The fundamental frequency is

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

. When length changes from 65 cm to 52 cm and tension increases by 21% (factor of 1.21), the new frequency is

f' = f \times \frac{L_{old}}{L_{new}} \times \sqrt{1.21} = 330 \times \frac{65}{52} \times 1.1 = 330 \times 1.25 \times 1.1 = 453.75 Hz ≈ 450 Hz

. Choice B incorrectly applies the tension factor. Choice C uses an incorrect length ratio. Choice D applies the factors incorrectly by multiplying instead of using the proper relationships.

Question 18

A standing wave pattern is established in a string of length 1.5 m fixed at both ends. The standing wave has 3 complete half-wavelengths along the string's length. If a small piece of paper placed at 0.75 m from one end remains stationary, what can be concluded about the wave?

The paper is at a node and the wavelength is 0.5 m
The paper is at an antinode and the wavelength is 1.0 m
The paper is at a node and the wavelength is 1.0 m (correct answer)
The paper is at an antinode and the wavelength is 0.5 m

Explanation: For 3 half-wavelengths in 1.5 m: λ/2 = 1.5/3 = 0.5 m, so λ = 1.0 m. Since the paper remains stationary, it must be at a node. For a string fixed at both ends, nodes occur at x = nλ/2 from one end. With λ = 1.0 m, nodes are at 0, 0.5, 1.0, and 1.5 m. The paper at 0.75 m is at the center, which is indeed a node position for this pattern.

Question 19

A string fixed at both ends has a fundamental frequency (first harmonic) of $f_0$ . What is the frequency of its third overtone?

$4f_0$ (correct answer)
$3f_0$
$5f_0$
$7f_0$

Explanation: For a string fixed at both ends, all integer harmonics are present. The fundamental frequency (1st harmonic) is

f_0

. Overtones are the frequencies higher than the fundamental. The first overtone is the 2nd harmonic (

2f_0

). The second overtone is the 3rd harmonic (

3f_0

). Therefore, the third overtone is the 4th harmonic, which has a frequency of

4f_0

Question 20

A standing wave is produced on a string of length L. The distance between the second node and the fifth node is measured to be 60 cm. What is the wavelength of the wave?

20 cm
30 cm
60 cm
40 cm (correct answer)

Explanation: In a standing wave, the distance between any two adjacent nodes is half a wavelength (λ/2). The distance between the second node and the fifth node spans three such segments (from node 2 to 3, 3 to 4, and 4 to 5). Therefore, the total distance is 3 × (λ/2). We are given this distance is 60 cm. So, 3λ/2 = 60 cm. Solving for λ gives λ = (2 × 60 cm) / 3 = 120 cm / 3 = 40 cm.