Oscillations & Waves
Q. A mass attached to a spring oscillates with a frequency of 2 Hz. What is the period of the oscillation?
A.
0.5 s
B.
1 s
C.
2 s
D.
4 s
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Solution
The period T is the reciprocal of frequency f. Thus, T = 1/f = 1/2 = 0.5 s.
Correct Answer: B — 1 s
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Q. A mass attached to a spring oscillates with a period of 2 seconds. What is its frequency? (2014)
A.
0.5 Hz
B.
1 Hz
C.
2 Hz
D.
4 Hz
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Solution
Frequency f = 1/T = 1/2 s = 0.5 Hz.
Correct Answer: A — 0.5 Hz
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Q. A mass-spring system oscillates with a frequency of 1 Hz. What is the angular frequency? (2023)
A.
2π rad/s
B.
π rad/s
C.
1 rad/s
D.
4π rad/s
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Solution
Angular frequency ω is related to frequency f by ω = 2πf. Therefore, for f = 1 Hz, ω = 2π(1) = 2π rad/s.
Correct Answer: A — 2π rad/s
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Q. A mass-spring system oscillates with a frequency of 5 Hz. What is the angular frequency? (2021)
A.
10π rad/s
B.
5π rad/s
C.
2π rad/s
D.
20π rad/s
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Solution
Angular frequency ω = 2πf = 2π × 5 Hz = 10π rad/s.
Correct Answer: A — 10π rad/s
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Q. A mass-spring system oscillates with an amplitude of 0.1 m. What is the maximum speed of the mass if the angular frequency is 20 rad/s?
A.
1 m/s
B.
2 m/s
C.
3 m/s
D.
4 m/s
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Solution
The maximum speed v_max is given by v_max = Aω, where A is the amplitude and ω is the angular frequency. Thus, v_max = 0.1 * 20 = 2 m/s.
Correct Answer: B — 2 m/s
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Q. A pendulum completes 20 oscillations in 40 seconds. What is its frequency? (2022)
A.
0.5 Hz
B.
1 Hz
C.
2 Hz
D.
4 Hz
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Solution
Frequency f = number of oscillations / time = 20 / 40 s = 0.5 Hz.
Correct Answer: B — 1 Hz
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Q. A pendulum swings with a maximum angle of 30 degrees. What is the approximate time period for small angles? (2022)
A.
1.0 s
B.
0.5 s
C.
2.0 s
D.
0.25 s
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Solution
For small angles, T ≈ 2π√(L/g). Assuming L = 1 m, T ≈ 2π√(1/9.8) ≈ 2.0 s.
Correct Answer: A — 1.0 s
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Q. A simple pendulum completes 20 oscillations in 40 seconds. What is its time period? (2021)
A.
1 s
B.
2 s
C.
0.5 s
D.
0.25 s
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Solution
Time period (T) = Total time / Number of oscillations = 40 s / 20 = 2 s.
Correct Answer: B — 2 s
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Q. A simple pendulum has a length of 1 m. What is its time period? (2023)
A.
1.0 s
B.
2.0 s
C.
0.5 s
D.
0.25 s
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Solution
The time period T of a simple pendulum is given by T = 2π√(L/g). For L = 1 m and g = 9.8 m/s², T = 2π√(1/9.8) ≈ 2.0 s.
Correct Answer: A — 1.0 s
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Q. A simple pendulum oscillates with a period of 2 seconds. What is the length of the pendulum? (2021)
A.
0.5 m
B.
1 m
C.
2 m
D.
4 m
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Solution
The period T of a simple pendulum is given by T = 2π√(L/g). Rearranging gives L = (T^2 * g) / (4π^2). Using g = 9.8 m/s² and T = 2 s, we find L = (2^2 * 9.8) / (4π^2) ≈ 1 m.
Correct Answer: B — 1 m
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Q. A sound wave travels at a speed of 340 m/s and has a frequency of 1700 Hz. What is its wavelength? (2021)
A.
0.2 m
B.
0.5 m
C.
2 m
D.
1 m
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Solution
Wavelength (λ) = Speed (v) / Frequency (f) = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer: A — 0.2 m
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Q. A sound wave travels through air at 343 m/s. If its frequency is 686 Hz, what is its wavelength? (2023)
A.
0.5 m
B.
1 m
C.
2 m
D.
3 m
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Solution
Wavelength λ = v/f = 343 m/s / 686 Hz = 0.5 m.
Correct Answer: A — 0.5 m
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Q. A sound wave travels through air at a speed of 340 m/s. If the frequency of the sound is 170 Hz, what is the wavelength?
A.
1 m
B.
2 m
C.
3 m
D.
4 m
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Solution
The wavelength λ is given by λ = v/f. Thus, λ = 340/170 = 2 m.
Correct Answer: B — 2 m
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Q. A sound wave travels through air at a speed of 340 m/s. If the wavelength is 0.85 m, what is the frequency? (2022)
A.
400 Hz
B.
300 Hz
C.
250 Hz
D.
200 Hz
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Solution
Frequency (f) = speed (v) / wavelength (λ) = 340 m/s / 0.85 m ≈ 400 Hz.
Correct Answer: A — 400 Hz
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Q. A sound wave travels through air at a speed of 340 m/s. If the wavelength is 0.85 m, what is the frequency of the sound wave? (2022)
A.
400 Hz
B.
300 Hz
C.
250 Hz
D.
500 Hz
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Solution
The frequency f can be calculated using the formula v = fλ. Rearranging gives f = v/λ = 340 m/s / 0.85 m ≈ 400 Hz.
Correct Answer: A — 400 Hz
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Q. A tuning fork produces a sound wave of frequency 440 Hz. What is the period of the wave? (2019)
A.
0.00227 s
B.
0.005 s
C.
0.01 s
D.
0.001 s
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Solution
Period T = 1/f = 1/440 Hz ≈ 0.00227 s.
Correct Answer: A — 0.00227 s
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Q. A tuning fork produces a sound wave of frequency 440 Hz. What is the time period of the sound wave?
A.
0.00227 s
B.
0.005 s
C.
0.01 s
D.
0.02 s
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Solution
The time period T is the reciprocal of frequency f. Thus, T = 1/f = 1/440 ≈ 0.00227 s.
Correct Answer: A — 0.00227 s
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Q. A tuning fork produces a sound wave of frequency 440 Hz. What is the time period of this sound wave? (2021)
A.
0.00227 s
B.
0.0045 s
C.
0.01 s
D.
0.005 s
Show solution
Solution
Time period (T) = 1/f = 1/440 Hz ≈ 0.00227 s.
Correct Answer: A — 0.00227 s
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Q. A tuning fork vibrates at a frequency of 440 Hz. What is the time period of the fork? (2020)
A.
0.00227 s
B.
0.0045 s
C.
0.01 s
D.
0.005 s
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Solution
Time period T = 1/f = 1/440 Hz ≈ 0.00227 s.
Correct Answer: A — 0.00227 s
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Q. A wave has a frequency of 10 Hz and a wavelength of 2 m. What is its speed? (2023)
A.
5 m/s
B.
10 m/s
C.
20 m/s
D.
15 m/s
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Solution
The speed v of a wave is given by v = fλ. Therefore, v = 10 Hz * 2 m = 20 m/s.
Correct Answer: C — 20 m/s
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Q. A wave has a frequency of 10 Hz and a wavelength of 5 m. What is its speed? (2020)
A.
50 m/s
B.
10 m/s
C.
2 m/s
D.
5 m/s
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Solution
Speed (v) = frequency (f) × wavelength (λ) = 10 Hz × 5 m = 50 m/s.
Correct Answer: A — 50 m/s
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Q. A wave has a frequency of 500 Hz and a wavelength of 0.6 m. What is the speed of the wave? (2022)
A.
300 m/s
B.
500 m/s
C.
600 m/s
D.
400 m/s
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Solution
Using the wave equation v = fλ, we find v = 500 Hz * 0.6 m = 300 m/s.
Correct Answer: A — 300 m/s
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Q. A wave has a frequency of 60 Hz and a wavelength of 3 m. What is its speed? (2016)
A.
180 m/s
B.
120 m/s
C.
60 m/s
D.
90 m/s
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Solution
Speed v = f × λ = 60 Hz × 3 m = 180 m/s.
Correct Answer: A — 180 m/s
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Q. A wave has a frequency of 60 Hz and a wavelength of 3 m. What is the speed of the wave?
A.
180 m/s
B.
120 m/s
C.
60 m/s
D.
30 m/s
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Solution
The speed v of a wave is given by v = fλ. Thus, v = 60 * 3 = 180 m/s.
Correct Answer: B — 120 m/s
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Q. A wave has a speed of 300 m/s and a frequency of 150 Hz. What is its wavelength? (2022)
A.
1 m
B.
2 m
C.
3 m
D.
4 m
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Solution
Wavelength λ can be calculated using the formula λ = v/f. Therefore, λ = 300 m/s / 150 Hz = 2 m.
Correct Answer: B — 2 m
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Q. A wave on a string has a speed of 50 m/s and a wavelength of 2 m. What is its frequency? (2018)
A.
25 Hz
B.
50 Hz
C.
100 Hz
D.
75 Hz
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Solution
Frequency f = v/λ = 50 m/s / 2 m = 25 Hz.
Correct Answer: A — 25 Hz
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Q. A wave on a string is described by the equation y(x, t) = 0.05 sin(2π(0.1x - 5t)). What is the amplitude of the wave?
A.
0.01 m
B.
0.05 m
C.
0.1 m
D.
0.2 m
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Solution
The amplitude of the wave is the coefficient in front of the sine function. Here, the amplitude is 0.05 m.
Correct Answer: B — 0.05 m
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Q. A wave traveling along a string has a frequency of 50 Hz and a wavelength of 2 m. What is the speed of the wave? (2020)
A.
25 m/s
B.
50 m/s
C.
100 m/s
D.
75 m/s
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Solution
Speed (v) = frequency (f) × wavelength (λ) = 50 Hz × 2 m = 100 m/s.
Correct Answer: C — 100 m/s
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Q. A wave travels in a medium with a speed of 300 m/s and has a frequency of 75 Hz. What is the wavelength? (2021)
A.
4 m
B.
2 m
C.
3 m
D.
1 m
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Solution
Wavelength λ = v/f = 300 m/s / 75 Hz = 4 m.
Correct Answer: A — 4 m
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Q. A wave travels through a medium with a speed of 500 m/s and has a wavelength of 2 m. What is its frequency? (2022)
A.
250 Hz
B.
500 Hz
C.
1000 Hz
D.
2000 Hz
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Solution
Using the wave equation v = fλ, we can find frequency f = v/λ = 500 m/s / 2 m = 250 Hz.
Correct Answer: A — 250 Hz
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