Q. A mass attached to a spring oscillates with a frequency of 1 Hz. What is the period of oscillation? (2023)
A.
1 s
B.
2 s
C.
0.5 s
D.
0.25 s
Show solution
Solution
Period (T) = 1 / frequency (f) = 1 / 1 Hz = 1 s.
Correct Answer:
A
— 1 s
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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
Show solution
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 frequency of 2 Hz. What is the period of oscillation? (2023)
A.
0.5 s
B.
1 s
C.
2 s
D.
4 s
Show solution
Solution
Period (T) = 1 / Frequency (f) = 1 / 2 Hz = 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
Show solution
Solution
Frequency f = 1/T = 1/2 s = 0.5 Hz.
Correct Answer:
A
— 0.5 Hz
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Q. A mass of 0.5 kg is attached to a spring with a spring constant of 200 N/m. What is the frequency of oscillation? (2021)
A.
0.5 Hz
B.
1 Hz
C.
2 Hz
D.
4 Hz
Show solution
Solution
The frequency f of a mass-spring system is given by f = (1/2π)√(k/m). Here, f = (1/2π)√(200/0.5) ≈ 2 Hz.
Correct Answer:
C
— 2 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
Show solution
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
Show solution
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
Show solution
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
Show solution
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 of length 1 m swings with a small amplitude. What is its approximate time period? (2023)
A.
1 s
B.
2 s
C.
0.5 s
D.
0.25 s
Show solution
Solution
Time period (T) = 2π√(L/g) ≈ 2π√(1/9.8) ≈ 2 s.
Correct Answer:
B
— 2 s
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Q. A pendulum swings with a maximum angle of 15 degrees. What is the approximate frequency of the pendulum if its length is 1 m? (2020)
A.
0.5 Hz
B.
1 Hz
C.
1.5 Hz
D.
2 Hz
Show solution
Solution
Frequency (f) = 1 / (2π√(L/g)) = 1 / (2π√(1/9.8)) ≈ 1 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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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 its frequency is 1700 Hz, what is its wavelength? (2022)
A.
0.2 m
B.
0.5 m
C.
1 m
D.
2 m
Show solution
Solution
Wavelength (λ) = speed (v) / frequency (f) = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer:
B
— 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
Show solution
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
Show solution
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
Show solution
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 sound wave travels through air at a speed of 343 m/s. If the frequency of the sound is 686 Hz, what is the wavelength? (2022)
A.
0.5 m
B.
1 m
C.
2 m
D.
3 m
Show solution
Solution
Using the wave speed formula v = f * λ, we can rearrange to find λ = v/f. Thus, λ = 343 m/s / 686 Hz = 0.5 m.
Correct Answer:
B
— 1 m
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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
Show solution
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 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 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
Show solution
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 wavelength of the sound in air (speed of sound = 340 m/s)? (2021)
A.
0.77 m
B.
0.85 m
C.
0.90 m
D.
1.00 m
Show solution
Solution
Wavelength λ is given by λ = v/f. Thus, λ = 340 m/s / 440 Hz ≈ 0.77 m.
Correct Answer:
A
— 0.77 m
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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
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 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
Show solution
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
Show solution
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 50 Hz and a speed of 340 m/s. What is its wavelength? (2021)
A.
6.8 m
B.
7.0 m
C.
7.5 m
D.
8.0 m
Show solution
Solution
Wavelength λ is given by λ = v/f. Thus, λ = 340 m/s / 50 Hz = 6.8 m.
Correct Answer:
A
— 6.8 m
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Showing 1 to 30 of 76 (3 Pages)
Oscillations & Waves MCQ & Objective Questions
Understanding the concepts of Oscillations & Waves is crucial for students preparing for school exams and competitive tests in India. This topic not only forms a significant part of the physics syllabus but also features prominently in various objective questions and MCQs. By practicing Oscillations & Waves MCQ questions, students can enhance their grasp of the subject and improve their chances of scoring better in exams.
What You Will Practise Here
Simple Harmonic Motion (SHM) and its characteristics
Wave properties: wavelength, frequency, amplitude, and speed
Types of waves: longitudinal and transverse waves
Sound waves and their applications
Principles of superposition and interference of waves
Standing waves and resonance phenomena
Key formulas related to oscillations and wave motion
Exam Relevance
The topic of Oscillations & Waves is integral to the curriculum of CBSE, State Boards, NEET, and JEE. Students can expect questions that test their understanding of fundamental concepts, as well as their ability to apply these concepts in problem-solving scenarios. Common question patterns include numerical problems, conceptual questions, and diagram-based queries, making it essential for students to be well-prepared with important Oscillations & Waves questions for exams.
Common Mistakes Students Make
Confusing the terms frequency and period
Misunderstanding the difference between longitudinal and transverse waves
Neglecting to apply the correct formulas in numerical problems
Overlooking the significance of units in wave calculations
FAQs
Question: What is simple harmonic motion?Answer: Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement from the equilibrium position.
Question: How do waves transfer energy?Answer: Waves transfer energy through the oscillation of particles in the medium, allowing energy to move from one point to another without the permanent displacement of the medium itself.
Ready to boost your understanding of Oscillations & Waves? Dive into our practice MCQs and test your knowledge today! Regular practice will not only solidify your concepts but also prepare you for success in your upcoming exams.
