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A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is
Options
(a) 4 Hz
(b) 3 Hz
(c) 2 Hz
(d) 1 Hz
Correct Answer:
1 Hz
Explanation:
a = 5 cm, vₘₐₓ = 31.4 cm/s
vₘₐₓ = ωa ⇒ 31.4 = 2πʋ × 5
⇒ 31.4 = 10 × 31.4 × ʋ
⇒ ʋ = 1 Hz
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Topics: Oscillations
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body of mass m=3.513 kg is moving along the x-axis with a speed of 5 m/s
- The half-life of a radioactive substance is 30 min. The time (in minutes) taken
- A simple pendulum has a time period T₁ when on the earth’s surface and T₂ When taken
- A ball is dropped from a high rise platform at t = 0 starting from rest
- Consider the following two statements about Kirchhoff’s law
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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