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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
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Monochromatic radiation emitted when electron on hydrogen atom
- When a wire of uniform cross-section a, length l and resistance R
- Three solids of masses m₁,m₂ and m₃ are connected with weightless string
- Find out the angle of projection, if range is 4 times of maximum height
- 300j work is done in sliding a 2kg block up an inclined plane of height 10m
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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