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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
- A beam of light contains two wavelengths 6500Å and 5200Å.
- A body is thrown vertically up with a velocity u.It passes three points A,B and C
- If X=3YZ², find dimensions of Y if X and Z are the dimensions of capacity
- A particle of mass ‘m’ is kept at rest at a height 3R from the surface of earth
- Work of 3.0×10⁻⁴ joule is required to be done in increasing the size of a soap film
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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