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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
Related Questions: - A microscope is having objective of focal length 1 cm and eye-piece of focal length 6 cm.
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Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A microscope is having objective of focal length 1 cm and eye-piece of focal length 6 cm.
- The ratio of longest wavelength corresponding to Lyman and Blamer series
- The earth is assumed to be a sphere of radius R. A platform is arranged at a height R
- A ray of light travelling in a transparent medium of refractive index µ, falls on surface
- The horizontal range and the maximum height of a projectile are equal
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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