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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
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If λ₁ and λ₂ are the wavelengths of the first members of the Lyman and Paschen
- Young’s modulus of perfectly rigid body material is
- The decreasing order of wavelength of infrared, microwave, ultraviolet and gamma
- A particle under the action of a constant force moves from rest upto 20 seconds
- A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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