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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
- The output of an AND gate is connected to both the inputs of a NOR gate,
- A particle of mass 1.96×10⁻¹⁵ kg is kept in equilibrium between two horizontal metal
- Suppose the kinetic energy of a body oscillating with amplitude A and at a distance
- In an L-C-R series resonant circuit, the capacitance is changed from C to 4C.
- The upper half of an inclined plane of inclination θ is perfectly smooth while lower
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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