| ⇦ | 
| ⇨ |
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: - Photocells convert
- Two particles A and B, move with constant velocities V₁ and V₂. At the initial
- The damping force on an oscillator is directly proportional to the velocity.
- A person wants a real image of his own, 3 times enlarged. Where should he stand
- The output of an AND gate is connected to both the inputs of a NOR gate,
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Photocells convert
- Two particles A and B, move with constant velocities V₁ and V₂. At the initial
- The damping force on an oscillator is directly proportional to the velocity.
- A person wants a real image of his own, 3 times enlarged. Where should he stand
- The output of an AND gate is connected to both the inputs of a NOR gate,
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply