| ⇦ | 
| ⇨ |
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:
- The period of oscillation of a mass M suspended from a spring of negligible mass is T
- A particle is dropped from a building. It takes 4 sec for the particle to reach the ground
- A ray of light is incident on a transparent glass slab of refractive index 1.62
- A sample of HCl gas is placed in an electric field of 5 x 10⁴ N/C
- A body of mass m slides down an incline and reaches the bottom
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