| ⇦ | 
| ⇨ |
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 source of light is placed at a distance of 50 cm from a photocell and the stopping
- For a cubic crystal structure which one of the following relations
- The number of beta particles emitted by a radioactive substance is twice the number
- If t₁ ̷ ₂ is the half-life of a substance, then t₃ ̷ ₄ is the time in which
- In potentiometer experiment, a cell of emf 1.25V gives balancing length of 30 cm
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A source of light is placed at a distance of 50 cm from a photocell and the stopping
- For a cubic crystal structure which one of the following relations
- The number of beta particles emitted by a radioactive substance is twice the number
- If t₁ ̷ ₂ is the half-life of a substance, then t₃ ̷ ₄ is the time in which
- In potentiometer experiment, a cell of emf 1.25V gives balancing length of 30 cm
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