| ⇦ | 
| ⇨ |
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 120 m long train is moving in a direction with speed 20 m/s.A train B moving
- The intensity of magnetisation of a bar magnet is 5×10⁴ Am⁻¹. The magnetic length
- A force F=ai+3j+6k is acting at a point r=2i-6j-12k. The value of α for
- Monochromatic light of wavelength 667 nm is produced by a helium neon laser
- If white light is used in Young’s double slit experiment
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A 120 m long train is moving in a direction with speed 20 m/s.A train B moving
- The intensity of magnetisation of a bar magnet is 5×10⁴ Am⁻¹. The magnetic length
- A force F=ai+3j+6k is acting at a point r=2i-6j-12k. The value of α for
- Monochromatic light of wavelength 667 nm is produced by a helium neon laser
- If white light is used in Young’s double slit experiment
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