| ⇦ | 
| ⇨ |
A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a/2 will be:
Options
(a) πa / T
(b) 3π²a / T
(c) πa√3 / T
(d) πa√3 / 2T
Correct Answer:
πa√3 / T
Explanation:
Speed v = ? √(a² – x²) , x = a / 2
v = ? √(a² – a² / 4) = ? √(3a² / 4) = 2π / T . a√3 / 2
πa√3 / T
Related Questions: - In a laser tube, all the photons
- If the velocity of sound in air is 350 m/s, thent he fundamental frequency
- Alternating current cannot be measured by D.C Ammeter because
- In common base circuit of a transistor, current amplification factor is 0.95.
- Two sources P and Q produce notes of frequency 660 Hz each.
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In a laser tube, all the photons
- If the velocity of sound in air is 350 m/s, thent he fundamental frequency
- Alternating current cannot be measured by D.C Ammeter because
- In common base circuit of a transistor, current amplification factor is 0.95.
- Two sources P and Q produce notes of frequency 660 Hz each.
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