AIPMT MAINS2009Physics-Oscillations
AIPMT MAINS 2009 Physics Simple Harmonic Motion MCQ Question
Type: MCQ-numerical-Medium-Class 11
A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of pendulum at x = a/2 will be:
A
πa/T
B
(3π²a)/T
C
(√3πa)/T
D
(√3π²a)/T
Correct Answer
Option C
Detailed Explanation
The speed of the pendulum at any point x is given by v = ω√(a² - x²), where ω = 2π/T. Substituting x = a/2 gives v = (√3πa)/T.
Found an issue with this question?