Question

The velocity of the bob of a second’s pendulum when it is 6 cm from its mean position and amplitude of 10 cm is:

• A. $8\pi$ cm/s
• B. $6\pi$ cm/s
• C. $4\pi$ cm/s
• D. $2\pi$ cm/s

Hint:

In simple harmonic motion, the velocity of the particle is maximum at the equilibrium position and zero at the extreme positions.

Answer 1

The Correct Option is A

Step 1: The velocity in simple harmonic motion is expressed as: \[ v = \omega \sqrt{A^2 - x^2} \] where \( A = 10 \) cm, \( x = 6 \) cm, and \( \omega = \frac{2\pi}{T} = \pi \) rad/s. 
Step 2: Substituting the given values: \[ v = \pi \sqrt{(10)^2 - (6)^2} = \pi \sqrt{100 - 36} = \pi \sqrt{64} = 8\pi. \] \bigskip