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:
Show Hint
- $8\pi$ cm/s
- $6\pi$ cm/s
- $4\pi$ cm/s
- $2\pi$ cm/s
The Correct Option is A
Solution and Explanation
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
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