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

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Question

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

Options

  • 8π cm/s

  • 6π cm/s

  • 4π cm/s

  • 2π cm/s

MCQ
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Solution

The velocity of bob of a second’s pendulum when it is 6 cm from its mean position and amplitude of 10 cm, is 8π cm/s.

Explanation:

The velocity of the bob in simple harmonic motion is given by:

`v = omega sqrt(A^2 - x^2)`

Where:

A = 10 cm (amplitude),

x = 6 cm (displacement),

`omega = (2pi)/T`, with T = 2 s (second's pendulum), so ω = π rad/s.

Substitute the values:

`v = pisqrt(10^2 - 6^2)`

= `pi sqrt(100 - 36)`

= `pi sqrt64`

Simplify:

v = π × 8

= 8 π cm/s.

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