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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
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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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