A body oscillates with a simple harmonic motion having amplitude 0.05 m. At a certain instant of time, its displacement is 0.01 m and acceleration is `1.0 m//s^(2)`. The period of oscillation is
A. 0.1 s
B. 0.2 s
C. `pi//10s`
D. `pi//5s`

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1 Answers

Correct Answer - D
Acceleration `=omega^(2)x`
`1=omega^(2)xx0.01`
`therefore omega^(2)=100 therefore omega=10" but "T=(2pi)/(omega)`
`therefore T=(pi)/(5)s`

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