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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 isA. 0.1 sB. 0.2 sC. `pi//10s`D. `pi//5s` |
Answer» 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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