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A simple harmonic motion is represented by `x = 12 sin (10 t + 0.6)` Find out the maximum acceleration, if displacement is measured in metres and time in seconds.A. `-1200 m`B. `-2000 m`C. `-2400 m`D. `7200 m`

Answer» Correct Answer - A
Given equation, `x = 12 sin (10 t + 0.6)`
On comparing with `x(t)=A sin (omega t+phi)`
We have,
Amplitude, A = 12 m
Angular frequency, `omega = 10 rads^(-1)`
`therefore` Maximum acceleration,
`a_(max)=-omega^(2)A`
`=-(10)^(2)xx12`
`= -1200ms^(-2)`


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