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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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