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A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is `pi m//s`. Amplitude of oscillation is `(cos 45^(@) = 1/(sqrt(2)))`A. `2sqrt2m`B. `4sqrt2m`C. `6sqrt2m`D. `8sqrt2m` |
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Answer» Correct Answer - D Displacement of the particle `=x=A sin omegat` Velocity of the particle `=v=(dx)/(dt)=A omega cos omegat` `v=pi m//s, T=16s, omega=(2pi)/(T)=(2pi)/(16)"rad/s"` `therefore pi=A xx (pi)/(8)xx cos""(pi)/(8)xx2` `therefore 1=(A)/(8)cos ""pi/4=(A)/(8).(1)/(sqrt2)` `therefore A=8sqrt2m` |
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