InterviewSolution
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InterviewSolution
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A particle executes simple harmonic motion with a period of `16s`. At time `t=2s`, the particle crosses the mean position while at `t=4s`, its velocity is `4ms^-1` amplitude of motion in metre isA. `sqrt(2) pi`B. `16 sqrt(2) pi`C. `24 sqrt(2) pi`D. `32 sqrt(2)//pi` |
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Answer» Correct Answer - D For simple harmonic motion, y = a sin `omega`t `therefore" "y = a sin ((2pi)/(T))t" "("at t"=2s)` `y_(1)=a sin [((2pi)/(16))xx2]=a sin((pi)/(4))=(a)/(sqrt(2))" "...(i)` At t = 4 s or after 2 s from mean position. `y_(1)=(a)/(sqrt(2))"and velocity"=4 ms^(-1)` `therefore` Velocity `= omega sqrt(a^(2)-y_(1)^(2))` or `4 = ((2pi)/(16))sqrt(a^(2)-(a^(2))/(2))" "["From Eq. (i)"]` or `4 = (pi)/(8) xx (a)/(sqrt(2))or a = (32 sqrt(2))/(pi)m` |
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