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A particle executes simple harmotic moti...

A particle executes simple harmotic motion about the point `x=0`. At time `t=0`, it has displacement `x=2cm` and zero uelocity. If the frequency of motion is `0.25 s_(-1)`, find (a) the period, (b) angular frequency, ( c) the amplitude, (d) maximum speed, (e) the displacement from the mean position at `t=3 s`and (f) the velocity at `t=3 s`.

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
(a) Period `T=(1)/(f)`
`=(1)/(0.25 s^(-1))=4 s`
(b) Angular frequency `omega=(2pi)/(T)`
`=(2pi)/(4)=(pi)/(2) rad//s`
`=1.57rad//s`
( c) Amplitude is the maximum displacement from mean position. Hence, `A=2-0=2cm`.
(d) Maximum speed `v_(max)=A omega`
`=2.(pi)/(2)=pi cm//s`
`=3.14 cm//s`
(e) At `t=0`, particle starts from extreme position.
`:. x=A cos omega t`
`=2 cos((pi)/(2)t)`
At `t=1sec`
`x=0`
(f) Velocity at `x=0`, is `v_(max)` i.e. `3.14cm//s`.
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