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- t=T12
- v=v02
- a=a02
- t=T8

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Solution

Verified by Toppr

here,x=A2=Asin(2πT)t

or, sin(2πT)t=12=sinπ6⇒t=T12

dxdt=2πTAcos(2πT)t

d2xdt2=−(2πT)2Asin(2πT)t

so, vmax=v0=(2πT)A,amax=a0=(2πT)2A

now at t=T/12, the velocity,v=dxdt=2πTAcos(2πT)T12

=2πTAcosπ6=2πTA√32

∴v=√3v0

at t=T/12, the acceleration,a=d2xdt2=(2πT)2Asin(2πT)T12

=2πTAsinπ6=2πTA12

∴a=a02

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