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A particle executing S.H.M. has angular ...

A particle executing S.H.M. has angular frequency `6.28 s^(-1)` and amplitude 10 cm find (a) the time period (b) the maximum speed ( c ) the maximum acceleration (d) the speed when the displacement is 6 cm from the mean postiton ( e ) the speed at `t=1//6`s assuming that the motion starts from rest at t=0

Text Solution

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a. Time period `=(2pi)/6.28s=1s`
b. Maimum speed `=Aomega=(0.1m)(6.28m^-1)`
`=0.628ms^-1`
c. Maximum acceleration `=Aomega^2` ltbr.gt `=(0.1m)(6.28s^-1)^2`
=4ms^-2`
d. `v=omegasqrt(A^2-x^2)=(6.28s^-1)sqrt((10cm)^2-(6cm)^2)`
`=50.2cms^-1`
c. At t=0 the velocity is zero i.e. the particle is at an extreme. The equation for displacement may be written as
`x=Acosomegat`
The velocity is ` v=-Aomegasinomegat`.
At `t=1/6s, v=-(0.1m)(6.28s^-1)sin(6.28/6)`
`=(-0.628ms^-1)sinpi/3`
`=-54.4cms^-1`
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