#### Question

The maximum velocity of a particle performing linear S.H.M. is 0.16 m/s. If its maximum acceleration is 0.64 m/s^{2}, calculate its period.

#### Solution

v_{max} = 0.16 m/s, a_{max} =0.64 m/s^{2}

Period (T) = ?

v_{max} = Aω and a_{max} = Aω^{2}

`a_max/v_max=omega=(2pi)/T`

`T=(2pixxv_max)/a_max`

`=(2xx3.14xx0.16)/0.64`

`= 1.571 s`

Period of particle is 1.571 s

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#### APPEARS IN

Solution The Maximum Velocity of a Particle Performing Linear S.H.M. is 0.16 m/s. If Its Maximum Acceleration is 0.64 m/s^2, Calculate Its Period. Concept: Differential Equation of Linear S.H.M..