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The period of a particle executing shm i...

The period of a particle executing shm is `2pi`. The total energy of the particle is 0.0786 J. After a time `pi//4s` the displacement is 0.2m. Calculate the amplitude and mass of the particle.

Text Solution

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Total energy = E = 0.0786 J
Period `=T = 2pi`
`omega = (2pi)/T = (2pi)/(2pi) =1 rad//s`
Displacement after a time t `= pi/4 s` is y = 0.2 m
`0.2 = a sin 1 xx (pi)/4`
`a= (0.2)/(sin ""pi/4)=0.2 xxsqrt2 = 0.283m`
Total energy `E=1/2 momega^2a^2=1/2xxmxx1^2xx0.283^2=0.04m`
`m = E/(0.04)=(0.0786)/(0.04)=1.96kg`
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