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The kinetic energy of a particle executing shm is 16 j when it is in its mean position If the amplitude of oscillation is 25 and the mass of particle is 5.12kg the time period of its oscillation is
Dear Pallavi
x=Xsinwt
v=dx/dt =Xwcoswt
at mean position particle has only kinetic energy
at mean position velocity is V=Xw
K.E=1/2 *m*V2
16= 1/2 m X2w2
32=5.12 (25)2 (2∏/T)2
now you can calculate T
i am agree with previous sol.
so ans.=52.34sec
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