In figure,k=100N/m,M=1kgandF=10N.Find the amplitude when block is in simple harmonic motion, when block is moving with velocity 2m/s.
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Solution
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F=kx0 ∴x0=Fk=10100=0.1 m
E=12mv2+12kx20 =12×1×(2)2+12×100×(0.1)2=2.5 J
From P to Q Work done by applied force=change in mechanic energy. ∴FA=EQ−EP ∴(10)A=12k(A+x0)2−2.5 =12×100(A+0.1)2−2.5 Solving this equation, we get A=0.2 m
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