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A particle of mass 0.1 kg executes SHM under a force F = (-10x) N. Speed of particle at mean position is 6 m/s. Then amplitude of oscillation is

A particle of mass 0.1 kg executes SHM under a force F = (-10x) N. Speed of particle at mean position is 6 m/s. Then amplitude of oscillation is

Grade:12th pass

1 Answers

Moazzam Ahmad
12 Points
3 years ago
According to the question
F = (-10x) N
compare it with F = -kx
We get  K = 10 ;
Speed of a particle is max  at mean position which is equal to = 6 ms-1
V(max) = Aw2  ( A is amplitude and w is angular velocity)
As we know w = (  K/M )½
Therefore w2 = ( K/M ) = (10/.1) = 100
Put above result in Vmax equation we get.
V(max) = A x 100
6 = A x 100
A = .06 m = 6 cm Ans Hope it helped you
                                      

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