InterviewSolution
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InterviewSolution
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One end of a spring of force constant K is fixed to a vertical wall and the other to a body of mass m resting on a smooth horizontal surface. There is another wall at a distance `x_0` from the body. The spring is then compressed by `3x_0` and released. The time taken to strike the wall from the instant of release is (given `sin^-1((1)/(3))=((pi)/(9))`)A. `(pi)/(6)sqrt((m)/(K))`B. `(2pi)/(3)sqrt((m)/(K))`C. `(pi)/(4)sqrt((m)/(K))`D. `(11pi)/(18)sqrt((m)/(K))` |
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Answer» Correct Answer - D When spring is compressed by `3x_0`. Amplitude `A=3x_0` The time taken from extreme compressed position to mean position `t_1=(T)/(4)` If time taken `(t_2)` from mean position to `x=x_0` is given `x=Asin(2pit_2)/(T)impliesx_0=3x_0sin(2pit_2)/(T)` `sin((2pit_2)/(T))` `T=(1)/(3)implies(2pit_2)/(T)=(pi)/(9)impliest_2=(T)/(18)` `t_1+t_2=(T)/(4)+(T)/(18)=(11)(18)T=(11)/(18)2pisqrt((m)/(K))=(11)/(9)pisqrt((m)/(K))` |
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