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InterviewSolution
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A body of mass `m` is released from a height h to a scale pan hung from a spring. The spring constant of the spring is `k`, the mass of the scale pan is negligible and the body does not bounce relative to the pan, then the amplitude of vibration isA. `(mg)/(k)sqrt(1-(2hk)/(mg))`B. `(mg)/(k)`C. `(mg)/(k)+(mg)/(k)sqrt(1+(2hk)/(mg))`D. `(mg)/(k)-(mg)/(k)sqrt(1-(2hk)/(mg))` |
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Answer» Correct Answer - C Decreases in potential energy of the mass when the pan gets lowered by distance `y` (due to mass hitting on the pan) `=mg(h+y)`, where `h` is the height through which the mass falls on the pan. Increases in elastic potential of the spring `=(1)/(2)ky^2` (according to low of conservation of energy) or `mg(h+y)=(1)/(2)ky^2` or `ky^2-2mgy-2mgh=0` `y=(2mg+-sqrt(4m^2g^2+8mghk))/(2k)` `=(mg)/(k)+-(mg)/(k)sqrt((1+(2hk)/(mg))` Velocity of the pan will be maximum at the time of collision and will be zero at the lowet position. Hence y should be the amplitude of oscillation. So, amplitude of vibration`=[(mg)/(k)+(mg)/(k)sqrt((1+(2hk)/(mg)))]` |
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