One end of a spring of force constant K is fixed to a vertical wall and other to a body of a mass M resting on  smooth horizontal surface. There is another wall at a distance X0 from the body. The Spring is then compressed by 2X0 and released. Find out the time taken to strike the wall.

Asked by Malavika Umesh | 21st May, 2015, 07:28: PM

Expert Answer:

begin mathsize 14px style Given space that space the space amplitude space of space the space motion comma space straight A space equals 2 straight x subscript 0 We space know space that space the space time space needed space to space cover space from space compressed space position space to space mean space position left parenthesis normal right parenthesis equals straight T over 4 space where space straight T space is space the space timeperiod space of space oscillation For space the space displacement space from space the space mean space position space to space straight x subscript 0 comma space the space time space taken comma straight t space can space be space obtained space from colon straight y equals straight A space sin space ωt where space straight y space is space the space displacement space from space mean space position comma space straight A space the space amplitude Therefore colon straight x subscript 0 equals 2 straight x subscript 0 space sin space ωt But space straight omega space equals fraction numerator 2 straight pi over denominator straight T end fraction space rightwards double arrow straight x subscript 0 equals 2 straight x subscript 0 space sin fraction numerator 2 straight pi over denominator straight T end fraction space straight t 1 half equals sin fraction numerator 2 straight pi over denominator straight T end fraction space straight t fraction numerator 2 straight pi over denominator straight T end fraction space straight t equals sin to the power of negative 1 end exponent left parenthesis 0.5 right parenthesis fraction numerator 2 straight pi over denominator straight T end fraction space straight t equals straight pi over 6 straight t equals straight T over 12 Therefore space the space time space taken space to space hit space the space wall space will space be equals straight T over 4 plus straight T over 12 equals open parentheses fraction numerator 3 plus 1 over denominator 12 end fraction close parentheses straight T space equals straight T over 3 The space time space taken space to space hit space the space wall space equals space straight T over 3 If space straight a space mass space straight M space is space suspended space from space straight a space spring space of space force space constant space straight K comma space then space t h e space t i m e space p e r i o d space w i l l space b e colon straight T equals 2 straight pi square root of straight M over straight K end root rightwards double arrow The space time space taken space to space hit space the space wall space equals space straight T over 3 equals fraction numerator 2 straight pi over denominator 3 end fraction square root of straight M over straight K end root The space time space taken space to space hit space the space wall space equals fraction numerator 2 straight pi over denominator 3 end fraction square root of straight M over straight K end root end style

Answered by Jyothi Nair | 22nd May, 2015, 09:47: AM

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