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A block of mass m is attached to the light spring of force constant k and released when it is in its natural length. Find amplitude of subsequent oscillations. |
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Answer» Solution :The maximum displacement of the spring in SUBSEQUENT motion will be `(2mg)/(k)` The equilibrium position of the SYSTEM will OCCUR at the extension of `(MG)/k` `therefore ` AMPLITUDE `=(2mg)/(k) - (mg)/(k) = (mg)/(k)` |
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