One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance $$x0$$ from the black. The spring is then compressed by $$2x0$$ and released. The time taken to strike the wall is

Question

One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance $$x0$$ from the black. The spring is then compressed by $$2x0$$ and released. The time taken to strike the wall is

Infinity Learn · Accepted Answer

The total time from A to CtAC $$=$$ $$tAB+tBC$$      $$=$$ $$(T4)+tBCwhere$$ T $$=$$ time period of oscillation of spring mass systemtBC can be obtained from, BC $$=$$ AB $$sin$$ $$(2$$$$π$$$$T)tBCPutting$$ BC AB $$=$$ $$12$$ we obtain tBC $$=$$ $$T12$$⇒tAC $$=$$ $$T4+T12$$ $$=$$ $$2$$$$π$$$$3mk$$.

One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance $x_{0}$ from the black. The spring is then compressed by 2 $x_{0}$ and released. The time taken to strike the wall is

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One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance x0 from the black. The spring is then compressed by 2x0 and released. The time taken to strike the wall is

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One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance $x_{0}$ from the black. The spring is then compressed by 2 $x_{0}$ and released. The time taken to strike the wall is