Figure shows a system consisting of a massless pulley, a spring of force constant k and a block of mass m. If the block is slightly displaced vertically down from its equilibrium position and then released, the period of its vertical oscillation is

Question

Infinity Learn · Accepted Answer

Let us assume that in equilibrium condition spring is $$x0$$ elongate from its natural lengthIn equilibrium $$T0$$ $$=$$ mg and $$kx0$$ $$=$$ $$2T0$$⇒$$kx0$$ $$=$$ $$2mg-------------(1)If$$ the mass m moves down a distance x from its equilibrium position then pulley will move down by $$x2$$. So the extra force in spring will be $$kx2$$. From $$figureFnet=$$ $$mg-T$$ $$=$$ $$mg-k2(x0+x2)Fnet$$ $$=$$ $$mg-kx02-kx4from$$ eq.(1), we getFnet $$=$$ $$-kx4-----------------(2)Now$$ compare eq.(2) with F $$=$$ $$-KSHMxthen$$ KSHM $$=$$ $$K4$$⇒ T $$=$$ $$2$$$$π$$mKSHM $$=$$ $$2$$$$π$$$$4mK$$ $$=$$ $$4$$$$π$$mK

Figure shows a system consisting of a massless pulley, a spring of force constant k and a block of mass m. If the block is slightly displaced vertically down from its equilibrium position and then released, the period of its vertical oscillation is

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