First slide

Simple hormonic motion

Question

A system shown in fig. (5) consists of mass less pulley,

a spring of force constant k and a block of mass m. If block is just slightly displaced vertically down from its equilibrium position and released. The time period of vertical oscillation is

Moderate

A

$T = π \sqrt{(\frac{m}{4 k})}$
B

$T = 2 π \sqrt{(\frac{m}{4 k})}$
C

$T = 2 π \sqrt{(\frac{m}{2 k})}$
D

$T = 2 π \sqrt{(\frac{m}{3 k})}$

Solution

If m moves downward through a distance y from its equilibrium position, then pulley will also move through a distance. So, the stretching of string will
be2y.
Tension
$T = F = k (2 y) Also mg = 2 T ∴ Restoring force, F^{'} = 2 T = 4 ky Now ma = - 4 k y or a = - 4 k y / m \frac{d^{2} y}{{dt}^{2}} = - (\frac{4 k}{m}) y or \frac{d^{2} y}{{dt}^{2}} + (\frac{4 k}{m}) y = 0 or \frac{d^{2} y}{{dt}^{2}} + ω^{2} y = 0 where ω = \sqrt{(\frac{4 k}{m})} This represents S . H . M . The time period is given by T = \frac{2 π}{ω} = 2 π \sqrt{(\frac{m}{4 k})}$

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