First slide

Simple hormonic motion

Question

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

Moderate

A

$2 π \sqrt{\frac{m}{K}}$
B

$π \sqrt{\frac{m}{4 K}}$
C

$π \sqrt{\frac{m}{K}}$
D

$4 π \sqrt{\frac{m}{K}}$

Solution

Let us assume that in equilibrium condition spring is $x_{0}$ elongate from its natural length
In equilibrium $T_{0} = mg$ and ${kx}_{0} = 2 T_{0}$
$\Rightarrow {kx}_{0} = 2 mg$ -------------(1)
If the mass m moves down a distance x from its equilibrium position then pulley will move down by $\frac{x}{2}$ . So the extra force in spring will be $\frac{kx}{2}$ . From figure
$F_{net} = mg - T = mg - \frac{k}{2} (x_{0} + \frac{x}{2})$
$F_{net} = mg - \frac{{kx}_{0}}{2} - \frac{kx}{4}$
from eq.(1), we get
$F_{net} = \frac{- kx}{4}$ -----------------(2)
Now compare eq.(2) with F = $- K_{SHM} x$
then $K_{SHM} = \frac{K}{4}$
$\Rightarrow T = 2 π \sqrt{\frac{m}{K_{SHM}}} = 2 π \sqrt{\frac{4 m}{K}} = 4 π \sqrt{\frac{m}{K}}$

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