First slide

Simple hormonic motion

Question

The frequency of oscillation of the springs shown in fig (4) will be

Easy

A

$\frac{1}{2 π} \sqrt{(\frac{k}{m})}$
B

$2 π \sqrt{\frac{k}{m}}$
C

$\frac{1}{2 π} \sqrt{[\frac{(k_{1} + k_{2}) m}{k_{1} k_{2}}]}$
D

$\frac{1}{2 π} \sqrt{[\frac{k_{1} k_{2}}{(k_{1} + k_{2}) m}]}$

Solution

Here $x_{1} = - F / k_{1} and x_{2} = - F / k_{2}$
Total extension $x = x_{1} + x_{2} = - F [\frac{1}{k_{1}} + \frac{1}{k_{2}}]$
we know that F = - k x
$k = (\frac{k_{1} k_{2}}{k_{1} + k_{2}}) Further T = 2 π \sqrt{(m / k)} = 2 π \sqrt{[\frac{m (k_{1} + k_{2})}{k_{1} k_{2}}]} n = \frac{1}{T} = \frac{1}{2 π} \sqrt{[\frac{k_{1} k_{2}}{m (k_{1} + k_{2})}]}$

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