First slide

Simple hormonic motion

Question

Two springs are connected to a block of mass M placed on a friction less surface as shown in fig. (1). If both the
springs have a spring constant k, the frequency of oscillation of the block is

Easy

A

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

$(\frac{1}{2 π}) \sqrt{(\frac{k}{2 M})}$
C

$(\frac{1}{2 π}) \sqrt{(\frac{2 k}{M})}$
D

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

Solution

For spring block system, the time period of oscillation is given by
$T = 2 π \sqrt{(\frac{M}{k_{eq}})}$
where $k_{eq}$ is equivalent spring constant of springs. Here
$\frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k} or k_{eq} = \frac{k}{2} T = 2 π \sqrt{[\frac{M}{(k / 2)}]}] = 2 π \sqrt{(\frac{2 M}{k})} or n = \frac{1}{T} = \frac{1}{2 π} \sqrt{(\frac{k}{2 M})}$

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