First slide

Simple hormonic motion

Question

Two identical springs of constant.k are connected in series and parallel as shown in fig. {g). A mass m is

suspended from them. The ratio of their frequencies of vertical oscillations will be

Easy

A

2 :1
B

1:1
C

1:2
D

4:1

Solution

$n = \frac{1}{2 π} \sqrt{(\frac{k}{m})} In series k = \frac{k_{1} k_{2}}{k_{1} + k_{2}} = \frac{k \cdot k}{2 k} = \frac{k}{2} in parallel k = k_{1} + k_{2} = 2 k n_{1} = \frac{1}{2 π} \sqrt{(\frac{k}{2 m})} and n_{2} = \frac{1}{2 π} \sqrt{(\frac{2 k}{m})}$

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