Two identical springs of constant ' k ' are connected in series and then in parallel. A mass ' m ' is suspended from them, the ratio of their frequencies of vertical oscillations w

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Two identical springs of constant ' k ' are connected in series and then in parallel. A mass ' m ' is suspended from them, the ratio of their frequencies of vertical oscillations will be

TestHub · Accepted Answer

For two identical springs of spring constant $k$ connected in series, the equivalent spring constant $k_{s}$ is given by:

$\frac{1}{k _{s}} = \frac{1}{k} + \frac{1}{k} \Rightarrow k_{s} = \frac{k}{2}$

The frequency of vertical oscillation in series is:

$f_{s} = \frac{1}{2 π} \frac{k _{s}}{m} = \frac{1}{2 π} \frac{k}{2 m}$

For the same springs connected in parallel, the equivalent spring constant $k_{p}$ is given by:

$k_{p} = k + k = 2 k$

The frequency of vertical oscillation in parallel is:

$f_{p} = \frac{1}{2 π} \frac{k _{p}}{m} = \frac{1}{2 π} \frac{2 k}{m}$

The ratio of their frequencies is:

$\frac{f _{s}}{f _{p}} = \frac{\frac{1}{2 π} \frac{k}{2 m}}{\frac{1}{2 π} \frac{2 k}{m}} = \frac{k}{2 m} \times \frac{m}{2 k} = \frac{1}{4} = \frac{1}{2}$

Thus, the ratio is $1 : 2$ .

Answer: 1:2

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