Question

A body is suspended from the two light springs separately, the periods of vertical oscillations are \[T_1\] and \[T_2\]. The same body is suspended from two springs connected in series first and then in parallel. The periods of oscillations in both cases respectively are ______.

Options

• \[\sqrt{\mathrm{T}_{1}\mathrm{T}_{2}},\sqrt{\frac{\mathrm{T}_{1}\mathrm{T}_{2}}{2}}\]

• \[\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2},\frac{\mathrm{T}_1\mathrm{T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}.\]

• \[\sqrt{\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{2}},\sqrt{\frac{\mathrm{T}_1\mathrm{T}_2}{2}}\]

• \[\sqrt{\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{3}},\frac{2\mathrm{~T}_1\mathrm{~T}_2}{\sqrt{\mathrm{~T}_1^2+\mathrm{T}_2^2}}\]

Answer 1

A body is suspended from the two light springs separately, the periods of vertical oscillations are \[T_1\] and \[T_2\]. The same body is suspended from two springs connected in series first and then in parallel. The periods of oscillations in both cases respectively are \[\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2},\frac{\mathrm{T}_1\mathrm{T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}.\].

Explanation:

Time period of oscillation

\[\mathrm{T}=2\pi\sqrt{\frac{m}{k}}\Rightarrow\mathrm{T}\alpha\frac{1}{\sqrt{K}}\]

For springs in series

Effective spring constant.

\[\frac{1}{\mathrm{K}_{\mathrm{series}}}=\frac{1}{\mathrm{K}_1}+\frac{1}{\mathrm{K}_2}\]

\[\mathrm{K}_3=\frac{\mathrm{k}_1\mathrm{k}_2}{\mathrm{k}_1+\mathrm{k}_2}\]

\[\therefore\] Time period: \[\mathrm{T_s=2\pi\sqrt{\frac{m}{K_s}}=2\pi\sqrt{\frac{m\left(k_1+k_2\right)}{k_1k_2}}}\]

\[\therefore\quad\mathrm{T}_\mathrm{S}=\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}\]

For springs in parallel

Effective spring constant

\[\mathrm{k_p=k_1+k_2}\]

\[\therefore\quad\frac{1}{\mathrm{T_P}^2}=\frac{1}{\mathrm{T_1}^2}+\frac{1}{\mathrm{T_2}^2}\Rightarrow\mathrm{T_P}\frac{\mathrm{T_1T_2}}{\sqrt{\mathrm{T_1}^2+\mathrm{T_2}^2}}\]