A particle connected to the end of a spring executes S.H.M. with period ' $\mathrm{T}_1$ '. While the…
A particle connected to the end of a spring executes S.H.M. with period ' $\mathrm{T}_1$ '. While the corresponding period for another spring is ' $T_2$ '. If the period of oscillation with two springs in series is ' $T$ ', then
$\mathrm{T}=\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}$
$\mathrm{T}=\sqrt{\mathrm{T}_2^2-\mathrm{T}_1^2}$
$\mathrm{T}=\mathrm{T}_1+\mathrm{T}_2$
$\mathrm{T}=\mathrm{T}_1-\mathrm{T}_2$
Solution
$\mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_1}} \quad \mathrm{~T}_2=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_2}}$
When the two springs are connected in series, the effective spring constant is given by $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$
$\begin{aligned}
& \therefore \mathrm{T}^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{k}}=4 \pi^2 \mathrm{~m}\left[\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2}\right]=4 \pi^2 \frac{\mathrm{m}}{\mathrm{k}_1}+4 \pi^2 \frac{\mathrm{m}}{\mathrm{k}_2} \\
& =\mathrm{T}_1^2+\mathrm{T}_2^2 \\
& \therefore \mathrm{T}=\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}
\end{aligned}$