Question

Let $T_1$ and $T_2$ be the time period of oscillation of two springs $A$ and $B$ when a mass $m$ is suspended from them separately. Now both the springs are connected in parallel and the same mass $m$ is suspended with them. If $T$ is time period of the parallel system, then

• A. $T=T_1+T_2$
• B. $T=\frac{T_1{T}_2}{T_1+T_2}$
• C. $T^2=T_1^2+T_2^2$
• D. $\frac{1}{T^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2}$

Answer 1

The correct option is D $\frac{1}{T^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2}$

Let the spring constant are $k_1$ and $k_2$.
The times period of the springs are
$T_1=2\pi \sqrt{\frac{m}{k_1}}$ ...(i)
$T_2=2\pi \sqrt{\frac{m}{k_2}}$ ...(ii)

When two springs are connected in parallel, equivalent stiffness
$k=k_1+k_2$
$T=2\pi \sqrt{\frac{m}{k}}$
$\Rightarrow T=2\pi \sqrt{\frac{m}{k_1+k_2}}$
$\Rightarrow (\frac{T}{2\pi }{)}^2=\frac{m}{k_1+k_2}$
$\Rightarrow (\frac{2\pi }{T}{)}^2=\frac{k_1}{m}+\frac{k_2}{m}$
From equation (i) and (ii)
$\Rightarrow \frac{1}{T^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2}$