Question

Let $T_1$ and $T_2$ be the time periods of two spring A and B when a mass m is suspended from them separately. Now both the springs are connected in parallel and same mass m is suspended with them. Now let T be the time period in this position. 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. $\frac{1}{T}=\frac{1}{T_1^2}+\frac{1}{T_2^2}$

Answer 1

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

$\begin{array}{c}{T}_1=2\pi \int \frac{m}{K_1}or{K}_1=\frac{A{\pi }^{2}m}{T_1^2}\\ K_2=\frac{A{\pi }^{2}m}{T_2^2}\end{array}$

In Parallel

$\begin{array}{c}K=K_1+K_2\\ \frac{1}{T}=\frac{1}{T_1^2}+\frac{1}{T_2^2}\end{array}$