Question

A block of mass m is suspended separately by two different springs have time period $t_1$ and $t_2$. If same mass is connected to parallel combination of both springs, then its time period is $T$. Then

• A. $T=t_1+t_2$
• B. $T^2=t_1^2+t_2^2$
• C. $T^{-1}=t_1^{-1}+t_2^{-1}$
• D. $T^{-2}=t_1^{-2}+t_2^{-2}$

Answer 1

The correct option is A $T=t_1+t_2$

Let the spring constant be $k_1$ and $k_2$

now, formula of time period of spring is given by

$T=2\pi \sqrt{m/k}$

now, $t_1=2\pi \sqrt{m/k_1}$

$t_1^2=\frac{4{\pi }^{2}m}{k_1}\Rightarrow k_1=4{\pi }^{2}m/t_1^2$----$\left(2\right)$

$t_2=2\pi \sqrt{m/k^{}}$

$t_2^2=\frac{4{\pi }^{2}m}{k_2}\Rightarrow k_2=4{\pi }^{2}m/t_2^2$---- $\left(2\right)$

now, both are connected in parallel then ;

$k_1+k_2\Rightarrow \frac{4{\pi }^2}{t_1^2}+\frac{4{\pi }^{2}m}{t_2^2}$

$4{\pi }^{2}m(\frac{1}{t_1^2}+\frac{1}{t_2^2})$

now, time period $=\sqrt{\frac{t_1^2.t_2^2}{t_1^2+t_2^2}}$