Question

A body at the end of a spring executes S.H.M. with a period $t_1,$ while the corresponding period for another spring is $t_2.$ If the period of oscillation with the two springs in series is $T$, then

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

Answer 1

The correct option is B $T^2=t_1^2+t_2^2$

Let mass of the particle be $m$ and $k$ represents the spring constant.
Time period of oscillation of a particle executing SHM $t=2\pi \sqrt{\frac{m}{k}}$
Thus, for first spring $t_1=2\pi \sqrt{\frac{m}{k_1}}$
$\Rightarrow k_1=\frac{4{\pi }^{2}n}{t_1^2}$
Now spring constant of the combination of two springs series $\frac{1}{k_c}=\frac{1}{k_1}+\frac{1}{k_2}$
Time period of the oscillation of combined springs $T=2\pi \sqrt{\frac{m}{k_c}}$
$\therefore k_c=\frac{4{\pi }^{2}m}{T^2}$
$\frac{k_1{k}_2}{k_1+k_2}=\frac{4{\pi }^{2}m}{T^2}$
$\frac{\frac{4{\pi }^{2}m}{t_1^2}\times \frac{4{\pi }^{2}m}{t_2^2}}{\frac{4{\pi }^{2}m}{t_1^2}+\frac{4{\pi }^{2}m}{t_2^2}}=\frac{4{\pi }^{2}m}{T^2}$
$\frac{\frac{1}{t_1^2\times t_2^2}}{\frac{1}{t_1^2}+\frac{1}{t_2^2}}=\frac{1}{T^2}\Rightarrow T^2=t_1^2+t_2^2$