Question

A particle at the end of a spring executes simple harmonic motion 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. $T^{-1}=t_1^{-1}+t_2^{-1}$
• D. $T^{-2}=t_1^{-2}+t_2^{-2}$

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 1st spring $t_1=2\pi \sqrt{\frac{m}{k_1}}$

$⟹k_1=\frac{4{\pi }^{2}m}{t_1^2}$

Similarly $k_2=\frac{4{\pi }^{2}m}{t_2^2}$

Now spring constant of the combination of two springs in series $\frac{1}{k_c}=\frac{1}{k_1}+\frac{1}{k_2}$

$⟹k_c=\frac{k_1{k}_2}{k_1+k_2}$

Time period of the oscillation of combined springs $T=2\pi \sqrt{\frac{m}{k_c}}$

$\therefore$ $k_c=\frac{4{\pi }^{2}m}{T^2}$

$\frac{k_1{k}_2}{k_1+k_2}=\frac{4{\pi }^{2}m}{T^2}$

$\frac{\frac{4{\pi }^{2}m}{t_1^2}\times \frac{4{\pi }^{2}m}{t_2^2}}{\frac{4{\pi }^{2}m}{t_1^2}+\frac{4{\pi }^{2}m}{t_2^2}}=\frac{4{\pi }^{2}m}{T^2}$

$\frac{\frac{1}{t_1^2{t}_2^2}}{\frac{1}{t_1^2}+\frac{1}{t_2^2}}=\frac{1}{T^2}⟹T^2=t_1^2+t_2^2$