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 parallel is $T$, then

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

Answer 1

Answer: (A)

The time period of spring pendulum
$T=2\pi \sqrt{\frac{m}{k}}$
where $k$ is the spring constant of the spring For the first spring
$T_1=2\pi \sqrt{\frac{m}{k_1}}$
or $T_1^2=\frac{4{\pi }^{2}m}{k_1}...\left(i\right)$
For second spring
$T_2=2\pi \sqrt{\frac{m}{k_2}}$
or $T_2^2=\frac{4{\pi }^{2}m}{k_2}...\left(ii\right)$
When these two springs are connected in parallel the effective spring constant is
$k_{eff}=k_1+k_2$
$\therefore$ The time period of oscillation is
$T=2\pi \sqrt{\frac{m}{k_{eff}}}=2\pi \sqrt{\frac{m}{k_1+k_2}}$
or $T^2=\frac{4{\pi }^{2}m}{k_1+k_2}$
or $\frac{1}{T^2}=\frac{k_1+k_2}{4{\pi }^{2}m}$
$=\frac{k_1}{4{\pi }^{2}m}+\frac{k_2}{4{\pi }^{2}m}$
or $\frac{1}{T^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2}$ (Using $(i)$ and $(ii)$)
or $T^{-2}=T_1^{-2}+T_2^{-2}$