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 $T_{1}^{2} = \frac{4\pi^{2}m}{k_{1}}\quad...\left(i\right)$
For second spring
$ T_{2} = 2\pi\sqrt{\frac{m}{k_{2}}} $
or $T_{2}^{2} = \frac{4\pi^{2}m}{k_{2}} \quad...\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 $T^{-2} = T_{1}^{-2} +T_{2}^{-2}$