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A mass is suspended separately by two different springs in successive order, then the time period is $T_1$ and $T_2$, respectively. If it is connected by both spring as shown in figure, then the time period $T_0$, so the correct reaction is:
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Verified Answer
The correct answer is:
$T_0^{-2}=T_1^{-2}+T_2^{-2}$
Time period of spring $T=2 \pi \sqrt{\frac{m}{\mathrm{~K}}}$
$\begin{aligned}
& \therefore K=4 \pi^2 \frac{m}{T^2} \\
& \Rightarrow K \propto \frac{1}{T^2}
\end{aligned}$
For parallel combination of spring
$K=K_1+K_2$
$\begin{aligned}
& \therefore \frac{1}{T_0^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2} \\
& \Rightarrow T_0^{-2}=T_1^{-2}+T_2^{-2}
\end{aligned}$
$\begin{aligned}
& \therefore K=4 \pi^2 \frac{m}{T^2} \\
& \Rightarrow K \propto \frac{1}{T^2}
\end{aligned}$
For parallel combination of spring
$K=K_1+K_2$
$\begin{aligned}
& \therefore \frac{1}{T_0^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2} \\
& \Rightarrow T_0^{-2}=T_1^{-2}+T_2^{-2}
\end{aligned}$
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