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Two spring of force constant $k_{1}$ and $k_{2}$ are configured as the figure given below
The angular frequency of this configuration is
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The angular frequency of this configuration is
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Verified Answer
The correct answer is:
$\sqrt{\frac{k_{1}+k_{2}}{m}}$
The equation of motion of this system about mean position is given by
$$
\begin{aligned}
& F &=k_{1} x+k_{2} x & \\
\Rightarrow & m \frac{d^{2} x}{d t^{2}} &=k_{1} x+k_{2} x \quad \Rightarrow \frac{d^{2} x}{d t^{2}}=x\left(\frac{k_{1}+k_{2}}{m}\right) \\
\because & a &=\frac{d^{2} x}{d t^{2}}=\omega^{2} x \\
\therefore & \omega^{2} &=\frac{k_{1}+k_{2}}{m} \text { or } \omega=\sqrt{\frac{k_{1}+k_{2}}{m}}
\end{aligned}
$$
$$
\begin{aligned}
& F &=k_{1} x+k_{2} x & \\
\Rightarrow & m \frac{d^{2} x}{d t^{2}} &=k_{1} x+k_{2} x \quad \Rightarrow \frac{d^{2} x}{d t^{2}}=x\left(\frac{k_{1}+k_{2}}{m}\right) \\
\because & a &=\frac{d^{2} x}{d t^{2}}=\omega^{2} x \\
\therefore & \omega^{2} &=\frac{k_{1}+k_{2}}{m} \text { or } \omega=\sqrt{\frac{k_{1}+k_{2}}{m}}
\end{aligned}
$$
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