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On a smooth inclined plane, a mass \(M\) is attached between two massless springs of force constant \(k\) each, as shown in the figure. The other ends of the springs are fixed to firm supports. The period of oscillation of the mass \(M\) is
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Verified Answer
The correct answer is:
\(2 \pi\left(\frac{M}{2 k}\right)^{\frac{1}{2}}\)
According to diagram given in question, it is clear that both springs are connected in series.
\(\therefore\) Equivalent spring constant \(k^{\prime}\) will be
\(k^{\prime}=\frac{k_1 k_2}{k_1+k_2}=\frac{k \cdot k}{k+k}=\frac{k}{2}\) ...(i)
Frequency of oscillation is given as
\(f=\frac{1}{2 \pi} \sqrt{\frac{k^{\prime}}{M}}\)
\(\therefore\) Time period of oscillation,
\(\begin{aligned}
T & =\frac{1}{f}=2 \pi \sqrt{\frac{M}{k^{\prime}}} \\
T & =2 \pi \sqrt{\frac{M}{k / 2}} \quad \text { [from Eq.(i)] } \\
\Rightarrow T & =2 \pi \sqrt{\frac{2 M}{k}}=2 \pi\left(\frac{M}{2 k}\right)^{1 / 2}
\end{aligned}\)
\(\therefore\) Equivalent spring constant \(k^{\prime}\) will be
\(k^{\prime}=\frac{k_1 k_2}{k_1+k_2}=\frac{k \cdot k}{k+k}=\frac{k}{2}\) ...(i)
Frequency of oscillation is given as
\(f=\frac{1}{2 \pi} \sqrt{\frac{k^{\prime}}{M}}\)
\(\therefore\) Time period of oscillation,
\(\begin{aligned}
T & =\frac{1}{f}=2 \pi \sqrt{\frac{M}{k^{\prime}}} \\
T & =2 \pi \sqrt{\frac{M}{k / 2}} \quad \text { [from Eq.(i)] } \\
\Rightarrow T & =2 \pi \sqrt{\frac{2 M}{k}}=2 \pi\left(\frac{M}{2 k}\right)^{1 / 2}
\end{aligned}\)
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