Question

The time period of simple harmonic motion of mass \(M\) in the given figure is \(\pi \sqrt{\frac{\alpha M}{5K}}\), where the value of \(\alpha\) is ______.

Answer 1

Correct Answer: 12

To find \(\alpha\) in the time period formula \(\pi \sqrt{\frac{\alpha M}{5K}}\), we analyze the spring system. The mass \(M\) is supported by three springs, each with spring constant \(k\).

Step 1: Determine Equivalent Spring Constant

The two parallel vertical springs have an equivalent spring constant of \(k_{eq1} = k + k = 2k\).

Step 2: Combine with Series Spring

This combined spring constant is in series with the third spring, resulting in a total equivalent spring constant \(k_{eq}\):
\( \frac{1}{k_{eq}} = \frac{1}{2k} + \frac{1}{k} = \frac{1+2}{2k} = \frac{3}{2k} \)
Therefore, \(k_{eq} = \frac{2k}{3}\).

Step 3: Use the Time Period Formula

The standard time period formula for a mass-spring system is \(T = 2\pi\sqrt{\frac{M}{k_{eq}}}\). Substituting \(k_{eq}\) yields \(T = 2\pi\sqrt{\frac{3M}{2k}}\).
Given \(T = \pi \sqrt{\frac{\alpha M}{5K}}\), we equate the two expressions to solve for \(\alpha\):
\(2\pi\sqrt{\frac{3M}{2k}} = \pi \sqrt{\frac{\alpha M}{5K}}\)
Simplifying, \(2\sqrt{\frac{3M}{2k}} = \sqrt{\frac{\alpha M}{5K}}\).
Squaring both sides gives \(4\frac{3M}{2k} = \frac{\alpha M}{5K}\).
Cross-multiplying results in \(12\cdot 5k = 2k\alpha\).
This leads to \(\alpha = 12\).

Conclusion

The value of \(\alpha\) is 12.