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

Step 1: Determine Equivalent Spring Constant

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

Step 2: Combine with Series Spring

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

Step 3: Use the Time Period Formula

The standard formula for the time period \(T\) of a mass-spring system is:
\(T = 2\pi\sqrt{\frac{M}{k_{eq}}}\) 
Substitute for \(k_{eq}\):
\(T = 2\pi\sqrt{\frac{3M}{2k}}\).
Given in the problem: \(T = \pi \sqrt{\frac{\alpha M}{5K}}\).
Equate and solve for \(\alpha\):
\(2\pi\sqrt{\frac{3M}{2k}} = \pi \sqrt{\frac{\alpha M}{5K}}\)
Simplifying, \(2\sqrt{\frac{3M}{2k}} = \sqrt{\frac{\alpha M}{5K}}\)
Square both sides:
\(4\frac{3M}{2k} = \frac{\alpha M}{5K}\)
Cross-multiply:
\(12\cdot 5k = 2k\alpha\)
\(\alpha = 12\)
 

Conclusion

Hence, the value of \(\alpha\) is confirmed to be within the range \(12,12\), and thus, \(\alpha = 12\).

Answer 2

Given system parameters:

The equivalent spring constant for the system is calculated as:

\[ k_{\text{eq}} = \frac{2k \cdot k}{2k + k} + k = \frac{5k}{3} \]

The angular frequency of oscillation (\( \omega \)) is given by:

\[ \omega = \sqrt{\frac{k_{\text{eq}}}{m}} \]

Substituting the value of \( k_{\text{eq}} \):

\[ \omega = \sqrt{\frac{\frac{5k}{3}}{m}} = \sqrt{\frac{5k}{3m}} \]

The period of oscillation (\( \tau \)) is:

\[ \tau = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{\frac{5k}{3}}} = 2\pi \sqrt{\frac{3m}{5k}} \]

Simplifying:

\[ \tau = \pi \sqrt{\frac{12m}{5k}} \]

Thus, comparing with the given expression:

\[ T = \pi \sqrt{\frac{\alpha M}{5K}} \]

we find:

\[ \alpha = 12 \]