Given that \( S_1 \) and \( S_2 \) are identical springs, they each have the same spring constant \( k \). When both springs are attached, they are in parallel, so the effective spring constant \( k_{\text{eq}} \) is:

\[
k_{\text{eq}} = k + k = 2k
\]

The frequency \( f \) of oscillation for mass \( m \) with effective spring constant \( k_{\text{eq}} = 2k \) is:

\[
f = \frac{1}{2\pi} \sqrt{\frac{2k}{m}}
\]

If One Spring is Removed

If one spring is removed, only one spring with constant \( k \) is left. The new frequency \( f' \) becomes:

\[
f' = \frac{1}{2\pi} \sqrt{\frac{k}{m}}
\]

Ratio of New Frequency to Original Frequency

\[
\frac{f'}{f} = \frac{\frac{1}{2\pi} \sqrt{\frac{k}{m}}}{\frac{1}{2\pi} \sqrt{\frac{2k}{m}}} = \frac{\sqrt{\frac{k}{m}}}{\sqrt{\frac{2k}{m}}} = \frac{1}{\sqrt{2}}
\]

Thus:

\[
f' = \frac{f}{\sqrt{2}}
\]

Answer: \( f' = \frac{f}{\sqrt{2}} \)

1. Left Side:
- Two springs with spring constants \( 2k \) and \( 2k \) are in series.
- The combined spring constant \( k_{\text{left}} \) for these two springs in series is:
\[
\frac{1}{k_{\text{left}}} = \frac{1}{2k} + \frac{1}{2k} = \frac{1}{k}
\]
\[
k_{\text{left}} = k
\]

2. Right Side:
- Two springs with spring constants \( k \) and \( 2k \) are in parallel.
- The combined spring constant \( k_{\text{right}} \) for these two springs in parallel is:
\[
k_{\text{right}} = k + 2k = 3k
\]

3. Combine Left and Right Sides:
- Since \( k_{\text{left}} \) and \( k_{\text{right}} \) are in parallel, the equivalent spring constant \( k_{\text{eq}} \) is:
\[
k_{\text{eq}} = k_{\text{left}} + k_{\text{right}} = k + 3k = 4k
\]

Step 2: Calculate the Frequency of Oscillation

The frequency \( f \) is given by:

\[
f = \frac{1}{2\pi} \sqrt{\frac{k_{\text{eq}}}{M}}
\]

Substitute \( k_{\text{eq}} = 4k \):

\[
f = \frac{1}{2\pi} \sqrt{\frac{4k}{M}}
\]

Final Answer

\[
f = \frac{1}{2\pi} \sqrt{\frac{4k}{M}}
\]

Given:

- Force \( F = 6.4 \, \text{N} \)
- Extension \( x = 0.1 \, \text{m} \)
- Time period \( T = \frac{\pi}{4} \, \text{s} \)

1. Find the spring constant \( k \):

\[
k = \frac{F}{x} = \frac{6.4}{0.1} = 64 \, \text{N/m}
\]

2. Use the formula for the time period of a mass-spring system:

\[
T = 2\pi \sqrt{\frac{m}{k}}
\]

Substitute \( T = \frac{\pi}{4} \) and \( k = 64 \):

\[
\frac{\pi}{4} = 2\pi \sqrt{\frac{m}{64}}
\]

3. Solve for \( m \):

\[
\frac{1}{8} = \sqrt{\frac{m}{64}}
\]

\[
\frac{1}{64} = \frac{m}{64}
\]

\[
m = 1 \, \text{kg}
\]

Answer: \( m = 1 \, \text{kg} \)

In Hollow Sphere (T) , Solid Spheres ( T1 and T2) Center of mass lie at center. Where as in Half filled sphere center of mass lie below center(T3).

 As length of pendulum will increase in T3. Time period is highest for it.

\[ T = 2\pi\sqrt{\frac{l}{g}} \]

\[ T_{1}= 2\pi\sqrt{\frac{1}{g}}= 2 sec \]

\[ T_{2}= 2\pi\sqrt{\frac{16}{g}}= 8 sec \]

\[ T = 2\pi\sqrt{\frac{l}{\sqrt{a^{2}+g^{2}}}} \]

\[ T = 2\pi\sqrt{\frac{0.5}{\sqrt{\left( \sqrt{3}g \right)^{2}+g^{2}}}} \]

\[ T = 2\pi\sqrt{\frac{0.5}{2g}} \]

\[ T = 2\pi\frac{1}{2\sqrt{g}} = 2\pi\frac{1}{2\pi}= 1 sec \]

\[ K.E= \frac{1}{2}K\left(A^{2}-x^{2} \right) \]

\[ P.E= \frac{1}{2}Kx^{2} \]

\[ \frac{1}{2}K\left(A^{2}-x^{2} \right)= \frac{1}{2}Kx^{2} \]

\[ x= A/\sqrt{2} \]

\[ K.E= \frac{1}{2}K\left(A^{2}-x^{2} \right) \]

\[ K.E= \frac{1}{2}K\left(A^{2}-\left( \frac{A}{2}\right)^{2} \right)= \frac{3}{8} K A^{2} \]

K.E/T.E =3/4

Spring Constant K = m ω² ⇒ 10 = 0.1 ω² ⇒ ω²= 100 ⇒ ω = 10 rad/sec

Speed is maximum at mean position 

Vmax= Aω

6= A × 10

A = 0.6 m

\[ U = \frac{1}{2}kx^{2} \]