\[ x= A sin\left( \omega t \right) \]

\[ \frac{A}{2}= A sin\left( \omega t \right) \]

\[ \frac{1}{2}= sin\left( \omega t \right) \]

\[ \frac{\Pi}{6}= \omega t =\frac{2\Pi}{T}t = \frac{2\Pi}{3}t \]

\[ t= \frac{1}{4}s \]

Equation of displacement x = A sin (ωt) so, acceleration a= -Aω sin(ωt) force will have same nature so,

\[ E_{1}= \frac{1}{2}Kx^{2} \]

\[ E_{2}= \frac{1}{2}Ky^{2} \]

\[  E= \frac{1}{2}K(x+y)^{2}= \frac{1}{2}Kx^{2} + \frac{1}{2}Ky^{2} + Kxy \]

1. Initial Time Period:
\[
T = 2\pi \sqrt{\frac{M}{k}}
\]

2. New Time Period:
\[
\frac{5T}{3} = 2\pi \sqrt{\frac{M + m}{k}}
\]

3. Divide the New Time Period by the Initial Time Period:

\[
\frac{\frac{5T}{3}}{T} = \sqrt{\frac{M + m}{M}}
\]

\[
\frac{5}{3} = \sqrt{\frac{M + m}{M}}
\]

4. Square Both Sides:

\[
\frac{25}{9} = \frac{M + m}{M}
\]

5. Solve for \( \frac{m}{M} \):

\[
\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}
\]

Answer:

\[
\frac{m}{M} = \frac{16}{9}
\]

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

\[ K.E= \frac{1}{2}\times 50\left( 10^{2}-5^{2} \right)/10^{4}= 0.19 J \]

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

K.E= 3E/4

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