To find the time required for a particle in Simple Harmonic Motion (SHM) to travel from \( x = A \) to \( x = \frac{A}{2} \), we can use the following steps:

1. Angular Frequency (\( \omega \)):
\[
\omega = \frac{2\pi}{T}
\]

2. Displacement in SHM:
The position \( x(t) \) in SHM can be described by:
\[
x(t) = A \cos(\omega t)
\]

3. Finding Time for Positions:
- For \( x = A \):
\[
A \cos(\omega t_1) = A;  \cos(\omega t_1) = 1 ; t_1 = 0
\]

- For \( x = \frac{A}{2} \):
\[
\frac{A}{2} = A \cos(\omega t_2 ; \cos(\omega t_2) = \frac{1}{2} ; \omega t_2 = \frac{\pi}{3}
\]
\[
t_2 = \frac{\pi}{3\omega} = \frac{\pi T}{6}
\]

4. Time Interval:
The time taken to travel from \( x = A \) to \( x = \frac{A}{2} \) is:
\[
\Delta t = t_2 - t_1 = \frac{\pi T}{6} - 0 = \frac{T}{6}
\]

Thus, the time required to travel from \( x = A \) to \( x = \frac{A}{2} \) is:
\[{\frac{T}{6}}
\]

In Simple Harmonic Motion (SHM), the relationship between displacement, velocity, and amplitude can be described using the following equations.

1. The maximum speed \( V_0 \) is given by:
\[
V_0 = \omega A
\]
where \( \omega \) is the angular frequency.

2. The speed \( v \) at a displacement \( x \) in SHM is given by the formula:
\[
v = \sqrt{V_0^2 - \left(\frac{\omega x}{\omega A}\right)^2}
\]
Simplifying this using \( \omega = \frac{V_0}{A} \):
\[
v = \sqrt{V_0^2 - \left(\frac{V_0}{A} \cdot x\right)^2}
\]
Substituting \( x = \frac{A}{2} \):
\[
v = \sqrt{V_0^2 - \left(\frac{V_0}{A} \cdot \frac{A}{2}\right)^2}
\]
\[
v = \sqrt{V_0^2 - \left(\frac{V_0}{2}\right)^2}
\]
\[
v = \sqrt{V_0^2 - \frac{V_0^2}{4}} = \sqrt{\frac{3V_0^2}{4}} = \frac{\sqrt{3}}{2} V_0
\]

Thus, the speed of the particle at displacement \( \frac{A}{2} \) is:
\[{\frac{\sqrt{3}}{2} V_0}
\]

For the SHM given by:

\[
y = A \cos(\omega t + \frac{\pi}{4})
\]

Solution:

1. Velocity: The velocity \( v \) is the derivative of \( y \) with respect to \( t \):
\[
v = \frac{dy}{dt} = -A \omega \sin(\omega t + \frac{\pi}{4})
\]

2. Maximum Speed: The speed will be maximum when \( \sin(\omega t + \frac{\pi}{4}) = \pm 1 \).

Therefore,
\[
\omega t + \frac{\pi}{4} = \frac{\pi}{2}
\]

3. Solving for \( t \):
\[
\omega t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}
\]

\[
t = \frac{\pi}{4\omega}
\]

Answer:
The speed will be maximum at \( t = \frac{\pi}{4\omega} \).

In simple harmonic motion (SHM), the velocity \( v \) and displacement \( x \) of a particle are related by:

\[
v = \pm \sqrt{\omega^2 A^2 - \omega^2 x^2}
\]

where:
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude.

 Solution:

1. Rearranging this equation, we get:
\[
\frac{v^2}{\omega^2 A^2} + \frac{x^2}{A^2} = 1
\]

2. This is the equation of an ellipse in the \( v \)-\( x \) plane.

Answer:
The graph of velocity as a function of displacement for a particle in SHM is an ellipse.

Given the SHM equation:

\[
x = 4 \cos(88t + \frac{\pi}{4})
\]

Solution:

1. Frequency: The general form of SHM is \( x = A \cos(\omega t + \phi) \), where \( \omega = 2 \pi f \) (angular frequency).

Here, \( \omega = 88 \).

\[
f = \frac{\omega}{2 \pi} = \frac{88}{2 \pi} = 14 \, \text{Hz}
\]

2. Initial Displacement: The initial displacement \( x_0 \) is the value of \( x \) at \( t = 0 \).

Substitute \( t = 0 \) into the equation:
\[
x_0 = 4 \cos\left(\frac{\pi}{4}\right) = 4 \cdot \frac{\sqrt{2}}{2} = 2 \sqrt{2} \, \text{m}
\]

Answer:
The frequency is  14 Hz, and the initial displacement is \( 2 \sqrt{2} \, \text{m} \).

Equation of y-t graph is : y  = A sin (ωt) differentiating v = A ω cos (ωt) and a= - Aω² sin (ωt) . So the graph is a function of - sin(ωt).

For a particle in simple harmonic motion (SHM), the force and acceleration are always directed towards the equilibrium position (center) and are opposite to the displacement.

Solution:

1. Displacement: Since the particle is 4 cm from \( B \) and moving towards \( A \), it is located closer to \( B \) on the negative side of the equilibrium point. This means the displacement is in the negative direction.

2. Velocity: Since the particle is moving towards \( A \) (the negative direction), its velocity is negative.

3. Acceleration and Force: In SHM, acceleration and force are always directed towards the equilibrium point. Since the particle is on the positive side of the equilibrium (closer to \( B \)), the acceleration and force are in the negative direction.

Conclusion:
The signs of velocity, acceleration, and force are all negative (–, –, –).

If equation of motion in shm is x = A sin ωt then eqn. of potential energy = ½k x²= ½kA² sin²ωt

\[ U = \frac{1}{2}K A^{2}\left( \frac{1- sin(2\omega t)}{2} \right) \]

Frequency becomes double so time period becomes half.

Phase difference 2φ= 120 degree

y = 3sin π/2 (50t - Φ)

\[ \frac{\partial y}{\partial t}= 3 \times \frac{\Pi}{2}\times 50 cos(π/2 (50t - Φ))\] 

Maximum speed = 75 π