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 (–, –, –).

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

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

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

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

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

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

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

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

a =- ω²x

comparing 

ω² = π² ⇒ ω =π ⇒ 2πυ= π ⇒ υ = 1/2 Hz