A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6s. Ast t=0 it is at position x=5 cm going towasrds positive x-direction. Write the eqwuation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t=4s.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the parameters of the simple harmonic motion (SHM)
- Amplitude (A) = 10 cm
- Time period (T) = 6 s
- Initial position (x₀) at t = 0 = 5 cm
- Direction of motion = positive x-direction

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of T:
\[ \omega = \frac{2\pi}{6} = \frac{\pi}{3} \text{ rad/s} \]

### Step 3: Write the general equation for displacement in SHM
The general equation for displacement x in SHM is:
\[ x(t) = A \sin(\omega t + \phi) \]
where A is the amplitude, ω is the angular frequency, and φ is the phase constant.

### Step 4: Determine the phase constant (φ)
At t = 0, the displacement x(0) = 5 cm. We can use this to find φ:
\[ x(0) = A \sin(\phi) = 5 \]
Substituting the value of A:
\[ 10 \sin(\phi) = 5 \implies \sin(\phi) = \frac{5}{10} = \frac{1}{2} \]
The value of φ that satisfies this equation and is consistent with the initial condition (moving towards positive x-direction) is:
\[ \phi = \frac{\pi}{6} \text{ rad} \]

### Step 5: Write the complete equation for displacement
Now substituting the values of A, ω, and φ into the displacement equation:
\[ x(t) = 10 \sin\left(\frac{\pi}{3} t + \frac{\pi}{6}\right) \]

### Step 6: Find the magnitude of the acceleration at t = 4 s
The acceleration in SHM is given by:
\[ a(t) = -A \omega^2 \sin(\omega t + \phi) \]
Substituting the known values:
\[ a(4) = -10 \left(\frac{\pi}{3}\right)^2 \sin\left(\frac{\pi}{3} \cdot 4 + \frac{\pi}{6}\right) \]
Calculating ω²:
\[ \omega^2 = \left(\frac{\pi}{3}\right)^2 = \frac{\pi^2}{9} \]
Now substituting this into the acceleration equation:
\[ a(4) = -10 \cdot \frac{\pi^2}{9} \sin\left(\frac{4\pi}{3} + \frac{\pi}{6}\right) \]
Calculating the angle:
\[ \frac{4\pi}{3} + \frac{\pi}{6} = \frac{8\pi}{6} + \frac{\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2} \]
Now, we know that:
\[ \sin\left(\frac{3\pi}{2}\right) = -1 \]
Thus,
\[ a(4) = -10 \cdot \frac{\pi^2}{9} \cdot (-1) = \frac{10\pi^2}{9} \text{ cm/s}^2 \]
The magnitude of the acceleration is:
\[ |a(4)| = \frac{10\pi^2}{9} \text{ cm/s}^2 \approx 10.95 \text{ cm/s}^2 \]

### Final Answer
The equation for the displacement x at time t is:
\[ x(t) = 10 \sin\left(\frac{\pi}{3} t + \frac{\pi}{6}\right) \text{ cm} \]
The magnitude of the acceleration of the particle at t = 4 s is approximately:
\[ |a(4)| \approx 10.95 \text{ cm/s}^2 \]