A particle executes simple harmonic motion with an amplitude of `10 cm` and time period `6 s`. At `t = 0` it is position `x = 5 cm` from mean postion and going towards positive `x-`direaction. Write the equation for the displacement `x` at time `L`. Find the magnitude of the acceleration of the particle at `t = 4 s`.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given parameters
- Amplitude (A) = 10 cm
- Time period (T) = 6 s
- Initial position (x) at t = 0 = 5 cm
- Direction of motion at t = 0 = 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 displacement \( x(t) \) in simple harmonic motion can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

### Step 4: Determine the phase constant (φ)
At \( t = 0 \), \( x(0) = 5 \) cm. Therefore:
\[ 5 = 10 \sin(\phi) \]
This simplifies to:
\[ \sin(\phi) = \frac{5}{10} = \frac{1}{2} \]
The angle \( \phi \) that satisfies this equation is:
\[ \phi = \frac{\pi}{6} \text{ (since the particle is moving in the positive direction)} \]

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

### Step 6: Find the displacement at \( t = 4 \) s
Now, substituting \( t = 4 \) s into the equation:
\[ x(4) = 10 \sin\left(\frac{\pi}{3} \cdot 4 + \frac{\pi}{6}\right) \]
Calculating the argument of the sine function:
\[ \frac{\pi}{3} \cdot 4 = \frac{4\pi}{3} \]
Now, adding \( \frac{\pi}{6} \):
\[ \frac{4\pi}{3} + \frac{\pi}{6} = \frac{8\pi}{6} + \frac{\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2} \]
Thus:
\[ x(4) = 10 \sin\left(\frac{3\pi}{2}\right) = 10 \cdot (-1) = -10 \text{ cm} \]

### Step 7: Find the acceleration at \( t = 4 \) s
The acceleration \( a \) in SHM is given by:
\[ a = -\omega^2 x \]
Substituting \( \omega = \frac{\pi}{3} \) and \( x = -10 \) cm:
\[ a = -\left(\frac{\pi}{3}\right)^2 \cdot (-10) = \frac{\pi^2}{9} \cdot 10 = \frac{10\pi^2}{9} \text{ cm/s}^2 \]
Calculating the numerical value:
\[ \frac{10\pi^2}{9} \approx \frac{10 \cdot 9.87}{9} \approx 10.96 \text{ cm/s}^2 \approx 11 \text{ cm/s}^2 \]

### Final Answer
The magnitude of the acceleration of the particle at \( t = 4 \) s is approximately \( 11 \text{ cm/s}^2 \).

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