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

Explanation

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 s = 5 cm
- Direction of motion: towards positive x-direction

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

### Step 3: Write the general equation for displacement in SHM
The standard equation for displacement (x) in simple harmonic motion is:
\[ x(t) = A \sin(\omega t + \phi) \]
Where:
- A = amplitude
- ω = angular frequency
- φ = phase constant

### Step 4: Determine the phase constant (φ)
At t = 0, we know that x = 5 cm. We can substitute this into the equation to find φ:
\[ 5 = 10 \sin\left(\frac{\pi}{3} \cdot 0 + \phi\right) \]
This simplifies to:
\[ 5 = 10 \sin(\phi) \]
Dividing both sides by 10:
\[ \sin(\phi) = 0.5 \]
The angle whose sine is 0.5 is:
\[ \phi = \frac{\pi}{6} \text{ (30 degrees)} \]

### Step 5: Write the final equation for displacement
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 displacement at t = 4 s
Substituting t = 4 s into the displacement equation:
\[ x(4) = 10 \sin\left(\frac{\pi}{3} \cdot 4 + \frac{\pi}{6}\right) \]
Calculating the argument:
\[ \frac{\pi}{3} \cdot 4 = \frac{4\pi}{3} \]
Adding the phase constant:
\[ \frac{4\pi}{3} + \frac{\pi}{6} = \frac{8\pi}{6} + \frac{\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2} \]
Now substituting back:
\[ x(4) = 10 \sin\left(\frac{3\pi}{2}\right) = 10 \cdot (-1) = -10 \text{ cm} \]

### Step 7: Calculate the acceleration at t = 4 s
The formula for acceleration in SHM is:
\[ a = -\omega^2 x \]
Substituting ω and x:
\[ a = -\left(\frac{\pi}{3}\right)^2 (-10) \]
Calculating ω²:
\[ \omega^2 = \frac{\pi^2}{9} \]
Thus,
\[ a = \frac{\pi^2}{9} \cdot 10 = \frac{10\pi^2}{9} \text{ cm/s}^2 \]
Calculating the numerical value:
\[ \approx \frac{10 \cdot 9.87}{9} \approx 10.98 \text{ cm/s}^2 \approx 11.11 \text{ cm/s}^2 \]

### Final Answer
The equation for displacement is:
\[ x(t) = 10 \sin\left(\frac{\pi}{3} t + \frac{\pi}{6}\right) \text{ cm} \]
The magnitude of the acceleration at t = 4 s is approximately:
\[ 11.11 \text{ cm/s}^2 \]