A particle moving with shm has a period 0.001 s and amplitude 0.5 cm. Find the acceleration, when it is 0.2 cm apart from its mean position.

Explanation

To solve the problem step by step, we will follow the procedure outlined in the video transcript.

### Step 1: Identify the Given Information
- Time period (T) = 0.001 s
- Amplitude (A) = 0.5 cm = 0.005 m (conversion to meters)
- Displacement from mean position (x) = 0.2 cm = 0.002 m (conversion to meters)

### Step 2: Calculate Angular Frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of T:
\[ \omega = \frac{2\pi}{0.001} = 2000\pi \, \text{rad/s} \]

### Step 3: Calculate the Magnitude of Acceleration (a)
The acceleration in simple harmonic motion (SHM) is given by:
\[ a = -\omega^2 x \]
Since we are interested in the magnitude, we can ignore the negative sign:
\[ a = \omega^2 x \]
Substituting the values of ω and x:
\[ a = (2000\pi)^2 \times 0.002 \]

### Step 4: Calculate ω²
Calculating ω²:
\[ \omega^2 = (2000\pi)^2 = 4000000\pi^2 \]

### Step 5: Substitute ω² into the Acceleration Formula
Now substituting ω² back into the acceleration formula:
\[ a = 4000000\pi^2 \times 0.002 \]
\[ a = 8000\pi^2 \]

### Step 6: Calculate the Numerical Value
Using the approximation \(\pi \approx \frac{22}{7}\):
\[ \pi^2 \approx \left(\frac{22}{7}\right)^2 = \frac{484}{49} \]
Now substituting this value:
\[ a = 8000 \times \frac{484}{49} \]
Calculating this gives:
\[ a \approx 8000 \times 9.8776 \approx 79020.8 \, \text{m/s}^2 \]

### Final Result
The acceleration when the particle is 0.2 cm apart from its mean position is approximately:
\[ a \approx 7.902 \times 10^4 \, \text{m/s}^2 \]