Time period of a particle executing SHM is 16s.At time `t = 2s`, it crosses the mean position . Its amplitude of motion is `( 32sqrt(2))/(pi) m`. Its velocity at `t = 4s` is

Options

A. `1 ms^(-1)`
B. ` 2ms^(-1)`
C. ` 4ms^(-1)`
D. `8 ms^(-1)`

Answer

To find the velocity of a particle executing simple harmonic motion (SHM) at a specific time, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Time period \( T = 16 \, \text{s} \)
- Amplitude \( A = \frac{32\sqrt{2}}{\pi} \, \text{m} \)
- Time \( t = 4 \, \text{s} \)

2. **Calculate Angular Frequency \( \omega \):**
The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{16} = \frac{\pi}{8} \, \text{rad/s} \]

3. **Determine the Phase Constant \( \phi \):**
At \( t = 2 \, \text{s} \), the particle crosses the mean position, which means the displacement \( x = 0 \). The equation of motion for SHM is:
\[ x(t) = A \sin(\omega t + \phi) \]
Setting \( x(2) = 0 \):
\[ 0 = \frac{32\sqrt{2}}{\pi} \sin\left(\frac{\pi}{8} \cdot 2 + \phi\right) \]
This implies:
\[ \sin\left(\frac{\pi}{4} + \phi\right) = 0 \]
Therefore, the argument must be a multiple of \( \pi \):
\[ \frac{\pi}{4} + \phi = n\pi \]
Let’s take \( n = 1 \):
\[ \frac{\pi}{4} + \phi = \pi \implies \phi = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \]

4. **Write the Displacement Equation:**
Now we can write the displacement equation:
\[ x(t) = \frac{32\sqrt{2}}{\pi} \sin\left(\frac{\pi}{8} t + \frac{3\pi}{4}\right) \]

5. **Find the Velocity Equation:**
The velocity \( v(t) \) is the derivative of displacement with respect to time:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]
Substituting the values:
\[ v(t) = \frac{32\sqrt{2}}{\pi} \cdot \frac{\pi}{8} \cos\left(\frac{\pi}{8} t + \frac{3\pi}{4}\right) \]
Simplifying:
\[ v(t) = 4\sqrt{2} \cos\left(\frac{\pi}{8} t + \frac{3\pi}{4}\right) \]

6. **Calculate Velocity at \( t = 4 \, \text{s} \):**
Now substitute \( t = 4 \):
\[ v(4) = 4\sqrt{2} \cos\left(\frac{\pi}{8} \cdot 4 + \frac{3\pi}{4}\right) = 4\sqrt{2} \cos\left(\frac{\pi}{2} + \frac{3\pi}{4}\right) \]
Simplifying the argument:
\[ \frac{\pi}{2} + \frac{3\pi}{4} = \frac{5\pi}{4} \]
Thus, we have:
\[ v(4) = 4\sqrt{2} \cos\left(\frac{5\pi}{4}\right) \]
Since \( \cos\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}} \):
\[ v(4) = 4\sqrt{2} \cdot \left(-\frac{1}{\sqrt{2}}\right) = -4 \, \text{m/s} \]

### Final Answer:
The velocity of the particle at \( t = 4 \, \text{s} \) is \( -4 \, \text{m/s} \).