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 the particle executing simple harmonic motion (SHM) at \( t = 4 \, \text{s} \), we can follow these steps:

### Step 1: Calculate the Angular Frequency
The angular frequency \( \omega \) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time period. Given \( T = 16 \, \text{s} \):
\[ \omega = \frac{2\pi}{16} = \frac{\pi}{8} \, \text{rad/s} \]

### Step 2: Determine the Phase Constant
At \( t = 2 \, \text{s} \), the particle crosses the mean position, which means the displacement \( x = 0 \). The equation for displacement in SHM is:
\[ x(t) = A \sin(\omega t + \phi) \]
Substituting \( x = 0 \) at \( t = 2 \, \text{s} \):
\[ 0 = \frac{32\sqrt{2}}{\pi} \sin\left(\frac{\pi}{8} \cdot 2 + \phi\right) \]
This implies:
\[ \sin\left(\frac{\pi}{4} + \phi\right) = 0 \]
Thus, \( \frac{\pi}{4} + \phi = n\pi \) for \( n \in \mathbb{Z} \). Choosing \( n = 0 \):
\[ \phi = -\frac{\pi}{4} \]

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

### Step 4: Calculate Displacement at \( t = 4 \, \text{s} \)
Substituting \( t = 4 \, \text{s} \):
\[ x(4) = \frac{32\sqrt{2}}{\pi} \sin\left(\frac{\pi}{8} \cdot 4 - \frac{\pi}{4}\right) \]
\[ = \frac{32\sqrt{2}}{\pi} \sin\left(\frac{\pi}{2} - \frac{\pi}{4}\right) = \frac{32\sqrt{2}}{\pi} \sin\left(\frac{\pi}{4}\right) \]
\[ = \frac{32\sqrt{2}}{\pi} \cdot \frac{\sqrt{2}}{2} = \frac{32}{\pi} \, \text{m} \]

### Step 5: Calculate Velocity at \( t = 4 \, \text{s} \)
The velocity \( v(t) \) in SHM is given by:
\[ v(t) = \omega \sqrt{A^2 - x^2} \]
Substituting the values:
\[ v(4) = \frac{\pi}{8} \sqrt{\left(\frac{32\sqrt{2}}{\pi}\right)^2 - \left(\frac{32}{\pi}\right)^2} \]
Calculating \( A^2 \) and \( x^2 \):
\[ A^2 = \left(\frac{32\sqrt{2}}{\pi}\right)^2 = \frac{2048}{\pi^2}, \quad x^2 = \left(\frac{32}{\pi}\right)^2 = \frac{1024}{\pi^2} \]
Thus:
\[ A^2 - x^2 = \frac{2048}{\pi^2} - \frac{1024}{\pi^2} = \frac{1024}{\pi^2} \]
Now substituting back into the velocity equation:
\[ v(4) = \frac{\pi}{8} \sqrt{\frac{1024}{\pi^2}} = \frac{\pi}{8} \cdot \frac{32}{\pi} = \frac{32}{8} = 4 \, \text{m/s} \]

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

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