This time period of a particle undergoing SHM is 16 s. It starts motion from the mean position. After 2 s, its velocity is 0.4 `ms^(-1)`. The amplitude is

Options

A. 1.44 m
B. 0.72 m
C. 2.88 m
D. 0.36 m

Explanation

To solve the problem step by step, we will use the concepts of Simple Harmonic Motion (SHM).

### Step 1: Identify the given parameters
- Time period (T) = 16 s
- Time elapsed (t) = 2 s
- Velocity (v) = 0.4 m/s

### 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:
\[ \omega = \frac{2\pi}{16} = \frac{\pi}{8} \text{ rad/s} \]

### Step 3: Write the expression for velocity in SHM
The velocity (v) of a particle in SHM is given by:
\[ v = A \omega \cos(\omega t) \]
where A is the amplitude.

### Step 4: Substitute the known values into the velocity equation
Substituting the values we have:
\[ 0.4 = A \left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{8} \cdot 2\right) \]
Calculating \(\cos\left(\frac{\pi}{4}\right)\):
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
So, we can rewrite the equation as:
\[ 0.4 = A \left(\frac{\pi}{8}\right) \left(\frac{1}{\sqrt{2}}\right) \]

### Step 5: Solve for the amplitude (A)
Rearranging the equation to solve for A:
\[ A = \frac{0.4 \cdot 8 \cdot \sqrt{2}}{\pi} \]

### Step 6: Calculate the value of A
Now substituting the values:
\[ A = \frac{0.4 \cdot 8 \cdot 1.414}{3.14} \approx \frac{4.52}{3.14} \approx 1.44 \text{ m} \]

### Final Answer
The amplitude \( A \) is approximately \( 1.44 \, \text{m} \).

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