A particle executes SHM, its time period is 16 s. If it passes through the centre of oscillation, then its velocity is 2 `ms^(-1)` at times 2 s. The amplitude will be

Options

A. 7.2 m
B. 4 cm
C. 6 cm
D. 0.72 m

Explanation

To find the amplitude of a particle executing simple harmonic motion (SHM) given its time period and velocity at a certain time, we can follow these steps:

### Step 1: Identify the given values
- Time period (T) = 16 s
- Velocity (v) at time (t) = 2 m/s
- Time (t) = 2 s

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

### Step 3: Use the velocity formula in SHM
The velocity (v) of a particle in SHM can be expressed as:
\[ v = A \omega \cos(\omega t) \]
Where:
- A is the amplitude
- ω is the angular frequency
- t is the time

### Step 4: Substitute known values into the velocity equation
At time t = 2 s, we have:
\[ 2 = A \left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{8} \cdot 2\right) \]
Calculating ωt:
\[ \omega t = \frac{\pi}{8} \cdot 2 = \frac{\pi}{4} \]
Now, we can find cos(π/4):
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Step 5: Substitute cos(π/4) back into the equation
Now substituting this back into the velocity equation:
\[ 2 = A \left(\frac{\pi}{8}\right) \left(\frac{1}{\sqrt{2}}\right) \]

### Step 6: Solve for amplitude (A)
Rearranging the equation to solve for A:
\[ A = \frac{2 \cdot 8 \cdot \sqrt{2}}{\pi} = \frac{16\sqrt{2}}{\pi} \]

### Step 7: Final calculation
To get a numerical approximation, we can calculate:
\[ A \approx \frac{16 \cdot 1.414}{3.14} \approx \frac{22.624}{3.14} \approx 7.2 \text{ m} \]

### Conclusion
The amplitude of the particle executing SHM is approximately \( \frac{16\sqrt{2}}{\pi} \) meters or about 7.2 m.

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