A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is `pi m//s`. Amplitude of oscillation is
`(cos 45^(@) = 1/(sqrt(2)))`

Options

A. `2sqrt(2)`
B. `4sqrt(2)`
C. `6sqrt(2)`
D. `8sqrt(2)`

Answer

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

### Step 1: Understand the given information
- The particle starts at the equilibrium position, which means its initial displacement \( x(0) = 0 \).
- The time period \( T = 16 \) seconds.
- After \( t = 2 \) seconds, the velocity \( v = \pi \) m/s.

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

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

### Step 4: Substitute the known values into the velocity equation
We know \( v = \pi \) m/s, \( \omega = \frac{\pi}{8} \) rad/s, and \( t = 2 \) seconds. Substituting these values:
\[ \pi = A \left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{8} \cdot 2\right) \]
This simplifies to:
\[ \pi = A \left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{4}\right) \]

### Step 5: Evaluate \( \cos\left(\frac{\pi}{4}\right) \)
Using the given information that \( \cos 45^\circ = \frac{1}{\sqrt{2}} \):
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Step 6: Substitute \( \cos\left(\frac{\pi}{4}\right) \) back into the equation
Now substituting this value back into our equation:
\[ \pi = A \left(\frac{\pi}{8}\right) \left(\frac{1}{\sqrt{2}}\right) \]

### Step 7: Solve for amplitude \( A \)
Now we can simplify and solve for \( A \):
\[ \pi = \frac{A \pi}{8\sqrt{2}} \]
Cancelling \( \pi \) from both sides (assuming \( \pi \neq 0 \)):
\[ 1 = \frac{A}{8\sqrt{2}} \]
Multiplying both sides by \( 8\sqrt{2} \):
\[ A = 8\sqrt{2} \text{ meters} \]

### Final Answer
The amplitude of the oscillation is \( A = 8\sqrt{2} \) meters.

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