The position and velocity of a particle executing simple harmonic motion at t = 0 are given by 3 cm and `8 cm s^(-1)` respectively . If the angular frequency of the particle is `2 "rad s" ^(-1)` , then the amplitude of oscillation (in cm) is

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to find the amplitude of a particle executing simple harmonic motion (SHM) given its position, velocity, and angular frequency at time \( t = 0 \).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Position \( x = 3 \, \text{cm} \)
- Velocity \( v = 8 \, \text{cm/s} \)
- Angular frequency \( \omega = 2 \, \text{rad/s} \)

2. **Write the Equation for Position in SHM:**
The position of a particle in SHM can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
At \( t = 0 \):
\[ x(0) = A \sin(\phi) \]
Therefore, we have:
\[ 3 = A \sin(\phi) \quad \text{(1)} \]

3. **Write the Equation for Velocity in SHM:**
The velocity of a particle in SHM is given by:
\[ v(t) = A \omega \cos(\omega t + \phi) \]
At \( t = 0 \):
\[ v(0) = A \omega \cos(\phi) \]
Therefore, we have:
\[ 8 = A \cdot 2 \cos(\phi) \quad \text{(2)} \]

4. **Rearranging Equation (2):**
From equation (2):
\[ A \cos(\phi) = \frac{8}{2} = 4 \quad \text{(3)} \]

5. **Square and Add Equations (1) and (3):**
Squaring both equations (1) and (3):
\[ (A \sin(\phi))^2 + (A \cos(\phi))^2 = 3^2 + 4^2 \]
This simplifies to:
\[ A^2 (\sin^2(\phi) + \cos^2(\phi)) = 9 + 16 \]
Using the identity \( \sin^2(\phi) + \cos^2(\phi) = 1 \):
\[ A^2 = 25 \]

6. **Solve for Amplitude \( A \):**
Taking the square root:
\[ A = \sqrt{25} = 5 \, \text{cm} \]

### Final Answer:
The amplitude of oscillation is \( 5 \, \text{cm} \).