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 will use the equations of motion for a particle executing simple harmonic motion (SHM). The position \( x \) and velocity \( v \) of the particle at time \( t = 0 \) are given, along with the angular frequency \( \omega \).

### Step 1: Write down the equations for SHM
The position of a particle in SHM can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

The velocity of the particle is given by the derivative of the position:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]

### Step 2: Substitute the known values at \( t = 0 \)
At \( t = 0 \):
- The position \( x(0) = 3 \, \text{cm} \)
- The velocity \( v(0) = 8 \, \text{cm/s} \)
- The angular frequency \( \omega = 2 \, \text{rad/s} \)

Substituting \( t = 0 \) into the position equation:
\[ x(0) = A \sin(\phi) = 3 \quad \text{(1)} \]

Substituting \( t = 0 \) into the velocity equation:
\[ v(0) = A \omega \cos(\phi) = 8 \quad \text{(2)} \]

### Step 3: Express \( A \) in terms of \( \phi \)
From equation (1):
\[ A \sin(\phi) = 3 \implies A = \frac{3}{\sin(\phi)} \quad \text{(3)} \]

From equation (2):
\[ A \cdot 2 \cos(\phi) = 8 \implies A \cos(\phi) = 4 \quad \text{(4)} \]

### Step 4: Substitute \( A \) from equation (3) into equation (4)
Substituting \( A \) from equation (3) into equation (4):
\[ \frac{3}{\sin(\phi)} \cos(\phi) = 4 \]
Multiplying both sides by \( \sin(\phi) \):
\[ 3 \cos(\phi) = 4 \sin(\phi) \]

### Step 5: Rearrange and use the Pythagorean identity
Rearranging gives:
\[ \frac{\cos(\phi)}{\sin(\phi)} = \frac{4}{3} \implies \tan(\phi) = \frac{3}{4} \]

Using the identity \( \sin^2(\phi) + \cos^2(\phi) = 1 \):
Let \( \sin(\phi) = 3k \) and \( \cos(\phi) = 4k \), where \( k \) is a constant.
Then:
\[ (3k)^2 + (4k)^2 = 1 \implies 9k^2 + 16k^2 = 1 \implies 25k^2 = 1 \implies k^2 = \frac{1}{25} \implies k = \frac{1}{5} \]

Thus:
\[ \sin(\phi) = 3k = \frac{3}{5}, \quad \cos(\phi) = 4k = \frac{4}{5} \]

### Step 6: Substitute back to find \( A \)
Now substitute \( \sin(\phi) \) back into equation (3):
\[ A = \frac{3}{\sin(\phi)} = \frac{3}{\frac{3}{5}} = 5 \, \text{cm} \]

### Final Answer
The amplitude of oscillation is:
\[ \boxed{5 \, \text{cm}} \]