A particle of mass `2g` is initially displaced through `2cm` and then released. The frictional force constant due to air on it is `12 xx 10^(-3)N//m`. The restoring force constant is `50 xx 10^(-3)N//m`. If it is in oscillatory motion, its time period is

Options

A. `pi sec`
B. `pi//2 sec`
C. `2pi sec`
D. `4pi sec`

Explanation

To find the time period of the oscillatory motion of the particle, we will follow these steps:

### Step 1: Convert the mass to kilograms
The mass of the particle is given as 2 grams. To convert this to kilograms:
\[ m = 2 \, \text{g} = 2 \times 10^{-3} \, \text{kg} \]

### Step 2: Identify the restoring force constant (k) and the frictional force constant (b)
The restoring force constant \( k \) is given as:
\[ k = 50 \times 10^{-3} \, \text{N/m} \]
The frictional force constant \( b \) is given as:
\[ b = 12 \times 10^{-3} \, \text{N/m} \]

### Step 3: Calculate the damping coefficient (γ)
The damping coefficient \( \gamma \) is calculated using the formula:
\[ \gamma = \frac{b}{2m} \]
Substituting the values:
\[ \gamma = \frac{12 \times 10^{-3}}{2 \times (2 \times 10^{-3})} = \frac{12 \times 10^{-3}}{4 \times 10^{-3}} = 3 \, \text{s}^{-1} \]

### Step 4: Calculate the natural frequency (ω₀)
The natural frequency \( \omega_0 \) is calculated using the formula:
\[ \omega_0 = \sqrt{\frac{k}{m}} \]
Substituting the values:
\[ \omega_0 = \sqrt{\frac{50 \times 10^{-3}}{2 \times 10^{-3}}} = \sqrt{25} = 5 \, \text{rad/s} \]

### Step 5: Calculate the damped frequency (ω)
The damped frequency \( \omega \) is given by:
\[ \omega = \sqrt{\omega_0^2 - \gamma^2} \]
Substituting the values:
\[ \omega = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \, \text{rad/s} \]

### Step 6: Calculate the time period (T)
The time period \( T \) of the damped oscillation is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{s} \]

### Final Answer
The time period of the oscillatory motion is:
\[ T = \frac{\pi}{2} \, \text{s} \]

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