The angular frequency of the damped oscillator is given by `omega=sqrt((k)/(m)-(r^(2))/(4m^(2)))` ,where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio `r^(2)//(m k)` is 8% ,the change in the time period compared to the undamped oscillator

Options

A. (a)increases by 1%
B. (b)increases by 8%
C. (c)decreased by 1%
D. (d)decrease by 8%

Explanation

To solve the problem step by step, we will analyze the given information and derive the required change in the time period of the damped oscillator compared to the undamped oscillator.

### Step 1: Understand the given formula for angular frequency
The angular frequency of a damped oscillator is given by:
\[ \omega = \sqrt{\frac{k}{m} - \frac{r^2}{4m^2}} \]
where:
- \( k \) is the spring constant,
- \( m \) is the mass of the oscillator,
- \( r \) is the damping constant.

### Step 2: Identify the undamped angular frequency
The angular frequency of an undamped oscillator (\( \omega_0 \)) is given by:
\[ \omega_0 = \sqrt{\frac{k}{m}} \]

### Step 3: Relate the time periods
The time period \( T \) of the damped oscillator is related to its angular frequency by:
\[ T = \frac{2\pi}{\omega} \]
For the undamped oscillator, the time period \( T_0 \) is:
\[ T_0 = \frac{2\pi}{\omega_0} \]

### Step 4: Substitute the expressions for angular frequencies
Substituting the expressions for \( \omega \) and \( \omega_0 \):
\[ T = \frac{2\pi}{\sqrt{\frac{k}{m} - \frac{r^2}{4m^2}}} \]
\[ T_0 = \frac{2\pi}{\sqrt{\frac{k}{m}}} \]

### Step 5: Find the change in time period
To find the change in time period \( T - T_0 \), we can express \( T \) in terms of \( T_0 \):
\[ T = T_0 \cdot \sqrt{\frac{\frac{k}{m}}{\frac{k}{m} - \frac{r^2}{4m^2}}} \]

### Step 6: Substitute the ratio given in the problem
We are given that:
\[ \frac{r^2}{mk} = 0.08 \quad \text{(8%)} \]
This implies:
\[ r^2 = 0.08mk \]

### Step 7: Substitute into the expression for \( T \)
Substituting \( r^2 \) into the expression for \( T \):
\[ T = T_0 \cdot \sqrt{\frac{\frac{k}{m}}{\frac{k}{m} - \frac{0.08mk}{4m^2}}} \]
\[ = T_0 \cdot \sqrt{\frac{\frac{k}{m}}{\frac{k}{m} - \frac{0.02k}{m}}} \]
\[ = T_0 \cdot \sqrt{\frac{\frac{k}{m}}{\frac{k}{m}(1 - 0.02)}} \]
\[ = T_0 \cdot \sqrt{\frac{1}{1 - 0.02}} = T_0 \cdot \sqrt{\frac{1}{0.98}} \]

### Step 8: Calculate the percentage change in time period
The percentage change in time period is given by:
\[ \frac{T - T_0}{T_0} \times 100 = \left(\sqrt{\frac{1}{0.98}} - 1\right) \times 100 \]
Calculating \( \sqrt{\frac{1}{0.98}} \):
\[ \sqrt{\frac{1}{0.98}} \approx 1.0102 \]
Thus,
\[ \frac{T - T_0}{T_0} \times 100 \approx (1.0102 - 1) \times 100 \approx 1.02\% \]

### Final Answer
The change in the time period compared to the undamped oscillator is approximately **1% increase**.