A student working with a spring mass system finds the values of mass, spring constant and damping constant in `CGS` unit as `90, 4.5xx10^(+3)` and `180` respectively. What will be the time period of the spring mass system ?

Options

A. `2pi//7`
B. `14pi`
C. `pi`
D. `2pi//sqrt(50)`

Explanation

To find the time period of a damped spring-mass system, we can use the following formula for the angular frequency (ω):

\[ \omega = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}} \]

Where:
- \(k\) is the spring constant,
- \(m\) is the mass,
- \(b\) is the damping constant.

Given values:
- Mass \(m = 90 \, \text{g}\)
- Spring constant \(k = 4.5 \times 10^3 \, \text{dyn/cm}\)
- Damping constant \(b = 180 \, \text{g/s}\)

### Step 1: Substitute the values into the formula

First, we need to calculate the term inside the square root:

\[ \omega = \sqrt{\frac{4.5 \times 10^3}{90} - \frac{180^2}{4 \times 90^2}} \]

### Step 2: Calculate \(\frac{k}{m}\)

Calculating \(\frac{k}{m}\):

\[ \frac{k}{m} = \frac{4.5 \times 10^3}{90} = \frac{4500}{90} = 50 \]

### Step 3: Calculate \(\frac{b^2}{4m^2}\)

Calculating \(\frac{b^2}{4m^2}\):

\[ \frac{b^2}{4m^2} = \frac{180^2}{4 \times 90^2} = \frac{32400}{4 \times 8100} = \frac{32400}{32400} = 1 \]

### Step 4: Substitute back into the equation for \(\omega\)

Now substituting back into the equation for \(\omega\):

\[ \omega = \sqrt{50 - 1} = \sqrt{49} = 7 \]

### Step 5: Calculate the time period \(T\)

The time period \(T\) is given by:

\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{7} \]

### Step 6: Final answer

Thus, the time period of the spring-mass system is:

\[ T \approx \frac{2 \times 3.14}{7} \approx 0.897 \, \text{seconds} \]

### Summary of Steps:
1. Write down the formula for angular frequency.
2. Substitute the given values into the formula.
3. Calculate \(\frac{k}{m}\).
4. Calculate \(\frac{b^2}{4m^2}\).
5. Substitute back to find \(\omega\).
6. Calculate the time period \(T\).