Two springs are made to oscillate simple harmonically due to the same mass individually. The time periods obtained are `T_1` and `T_2`. If both the springs are connected in series and then made to oscillate by the same mass, the resulting time period will be

Options

A. `T_1+T_2`
B. `(T_1T_2)/(T_1+T_2)`
C. `sqrt(T_1^2+T_2^2)`
D. `(T_1+T_2)/(2)`

Explanation

To solve the problem of finding the time period when two springs are connected in series and oscillating with the same mass, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Time Period of a Spring**:
The time period \( T \) of a mass \( m \) attached to a spring with spring constant \( k \) is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

2. **Time Periods of Individual Springs**:
For the first spring with spring constant \( k_1 \), the time period \( T_1 \) is:
\[ T_1 = 2\pi \sqrt{\frac{m}{k_1}} \]
For the second spring with spring constant \( k_2 \), the time period \( T_2 \) is:
\[ T_2 = 2\pi \sqrt{\frac{m}{k_2}} \]

3. **Express Spring Constants in Terms of Time Periods**:
Rearranging the above equations gives us:
\[ k_1 = \frac{4\pi^2 m}{T_1^2} \]
\[ k_2 = \frac{4\pi^2 m}{T_2^2} \]

4. **Equivalent Spring Constant for Springs in Series**:
When two springs are connected in series, the equivalent spring constant \( k_{\text{eq}} \) is given by:
\[ \frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} \]
Substituting the expressions for \( k_1 \) and \( k_2 \):
\[ \frac{1}{k_{\text{eq}}} = \frac{T_1^2}{4\pi^2 m} + \frac{T_2^2}{4\pi^2 m} \]
Simplifying gives:
\[ \frac{1}{k_{\text{eq}}} = \frac{T_1^2 + T_2^2}{4\pi^2 m} \]
Therefore, we can express \( k_{\text{eq}} \) as:
\[ k_{\text{eq}} = \frac{4\pi^2 m}{T_1^2 + T_2^2} \]

5. **Finding the Time Period for the Series Connection**:
Now, using the formula for the time period with the equivalent spring constant:
\[ T = 2\pi \sqrt{\frac{m}{k_{\text{eq}}}} \]
Substituting for \( k_{\text{eq}} \):
\[ T = 2\pi \sqrt{\frac{m}{\frac{4\pi^2 m}{T_1^2 + T_2^2}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{T_1^2 + T_2^2}{4\pi^2}} = \sqrt{T_1^2 + T_2^2} \]

### Final Result:
Thus, the time period when both springs are connected in series is:
\[ T = \sqrt{T_1^2 + T_2^2} \]