A body at the end of a spring executes SHM with a period `t_(1),` while the corresponding period for another spring is `t_(2),` If the period of oscillation with the two spring in series is T, then

Options

A. `T=t_(1)+t_(2)`
B. `T^(2)=t_(1)^(2)+t_(2)^(2)`
C. `(1)/(T)= (1)/(t_(1))+(1)/(t_(2))`
D. `(1)/(T^(2))=(1)/(t_(1)^(2))+(1)/(t_(2)^(2))`

Answer

To solve the problem, we need to find the period of oscillation \( T \) when two springs with periods \( t_1 \) and \( t_2 \) are connected in series. We will use the relationship between the time period and the spring constant.

### Step-by-Step Solution:

1. **Understanding the Time Period of a Spring**:
The time period \( t \) of a mass-spring system is given by the formula:
\[ t = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass and \( k \) is the spring constant.

2. **Relating Time Period to Spring Constant**:
Squaring both sides of the equation gives:
\[ t^2 = 4\pi^2 \frac{m}{k} \]
Rearranging this, we find:
\[ k = \frac{4\pi^2 m}{t^2} \]

3. **Finding the Spring Constants**:
For the two springs with periods \( t_1 \) and \( t_2 \):
- For spring 1:
\[ k_1 = \frac{4\pi^2 m}{t_1^2} \]
- For spring 2:
\[ k_2 = \frac{4\pi^2 m}{t_2^2} \]

4. **Calculating the Equivalent Spring Constant**:
When the two springs are connected in series, the equivalent spring constant \( k_{eq} \) is given by:
\[ \frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} \]
Substituting the values of \( k_1 \) and \( k_2 \):
\[ \frac{1}{k_{eq}} = \frac{t_1^2}{4\pi^2 m} + \frac{t_2^2}{4\pi^2 m} \]
This simplifies to:
\[ \frac{1}{k_{eq}} = \frac{t_1^2 + t_2^2}{4\pi^2 m} \]
Therefore, the equivalent spring constant is:
\[ k_{eq} = \frac{4\pi^2 m}{t_1^2 + t_2^2} \]

5. **Finding the Period of the Combined System**:
Now, using the formula for the time period with the equivalent spring constant:
\[ T = 2\pi \sqrt{\frac{m}{k_{eq}}} \]
Substituting for \( k_{eq} \):
\[ T = 2\pi \sqrt{\frac{m}{\frac{4\pi^2 m}{t_1^2 + t_2^2}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{t_1^2 + t_2^2}{4\pi^2}} = \sqrt{\frac{t_1^2 + t_2^2}{4}} = \frac{1}{2} \sqrt{t_1^2 + t_2^2} \]

### Final Answer:
The period of oscillation \( T \) when the two springs are connected in series is:
\[ T = \frac{1}{2} \sqrt{t_1^2 + t_2^2} \]