When a body is suspended from two light springs separately, the periods of vertical oscillations are `T_1` and `T_2`. When the same body is suspended from the two spring connected in series, the period will be

Options

A. `T=T_1+T_2`
B. `1/T=1/T_1+1/T_2`
C. `T^2=T_1^2+T_2^2`
D. `1/T^2=1/T_1^2+1/T^2`

Answer

To find the period of vertical oscillations when a body is suspended from two springs connected in series, we can follow these steps:

### Step 1: Understand the Period of Oscillation for Individual Springs
The period of oscillation for a mass \( m \) suspended from a spring with spring constant \( k \) is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
For the first spring with spring constant \( k_1 \), the period is:
\[ T_1 = 2\pi \sqrt{\frac{m}{k_1}} \]
For the second spring with spring constant \( k_2 \), the period is:
\[ T_2 = 2\pi \sqrt{\frac{m}{k_2}} \]

### Step 2: Determine the Effective Spring Constant for Springs in Series
When two springs are connected in series, the effective spring constant \( k_s \) can be calculated using the formula:
\[ \frac{1}{k_s} = \frac{1}{k_1} + \frac{1}{k_2} \]
This can be rearranged to find \( k_s \):
\[ k_s = \frac{k_1 k_2}{k_1 + k_2} \]

### Step 3: Calculate the Period for the Springs in Series
Now, we can find the period of oscillation \( T_s \) when the body is suspended from the two springs in series:
\[ T_s = 2\pi \sqrt{\frac{m}{k_s}} = 2\pi \sqrt{\frac{m}{\frac{k_1 k_2}{k_1 + k_2}}} \]
This simplifies to:
\[ T_s = 2\pi \sqrt{\frac{m(k_1 + k_2)}{k_1 k_2}} \]

### Step 4: Relate \( T_s \) to \( T_1 \) and \( T_2 \)
Now, we can express \( T_s \) in terms of \( T_1 \) and \( T_2 \):
- From the expressions for \( T_1 \) and \( T_2 \):
\[ T_1^2 = 4\pi^2 \frac{m}{k_1} \quad \text{and} \quad T_2^2 = 4\pi^2 \frac{m}{k_2} \]
- Therefore, we can write:
\[ T_s^2 = 4\pi^2 \frac{m(k_1 + k_2)}{k_1 k_2} \]
- This can be rewritten as:
\[ T_s^2 = \frac{T_1^2 k_2 + T_2^2 k_1}{k_1 k_2} \]

### Conclusion
Thus, the period of oscillation when the body is suspended from the two springs connected in series is given by:
\[ T_s^2 = T_1^2 + T_2^2 \]

### Final Answer
The period of oscillation when the body is suspended from the two springs connected in series is:
\[ T_s = \sqrt{T_1^2 + T_2^2} \]