When a block of mass m is suspended separately by two different springs have time period `t_(1)" and "t_(2)`. If same mass is connected to parallel combination of both springs, then its time period is given by :-

Options

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

Answer

To solve the problem, we need to find the time period when a block of mass m is suspended by two different springs connected in parallel. We will derive the time period step by step.

### Step-by-Step Solution:

1. **Understanding 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. **Finding Spring Constants \( k_1 \) and \( k_2 \)**:
For the first spring with time period \( t_1 \):
\[ t_1 = 2\pi \sqrt{\frac{m}{k_1}} \]
Squaring both sides, we get:
\[ t_1^2 = 4\pi^2 \frac{m}{k_1} \implies k_1 = \frac{4\pi^2 m}{t_1^2} \]

For the second spring with time period \( t_2 \):
\[ t_2 = 2\pi \sqrt{\frac{m}{k_2}} \]
Squaring both sides, we get:
\[ t_2^2 = 4\pi^2 \frac{m}{k_2} \implies k_2 = \frac{4\pi^2 m}{t_2^2} \]

3. **Finding the Equivalent Spring Constant \( k_{eq} \)**:
When the two springs are connected in parallel, the equivalent spring constant \( k_{eq} \) is the sum of the individual spring constants:
\[ k_{eq} = k_1 + k_2 \]
Substituting the values of \( k_1 \) and \( k_2 \):
\[ k_{eq} = \frac{4\pi^2 m}{t_1^2} + \frac{4\pi^2 m}{t_2^2} \]
Factoring out \( 4\pi^2 m \):
\[ k_{eq} = 4\pi^2 m \left( \frac{1}{t_1^2} + \frac{1}{t_2^2} \right) \]

4. **Finding the Time Period for the Parallel Combination**:
The time period \( T_{eq} \) for the equivalent spring constant is given by:
\[ T_{eq} = 2\pi \sqrt{\frac{m}{k_{eq}}} \]
Substituting \( k_{eq} \):
\[ T_{eq} = 2\pi \sqrt{\frac{m}{4\pi^2 m \left( \frac{1}{t_1^2} + \frac{1}{t_2^2} \right)}} \]
Simplifying:
\[ T_{eq} = 2\pi \sqrt{\frac{1}{4\pi^2 \left( \frac{1}{t_1^2} + \frac{1}{t_2^2} \right)}} \]
\[ T_{eq} = \frac{1}{2\pi} \sqrt{\frac{1}{\frac{1}{t_1^2} + \frac{1}{t_2^2}}} \]
\[ T_{eq} = \sqrt{\frac{t_1^2 t_2^2}{t_1^2 + t_2^2}} \]

5. **Final Result**:
Thus, the time period \( T_{eq} \) for the mass connected to the parallel combination of both springs is given by:
\[ T_{eq} = \frac{t_1 t_2}{\sqrt{t_1^2 + t_2^2}} \]