A body executes SHM under the influence of one force and has a period `T_1` seconds and the same body executes SHM with period `T_2` seconds when under the influence of another force. When both forces act simultaneously and in the same direction, then the time period of the same body (in seconds) is:

Options

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

Explanation

To solve the problem, we need to find the time period of a body executing Simple Harmonic Motion (SHM) when two forces act on it simultaneously. Here’s the step-by-step solution:

### Step 1: Understand the Forces Acting on the Body
When a body is under the influence of two forces, the forces can be represented as:
- Force 1: \( F_1 = k_1 x \)
- Force 2: \( F_2 = k_2 x \)

Where \( k_1 \) and \( k_2 \) are the force constants for each force, and \( x \) is the displacement from the mean position.

### Step 2: Determine the Time Periods for Each Force
The time period \( T \) of a body executing SHM is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Where \( m \) is the mass of the body and \( k \) is the effective spring constant.

For the first force:
\[ T_1 = 2\pi \sqrt{\frac{m}{k_1}} \]
For the second force:
\[ T_2 = 2\pi \sqrt{\frac{m}{k_2}} \]

### Step 3: Express the Spring Constants in Terms of Time Periods
Rearranging the formulas for \( k_1 \) and \( k_2 \):
\[ k_1 = \frac{4\pi^2 m}{T_1^2} \]
\[ k_2 = \frac{4\pi^2 m}{T_2^2} \]

### Step 4: Find the Effective Spring Constant When Both Forces Act
When both forces act simultaneously, the effective spring constant \( k_{\text{net}} \) is the sum of the individual spring constants:
\[ k_{\text{net}} = k_1 + k_2 \]

Substituting the expressions for \( k_1 \) and \( k_2 \):
\[ k_{\text{net}} = \frac{4\pi^2 m}{T_1^2} + \frac{4\pi^2 m}{T_2^2} \]

### Step 5: Calculate the Time Period for the Combined System
Now, using the effective spring constant to find the new time period \( T_{\text{net}} \):
\[ T_{\text{net}} = 2\pi \sqrt{\frac{m}{k_{\text{net}}}} \]
Substituting for \( k_{\text{net}} \):
\[ T_{\text{net}} = 2\pi \sqrt{\frac{m}{\frac{4\pi^2 m}{T_1^2} + \frac{4\pi^2 m}{T_2^2}}} \]
This simplifies to:
\[ T_{\text{net}} = 2\pi \sqrt{\frac{m}{4\pi^2 m \left(\frac{1}{T_1^2} + \frac{1}{T_2^2}\right)}} \]
The mass \( m \) cancels out:
\[ T_{\text{net}} = 2\pi \sqrt{\frac{1}{4\pi^2 \left(\frac{1}{T_1^2} + \frac{1}{T_2^2}\right)}} \]
\[ T_{\text{net}} = \sqrt{\frac{T_1^2 T_2^2}{T_1^2 + T_2^2}} \]

### Final Result
Thus, the time period when both forces act simultaneously is:
\[ T_{\text{net}} = \sqrt{\frac{T_1^2 T_2^2}{T_1^2 + T_2^2}} \]