If a particle executres S.H.M. with time period 4 s when magnitude of resotring force constant is5 `N//m` and with timper period2 s when magnitude restoring force constant is ` 20 N //m`, then time period under the combined action of two forces willl be

Options

A. `4 sqrt( 5) s`
B. `2 sqrt(5) s`
C. ` ( 2) /(sqrt(5)) s `
D. ` ( 4) /( sqrt(5))s`

Answer

To solve the problem, we need to find the time period of a particle executing Simple Harmonic Motion (SHM) under the combined action of two restoring forces. The given data is:

1. Time period \( T_1 = 4 \, \text{s} \) with restoring force constant \( k_1 = 5 \, \text{N/m} \)
2. Time period \( T_2 = 2 \, \text{s} \) with restoring force constant \( k_2 = 20 \, \text{N/m} \)

### Step 1: Relate time period to restoring force constant
The time period \( T \) of a particle in SHM is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass of the particle and \( k \) is the restoring force constant.

### Step 2: Set up equations for both scenarios
From the formula, we can express the mass \( m \) in terms of \( T \) and \( k \):
1. For the first case:
\[ T_1 = 2\pi \sqrt{\frac{m}{k_1}} \implies m = \frac{T_1^2 k_1}{4\pi^2} \]
Substituting \( T_1 = 4 \, \text{s} \) and \( k_1 = 5 \, \text{N/m} \):
\[ m = \frac{(4)^2 \cdot 5}{4\pi^2} = \frac{16 \cdot 5}{4\pi^2} = \frac{20}{\pi^2} \]

2. For the second case:
\[ T_2 = 2\pi \sqrt{\frac{m}{k_2}} \implies m = \frac{T_2^2 k_2}{4\pi^2} \]
Substituting \( T_2 = 2 \, \text{s} \) and \( k_2 = 20 \, \text{N/m} \):
\[ m = \frac{(2)^2 \cdot 20}{4\pi^2} = \frac{4 \cdot 20}{4\pi^2} = \frac{20}{\pi^2} \]

### Step 3: Combine the restoring forces
The combined restoring force constant \( k_{\text{combined}} \) when both forces act together is:
\[ k_{\text{combined}} = k_1 + k_2 = 5 + 20 = 25 \, \text{N/m} \]

### Step 4: Find the combined time period
Using the combined restoring force constant, we can find the time period \( T_0 \):
\[ T_0 = 2\pi \sqrt{\frac{m}{k_{\text{combined}}}} \]
Substituting \( m = \frac{20}{\pi^2} \) and \( k_{\text{combined}} = 25 \):
\[ T_0 = 2\pi \sqrt{\frac{\frac{20}{\pi^2}}{25}} = 2\pi \sqrt{\frac{20}{25\pi^2}} = 2\pi \cdot \frac{\sqrt{20}}{5\pi} = \frac{2\sqrt{20}}{5} \]
Simplifying further:
\[ T_0 = \frac{2\sqrt{4 \cdot 5}}{5} = \frac{4\sqrt{5}}{5} \]

### Step 5: Final calculation
To express \( T_0 \) in terms of \( T_1 \):
\[ T_0 = T_1 \cdot \frac{1}{\sqrt{5}} = 4 \cdot \frac{1}{\sqrt{5}} = \frac{4}{\sqrt{5}} \, \text{s} \]

Thus, the time period under the combined action of the two forces is:
\[ \boxed{\frac{4}{\sqrt{5}} \, \text{s}} \]