Frequency of oscillation of a body is 5 Hz when a force `F_(1)` is applied and 12 Hz when another force `F_(2)` is applied. If both forces `F_(1)` and `F_(2)` are applied together, then frequency of oscillation of the body will be

Options

A. 13 Hz
B. 169 Hz
C. 62 Hz
D. 52 Hz

Explanation

To solve the problem, we need to find the frequency of oscillation when both forces \( F_1 \) and \( F_2 \) are applied together. The frequencies given for the individual forces are 5 Hz and 12 Hz, respectively.

### Step-by-Step Solution:

1. **Understanding the relationship between frequency and force constant**:
The frequency of oscillation \( f \) of a mass \( m \) attached to a spring (or any oscillatory system) is given by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
where \( k \) is the spring constant (or force constant).

2. **Setting up equations for the given frequencies**:
For the first force \( F_1 \) (frequency = 5 Hz):
\[ 5 = \frac{1}{2\pi} \sqrt{\frac{k_1}{m}} \implies k_1 = (5 \cdot 2\pi)^2 m = 100\pi^2 m \]

For the second force \( F_2 \) (frequency = 12 Hz):
\[ 12 = \frac{1}{2\pi} \sqrt{\frac{k_2}{m}} \implies k_2 = (12 \cdot 2\pi)^2 m = 576\pi^2 m \]

3. **Finding the combined spring constant**:
When both forces are applied together, the effective spring constant \( k \) is the sum of the individual spring constants:
\[ k = k_1 + k_2 = 100\pi^2 m + 576\pi^2 m = 676\pi^2 m \]

4. **Calculating the new frequency**:
Now, we can find the new frequency \( f \) when both forces are applied:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \frac{1}{2\pi} \sqrt{\frac{676\pi^2 m}{m}} = \frac{1}{2\pi} \sqrt{676\pi^2} = \frac{1}{2\pi} \cdot 26\pi = \frac{26}{2} = 13 \text{ Hz} \]

### Final Answer:
The frequency of oscillation of the body when both forces \( F_1 \) and \( F_2 \) are applied together is **13 Hz**.

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