Two tuning forks produce `4` beats per seconds when they are sounded together. Now both the forks are moved towards the observer at same speed `(u)`. The beat frequency now becomes `5 Hz`. If the observer also beings to run with speed `u` towards both the forks, then beat frequency is.

Options

A. A. `2 Hz`
B. B. `4 Hz`
C. C. `6 Hz`
D. D. `8 Hz`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Conditions
Two tuning forks produce beats when sounded together. The beat frequency is given as 4 Hz, which means the difference in their frequencies is:
\[ |f_1 - f_2| = 4 \, \text{Hz} \]

### Step 2: Analyze the Case When Both Forks Move Towards the Observer
When both tuning forks are moving towards the observer at speed \( u \), the new beat frequency becomes 5 Hz. The observed frequencies of the tuning forks can be expressed as:
\[ f_1' = \frac{v}{v - u} f_1 \]
\[ f_2' = \frac{v}{v - u} f_2 \]
The new beat frequency can be expressed as:
\[ |f_1' - f_2'| = 5 \, \text{Hz} \]

### Step 3: Substitute the Frequencies into the Beat Frequency Equation
Using the expressions for \( f_1' \) and \( f_2' \):
\[ \left| \frac{v}{v - u} f_1 - \frac{v}{v - u} f_2 \right| = 5 \]
This simplifies to:
\[ \frac{v}{v - u} |f_1 - f_2| = 5 \]
Substituting \( |f_1 - f_2| = 4 \):
\[ \frac{v}{v - u} \cdot 4 = 5 \]

### Step 4: Solve for \( u \)
Rearranging the equation gives:
\[ 4v = 5(v - u) \]
Expanding and rearranging:
\[ 4v = 5v - 5u \]
\[ 5u = v \]
\[ u = \frac{v}{5} \]

### Step 5: Analyze the Case When the Observer Also Moves Towards the Forks
Now, the observer is also moving towards the tuning forks at speed \( u \). The new observed frequencies become:
\[ f_1'' = \frac{v + u}{v - u} f_1 \]
\[ f_2'' = \frac{v + u}{v - u} f_2 \]
The new beat frequency is:
\[ |f_1'' - f_2''| = \left| \frac{v + u}{v - u} f_1 - \frac{v + u}{v - u} f_2 \right| = \frac{v + u}{v - u} |f_1 - f_2| \]
Substituting \( |f_1 - f_2| = 4 \):
\[ |f_1'' - f_2''| = \frac{v + u}{v - u} \cdot 4 \]

### Step 6: Substitute \( u \) into the Beat Frequency Equation
Substituting \( u = \frac{v}{5} \):
\[ |f_1'' - f_2''| = \frac{v + \frac{v}{5}}{v - \frac{v}{5}} \cdot 4 \]
This simplifies to:
\[ |f_1'' - f_2''| = \frac{\frac{6v}{5}}{\frac{4v}{5}} \cdot 4 = \frac{6}{4} \cdot 4 = 6 \, \text{Hz} \]

### Final Answer
Thus, the final beat frequency when the observer also runs towards the tuning forks is:
\[ \boxed{6 \, \text{Hz}} \]