A tuning fork `P` of unknows frequency gives `7` beats in `2` seconds with another tuning fork `Q`. When `Q` runs towards a wall with a speed of `5 m//s` it gives `5` beats per second with its echo. On loading `P` with wax, it gives `5` beats per second with `Q` . What is the frequency of `P`? Assume speed of sound = `332 m//s` .

Explanation

To find the frequency of tuning fork P, we will follow these steps:

### Step 1: Determine the beat frequency between P and Q
From the problem, we know that tuning fork P gives 7 beats in 2 seconds with tuning fork Q.

**Beat Frequency (f_b)** is calculated as:
\[ f_b = \frac{\text{Number of beats}}{\text{Time}} = \frac{7}{2} = 3.5 \text{ Hz} \]

### Step 2: Analyze the scenario when Q runs towards a wall
When tuning fork Q runs towards a wall at a speed of 5 m/s, it produces 5 beats per second with its echo. The frequency of the sound heard by the wall (the observer) will be affected by the Doppler effect.

Using the Doppler effect formula for a source moving towards a stationary observer:
\[ f' = f \left( \frac{v + v_o}{v - v_s} \right) \]
Where:
- \( f' \) is the frequency heard by the observer (the echo),
- \( f \) is the frequency of Q,
- \( v \) is the speed of sound (332 m/s),
- \( v_o \) is the speed of the observer (0 m/s, since the wall is stationary),
- \( v_s \) is the speed of the source (5 m/s).

The beat frequency with the echo is given as 5 Hz. Therefore, we can write:
\[ f' - f = 5 \text{ Hz} \]

### Step 3: Set up the equations
From the above, we can express:
\[ f' = f + 5 \]
Substituting into the Doppler effect equation:
\[ f + 5 = f \left( \frac{332 + 0}{332 - 5} \right) \]
This simplifies to:
\[ f + 5 = f \left( \frac{332}{327} \right) \]

### Step 4: Solve for frequency f (frequency of Q)
Rearranging gives:
\[ f + 5 = \frac{332f}{327} \]
Multiplying through by 327 to eliminate the fraction:
\[ 327f + 5 \times 327 = 332f \]
\[ 327f + 1635 = 332f \]
\[ 332f - 327f = 1635 \]
\[ 5f = 1635 \]
\[ f = \frac{1635}{5} = 327 \text{ Hz} \]

### Step 5: Determine the frequency of P
Now we know the frequency of Q, we can find the frequency of P using the beat frequency we calculated earlier:
\[ |f_Q - f_P| = 3.5 \text{ Hz} \]
Substituting \( f_Q = 327 \text{ Hz} \):
\[ 327 - f_P = 3.5 \]
Solving for \( f_P \):
\[ f_P = 327 - 3.5 = 323.5 \text{ Hz} \]

### Final Answer
The frequency of tuning fork P is approximately **323.5 Hz**.

---