To solve the problem, we need to determine the original frequency of tuning fork P, given the frequency of tuning fork Q and the information about the beats produced.
### Step-by-Step Solution:
1. **Identify Given Information:**
- Frequency of tuning fork Q, \( \nu_Q = 250 \, \text{Hz} \)
- Initial beats per second = 4 beats/s
- Beats per second after filing prong of fork P = 2 beats/s
2. **Understanding Beats:**
- The number of beats produced is given by the absolute difference in frequencies of the two tuning forks.
- Initially, we have:
\[
|\nu_P - \nu_Q| = 4
\]
- After filing the prong of fork P, we have:
\[
|\nu_P' - \nu_Q| = 2
\]
- Here, \( \nu_P' \) is the new frequency of fork P after filing.
3. **Setting Up Equations:**
- From the first equation, we can write two possible cases:
1. \( \nu_P - \nu_Q = 4 \)
2. \( \nu_Q - \nu_P = 4 \)
- From the second equation, after filing:
1. \( \nu_P' - \nu_Q = 2 \)
2. \( \nu_Q - \nu_P' = 2 \)
4. **Finding \( \nu_P' \):**
- Since filing increases the frequency, we can assume:
\[
\nu_P' > \nu_P
\]
- Thus, we will consider:
\[
\nu_P' - \nu_Q = 2 \implies \nu_P' = 2 + \nu_Q = 2 + 250 = 252 \, \text{Hz}
\]
5. **Finding \( \nu_P \):**
- Now substituting \( \nu_P' \) back into the equations:
- From the first case of beats:
\[
\nu_P - 250 = 4 \implies \nu_P = 254 \, \text{Hz}
\]
- From the second case of beats:
\[
250 - \nu_P = 4 \implies \nu_P = 246 \, \text{Hz}
\]
6. **Determining the Valid Frequency:**
- We have two possible frequencies for \( \nu_P \):
- \( \nu_P = 254 \, \text{Hz} \) which gives \( \nu_P' = 252 \, \text{Hz} \) (valid since \( \nu_P' < \nu_P \))
- \( \nu_P = 246 \, \text{Hz} \) which gives \( \nu_P' = 248 \, \text{Hz} \) (valid since \( \nu_P' > \nu_P \))
- Since filing increases the frequency, the only valid case is:
\[
\nu_P = 246 \, \text{Hz}
\]
### Final Answer:
The original frequency of tuning fork P is \( \nu_P = 246 \, \text{Hz} \).
