A tuning fork A makes 4 beats per second with a fork B of frequency 256 Hz. A is filed and the beats occur at shorter interval, find its original frequency.

Answer

To solve the problem step by step, we will analyze the information given and apply the concepts of wave motion and beats.

### Step 1: Understand the concept of beats
When two tuning forks of slightly different frequencies are sounded together, they produce a phenomenon known as beats. The beat frequency is equal to the absolute difference between the frequencies of the two tuning forks.

### Step 2: Identify the frequencies involved
We are given:
- Frequency of tuning fork B (fB) = 256 Hz
- Beat frequency = 4 beats per second

### Step 3: Determine the possible frequencies of tuning fork A (fA)
The beat frequency indicates that tuning fork A can either be above or below the frequency of tuning fork B. Thus, we can express the frequency of tuning fork A as:
- fA = fB + beat frequency = 256 Hz + 4 Hz = 260 Hz
- fA = fB - beat frequency = 256 Hz - 4 Hz = 252 Hz

So, the possible frequencies for tuning fork A before filing are 260 Hz or 252 Hz.

### Step 4: Analyze the effect of filing on tuning fork A
When a tuning fork is filed, its frequency increases. Since we are told that after filing, the beats occur at shorter intervals, this implies that the frequency of A has increased.

### Step 5: Determine the original frequency of tuning fork A
Since filing increases the frequency and the beats occur at shorter intervals, we conclude that the original frequency of tuning fork A must have been the higher value:
- Original frequency of tuning fork A (fA) = 260 Hz

### Conclusion
Thus, the original frequency of tuning fork A is **260 Hz**.

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