Two tuning forks when sounded together produced `4beats//sec`. The frequency of one fork is 256. The number of beats heard increases when the fork of frequency 256 is loaded with wax. The frequency of the other fork is

Options

A. 504
B. 520
C. 260
D. 252

Answer

To find the frequency of the other tuning fork, we can follow these steps:

### Step 1: Understand the concept of beats
When two tuning forks are sounded together, the phenomenon of beats occurs due to the interference of sound waves from the two sources. The beat frequency is given by the absolute difference between the frequencies of the two tuning forks.

### Step 2: Set up the equation for beats
Let the frequency of the other tuning fork be \( f \). Given that the frequency of one fork is \( 256 \) Hz, the beat frequency is given as \( 4 \) beats per second. Therefore, we can write the equation for the beat frequency as:
\[ |f - 256| = 4 \]

### Step 3: Solve the equation
This absolute value equation can be split into two cases:
1. \( f - 256 = 4 \)
2. \( f - 256 = -4 \)

#### Case 1:
\[ f - 256 = 4 \implies f = 256 + 4 = 260 \text{ Hz} \]

#### Case 2:
\[ f - 256 = -4 \implies f = 256 - 4 = 252 \text{ Hz} \]

Thus, the possible frequencies for the other tuning fork are \( 260 \) Hz and \( 252 \) Hz.

### Step 4: Analyze the effect of loading the fork with wax
When the fork with frequency \( 256 \) Hz is loaded with wax, its frequency decreases. This means that the frequency will be less than \( 256 \) Hz, leading to an increase in the beat frequency.

### Step 5: Determine the correct frequency
Since loading the fork with wax decreases its frequency, we need to check which of the two frequencies (260 Hz or 252 Hz) will lead to an increased beat frequency when the frequency of the 256 Hz fork decreases.

- If the frequency of the fork becomes less than \( 256 \) Hz (for example, let’s say it becomes \( 255 \) Hz), then:
- If \( f = 260 \) Hz, the beat frequency would be:
\[ |260 - 255| = 5 \text{ beats/sec} \]
- If \( f = 252 \) Hz, the beat frequency would be:
\[ |252 - 255| = 3 \text{ beats/sec} \]

Since the beat frequency increases when the fork is loaded with wax, the frequency of the other fork must be \( 260 \) Hz.

### Final Answer:
The frequency of the other tuning fork is \( 260 \) Hz.

---