A tuning fork arrangement (pair) produces `4 beats//sec` with one fork of frequency `288cps`. A little wax is placed on the unknown fork and it then produces `2 beats//sec`. The frequency of the unknown fork is

Options

A. 286 cps
B. 284 cps
C. 292 cps
D. 290 cps

Explanation

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

### Step 1: Understand the concept of beats
When two tuning forks of different frequencies are struck together, they produce beats. The number of beats per second is equal to the absolute difference between their frequencies.

### Step 2: Set up the equation for the first scenario
Let the frequency of the unknown fork be \( F' \). According to the problem, when the unknown fork is struck with the known fork of frequency \( F = 288 \, \text{cps} \), they produce \( 4 \, \text{beats/sec} \). Therefore, we can write:

\[ |F' - 288| = 4 \]

This gives us two possible equations:

1. \( F' - 288 = 4 \)
2. \( 288 - F' = 4 \)

### Step 3: Solve the equations
**From the first equation:**
\[ F' - 288 = 4 \implies F' = 292 \, \text{cps} \]

**From the second equation:**
\[ 288 - F' = 4 \implies F' = 284 \, \text{cps} \]

### Step 4: Analyze the effect of adding wax
When wax is added to the unknown fork, the beat frequency decreases to \( 2 \, \text{beats/sec} \). This means that the frequency of the unknown fork must have decreased.

Now, we set up another equation for the new scenario:

\[ |F' - 288| = 2 \]

This gives us two new possible equations:

1. \( F' - 288 = 2 \)
2. \( 288 - F' = 2 \)

### Step 5: Solve the new equations
**From the first equation:**
\[ F' - 288 = 2 \implies F' = 290 \, \text{cps} \]

**From the second equation:**
\[ 288 - F' = 2 \implies F' = 286 \, \text{cps} \]

### Step 6: Determine the correct frequency
Now we have four potential frequencies from both scenarios:
- From the first scenario: \( 292 \, \text{cps} \) and \( 284 \, \text{cps} \)
- From the second scenario: \( 290 \, \text{cps} \) and \( 286 \, \text{cps} \)

Since adding wax decreases the frequency, we can conclude that the original frequency of the unknown fork must have been greater than \( 288 \, \text{cps} \). Thus, the only feasible solution is:

\[ F' = 292 \, \text{cps} \]

### Final Answer
The frequency of the unknown fork is \( 292 \, \text{cps} \).

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