Two identical tuning forks vibrate at 256 Hz. One of them is then loaded with a drop of wax, after which 6 beats per second are heard. The period of the loaded tuning fork is

Options

A. 0.006 s
B. 0.005 s
C. 0.004 s
D. 0.00 3 s

Answer

To solve the problem, we need to find the period of the loaded tuning fork after a drop of wax is applied to it. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have two identical tuning forks, both vibrating at a frequency of 256 Hz. When one fork is loaded with wax, it vibrates at a lower frequency, and we hear 6 beats per second when both forks are sounded together.

### Step 2: Identify the Frequencies
Let:
- \( f_1 \) = frequency of the first tuning fork (unloaded) = 256 Hz
- \( f_2 \) = frequency of the second tuning fork (loaded with wax)

The beat frequency \( f_b \) is given by the absolute difference between the two frequencies:
\[ f_b = |f_1 - f_2| \]
Given that \( f_b = 6 \) Hz, we can write:
\[ |256 - f_2| = 6 \]

### Step 3: Solve for \( f_2 \)
This absolute value equation can be split into two cases:
1. \( 256 - f_2 = 6 \)
2. \( f_2 - 256 = 6 \)

**Case 1:**
\[ 256 - f_2 = 6 \implies f_2 = 256 - 6 = 250 \text{ Hz} \]

**Case 2:**
\[ f_2 - 256 = 6 \implies f_2 = 256 + 6 = 262 \text{ Hz} \]

Since the wax loading decreases the frequency, we take the first case:
\[ f_2 = 250 \text{ Hz} \]

### Step 4: Calculate the Period of the Loaded Tuning Fork
The period \( T \) of a tuning fork is the reciprocal of its frequency:
\[ T = \frac{1}{f_2} \]
Substituting the value of \( f_2 \):
\[ T = \frac{1}{250} \text{ seconds} \]

### Step 5: Final Calculation
Calculating the period:
\[ T = 0.004 \text{ seconds} \]

### Conclusion
The period of the loaded tuning fork is \( 0.004 \) seconds.

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