A tuning fork gives 5 beats with another tuning fork of frequency 100 Hz . When the first tuning fork is loaded with wax, then the number of beats remains unchanged, then what will be the frequency of the first tuning fork

Options

A. 95 Hz
B. 100Hz
C. 105 Hz
D. 110 Hz

Answer

To solve the problem step by step, we need to analyze the information given and apply the concept of beats in sound waves.

### Step 1: Understand the concept of beats
Beats are produced when two sound waves of slightly different frequencies interfere with each other. The frequency of the beats is given by the absolute difference between the two frequencies.

### Step 2: Set up the known values
We know:
- Frequency of the second tuning fork (new 2) = 100 Hz
- Number of beats = 5 beats per second

From the information, we can express the relationship as:
\[ \text{Number of beats} = | \text{new 1} - \text{new 2} | = 5 \text{ Hz} \]

### Step 3: Establish possible cases for new 1
Let’s denote the frequency of the first tuning fork as new 1. The equation for beats can be expressed in two cases:
1. Case 1: \( \text{new 1} - \text{new 2} = 5 \)
2. Case 2: \( \text{new 2} - \text{new 1} = 5 \)

### Step 4: Solve for new 1 in both cases
**Case 1:**
\[ \text{new 1} - 100 = 5 \]
\[ \text{new 1} = 105 \text{ Hz} \]

**Case 2:**
\[ 100 - \text{new 1} = 5 \]
\[ \text{new 1} = 95 \text{ Hz} \]

### Step 5: Analyze the effect of loading the first tuning fork with wax
When the first tuning fork is loaded with wax, its frequency decreases. The problem states that the number of beats remains unchanged at 5 beats per second.

### Step 6: Determine the implications of frequency change
If the frequency of the first tuning fork decreases (from either 105 Hz or 95 Hz), we need to check which case allows the number of beats to remain unchanged:
- If new 1 = 105 Hz and decreases, it will eventually become less than 100 Hz, leading to an increase in the number of beats.
- If new 1 = 95 Hz and decreases, it would still be less than 100 Hz, which would also lead to an increase in the number of beats.

Since the number of beats remains unchanged, the only feasible frequency for new 1 that satisfies this condition is:
\[ \text{new 1} = 95 \text{ Hz} \]

### Final Answer
The frequency of the first tuning fork is **95 Hz**.

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