When a tuning fork of frequency 341 is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first fork, the number of beats is two per second. The natural frequency of the second tuning fork is

Options

A. `334`
B. `339`
C. `343`
D. `347`

Explanation

To find the natural frequency of the second tuning fork (NB), we can follow these steps:

### Step 1: Understand the problem
We have a tuning fork with a known frequency (NA) of 341 Hz and another tuning fork with an unknown frequency (NB). When both forks are sounded together, they produce beats. The number of beats per second indicates how close the frequencies are to each other.

### Step 2: Analyze the first scenario
In the first scenario, when the two tuning forks are sounded together, we hear 6 beats per second. This means the difference in frequencies between the two tuning forks is 6 Hz. Therefore, we can write:
\[ |NA - NB| = 6 \]
This gives us two possible equations:
1. \( NA - NB = 6 \) (if NA is greater than NB)
2. \( NB - NA = 6 \) (if NB is greater than NA)

### Step 3: Analyze the second scenario
In the second scenario, when the second tuning fork is loaded with wax, its frequency decreases. The number of beats heard is now 2 beats per second. This means the new frequency of the second tuning fork (let's denote it as NB') is now closer to the frequency of the first tuning fork (NA). Thus, we can write:
\[ |NA - NB'| = 2 \]
Since loading with wax decreases the frequency of the second tuning fork, we have:
\[ NB' = NB - k \] (where k is the decrease in frequency due to the wax)

### Step 4: Set up the equations
From the first scenario, we can express NB in terms of NA:
1. If \( NA - NB = 6 \):
\[ NB = NA - 6 \]
2. If \( NB - NA = 6 \):
\[ NB = NA + 6 \]

From the second scenario, we can express NB':
\[ NB' = NB - k \]

We know that:
\[ |NA - NB'| = 2 \]

### Step 5: Solve for NB
Assuming the first case where \( NB = NA - 6 \):
1. Substitute into the second scenario:
\[ |NA - (NA - 6 - k)| = 2 \]
\[ |6 + k| = 2 \]
This gives us two cases:
- Case 1: \( 6 + k = 2 \) → \( k = -4 \) (not possible since k is a decrease)
- Case 2: \( 6 + k = -2 \) → \( k = -8 \) (not possible since k is a decrease)

Now, assuming the second case where \( NB = NA + 6 \):
1. Substitute into the second scenario:
\[ |NA - (NA + 6 - k)| = 2 \]
\[ |-6 + k| = 2 \]
This gives us two cases:
- Case 1: \( -6 + k = 2 \) → \( k = 8 \)
- Case 2: \( -6 + k = -2 \) → \( k = 4 \)

Since the frequency decreases with wax, we take \( k = 4 \).

### Step 6: Calculate NB
Now we can find NB:
\[ NB = NA + 6 \]
\[ NB = 341 + 6 = 347 \text{ Hz} \]

### Conclusion
The natural frequency of the second tuning fork (NB) is **347 Hz**.

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