The natural frequency of a tuning fork `P` is `432`Hz. `3` beats/s are produced when tuning fork `P` and another tuning fork `Q` are sounded together. If P is loaded with wax, the number of beats increases to `5` beats/s. The frequency of Q is

Options

A. `429Hz`
B. `435Hz`
C. `437Hz`
D. `427Hz`

Explanation

To find the frequency of tuning fork Q, we can analyze the information given in the problem step by step.

### Step 1: Understand the concept of beats
Beats occur when two sound waves of slightly different frequencies interfere with each other. The number of beats per second is equal to the absolute difference between the two frequencies.

### Step 2: Set up the equations
Let the frequency of tuning fork Q be \( f_Q \). From the information given:
1. The natural frequency of tuning fork P, \( f_P = 432 \) Hz.
2. When P and Q are sounded together, the number of beats is 3 beats/s. This gives us the equation:
\[ |f_Q - f_P| = 3 \]
This can be expressed as:
\[ f_Q - 432 = 3 \quad \text{or} \quad 432 - f_Q = 3 \]

### Step 3: Solve the first equation
From \( f_Q - 432 = 3 \):
\[ f_Q = 432 + 3 = 435 \text{ Hz} \]

### Step 4: Solve the second equation
From \( 432 - f_Q = 3 \):
\[ f_Q = 432 - 3 = 429 \text{ Hz} \]

### Step 5: Analyze the second condition
When tuning fork P is loaded with wax, the number of beats increases to 5 beats/s. This means:
\[ |f_Q - f_P'| = 5 \]
where \( f_P' \) is the new frequency of tuning fork P after being loaded with wax. Since loading with wax decreases the frequency, we can express it as:
\[ f_P' < 432 \text{ Hz} \]

### Step 6: Set up the new equations
We can express the new frequency of P as \( f_P' = 432 - x \), where \( x \) is the decrease in frequency due to the wax. Thus, the equation becomes:
\[ |f_Q - (432 - x)| = 5 \]

### Step 7: Substitute the values of \( f_Q \)
We can substitute the two possible values of \( f_Q \) (435 Hz and 429 Hz) into the new equation.

1. For \( f_Q = 435 \):
\[ |435 - (432 - x)| = 5 \]
This simplifies to:
\[ |3 + x| = 5 \]
This gives two cases:
- \( 3 + x = 5 \) → \( x = 2 \)
- \( 3 + x = -5 \) → No valid solution since \( x \) cannot be negative.

2. For \( f_Q = 429 \):
\[ |429 - (432 - x)| = 5 \]
This simplifies to:
\[ |x - 3| = 5 \]
This gives two cases:
- \( x - 3 = 5 \) → \( x = 8 \)
- \( x - 3 = -5 \) → \( x = -2 \) (not valid)

### Step 8: Conclusion
Since \( x \) must be a positive value representing the decrease in frequency, we can conclude that the only valid frequency for tuning fork Q is:
\[ f_Q = 435 \text{ Hz} \]

### Final Answer
The frequency of tuning fork Q is **435 Hz**.

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