A tuning fork and an organ pipe at temperature `88^@C` produce 5 beats per second. When the temperature of the air column is decreased to `51^@C` the two produce 1 beat per sec. What is the frequency of the tuning fork?

Options

A. f’=81Hz
B. f'= 71Hz
C. f’=110 Hz
D. f'= 55 Hz

Answer

To solve the problem, we need to find the frequency of the tuning fork given the conditions of beat frequency at two different temperatures. Let's break down the solution step by step.

### Step 1: Define Variables
Let \( f \) be the frequency of the tuning fork and \( f_a \) be the frequency of the organ pipe at the respective temperatures.

### Step 2: Understand Beat Frequency
The beat frequency is the absolute difference between the frequencies of the two sources:
- At \( 88^\circ C \), the beat frequency is 5 Hz:
\[ |f - f_a| = 5 \]
- At \( 51^\circ C \), the beat frequency is 1 Hz:
\[ |f - f_a'| = 1 \]

### Step 3: Relate Frequencies to Temperature
The frequency of the organ pipe is related to the temperature as:
\[ f_a \propto \sqrt{T} \]
where \( T \) is the absolute temperature in Kelvin. The temperatures in Kelvin are:
- \( T_1 = 273 + 88 = 361 \, K \)
- \( T_2 = 273 + 51 = 324 \, K \)

### Step 4: Set Up the Frequency Ratios
Using the proportionality of frequencies to the square root of temperature, we can express the frequencies at the two temperatures:
\[ \frac{f_a}{f_a'} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{361}{324}} = \frac{19}{18} \]
This gives us:
\[ f_a = \frac{19}{18} f_a' \]

### Step 5: Substitute Frequencies
From the beat frequencies, we can express \( f_a \) and \( f_a' \):
1. At \( 88^\circ C \):
\[ f_a = f + 5 \]
2. At \( 51^\circ C \):
\[ f_a' = f - 1 \]

### Step 6: Substitute into the Ratio
Substituting \( f_a' = f - 1 \) into the ratio:
\[ f + 5 = \frac{19}{18}(f - 1) \]

### Step 7: Solve the Equation
Multiply both sides by 18 to eliminate the fraction:
\[ 18(f + 5) = 19(f - 1) \]
Expanding both sides:
\[ 18f + 90 = 19f - 19 \]
Rearranging gives:
\[ 90 + 19 = 19f - 18f \]
\[ f = 109 \, Hz \]

### Step 8: Find the Frequency of the Organ Pipe
Now, substituting back to find \( f_a \):
\[ f_a = f + 5 = 109 + 5 = 114 \, Hz \]
And for \( f_a' \):
\[ f_a' = f - 1 = 109 - 1 = 108 \, Hz \]

### Step 9: Calculate the Final Frequency
Now we can check the values:
Using \( f_a = \frac{19}{18} f_a' \):
\[ f_a = \frac{19}{18} \times 108 = 114 \, Hz \]

### Conclusion
Thus, the frequency of the tuning fork is:
\[ \boxed{109 \, Hz} \]