Question

Fifty six tuning forks are arranged in a series such that each fork gives 6 beats per second with the preceding one. Assuming the frequency of the last fork to be double of the first fork, the frequency of the last fork should be

• A. 220 Hz
• B. 330 Hz
• C. 440 Hz
• D. 660 Hz

Hint:

Beat frequency = difference between frequencies of two forks.

Answer 1

The Correct Option is D

To solve the problem of determining the frequency of the last tuning fork, we need to understand the relationship between the tuning forks and the number of beats they produce.

1. We have a total of 56 tuning forks arranged in a series.
2. Each fork gives 6 beats per second with the preceding one. This implies that the frequency difference between successive forks is 6 Hz.
3. Let the frequency of the first fork be \( f_1 \).
4. According to the problem, the frequency of the last fork (56th fork) is double the frequency of the first fork. Let the frequency of the last fork be \( f_{56} = 2f_1 \).
5. The frequency relationship between the forks can be expressed as: \(f_n = f_1 + (n-1) \cdot 6 \, \text{Hz}\) , where \( n \) is the fork number ranging from 1 to 56.
6. Thus, the frequency of the last (56th) fork is: \(f_{56} = f_1 + (56-1) \cdot 6\) , leading to: \(f_{56} = f_1 + 330\) .
7. Given that \( f_{56} = 2f_1 \), we equate the expressions: \(2f_1 = f_1 + 330\) .
8. Solving for \( f_1 \): \(2f_1 - f_1 = 330\) , hence: \(f_1 = 330\) .
9. Now, substitute \( f_1 \) back into the expression for \( f_{56} \): \(f_{56} = 2 \times 330 = 660 \, \text{Hz}\) .
10. Thus, the frequency of the last fork is 660 Hz.

The calculations confirm that the correct answer is 660 Hz.