Question

A set of \(14\) tuning forks is arranged in a series of increasing frequencies. Each fork produces '\(x\)' beats per second with the preceding fork and the last fork is an octave of the first fork. If the seventh fork has the frequency of \(114\) Hz, the value of \(x\) is

• A. \(4\)
• B. \(5\)
• C. \(6\)
• D. \(7\)

Hint:

Frequencies form an arithmetic progression; the last is twice the first.

Answer 1

The Correct Option is C

Step 1: Arithmetic Progression:
Let the first fork have frequency \(f_1\). Each next fork is \(x\) Hz higher, so \(f_n=f_1+(n-1)x\).

Step 2: Use the Octave Condition:
The 14th fork is an octave of the first: \(f_{14}=2f_1\), so \(f_1+13x=2f_1\Rightarrow f_1=13x\).

Step 3: Use the 7th Fork:
\(f_7=f_1+6x=13x+6x=19x=114\) Hz.

Step 4: Solve:
\[ x=\frac{114}{19}=6\ \text{Hz} \]
Check: \(f_1=78\) Hz and \(f_{14}=78+78=156\) Hz, which is double. So (C) is correct.

Final Answer:
The value of \(x\) is 6, option (C). \[ \boxed{\text{(C) } 6} \]