Question

There are $26$ tuning forks arranged in the decreasing order of their frequencies. Each tuning fork gives $3$ beats with the next. The first one is octave of the last. What is the frequency of 18th tuning fork?

• A. 100 Hz
• B. 99 Hz
• C. 96 Hz
• D. 103 Hz

Answer 1

Answer: (B)

99 Hz

Answer 2

Each tuning forck gives 3 beats with the next, so the difference in the frequencies of two consecutive forks is 3 . $\therefore f_{26} =f_{1}+(n-1) \times-3$ $f =2 f+(26-1) \times-3$ $f =75\, Hz .$ $\therefore$ Frequency of 18 th tuning fork $f_{18} =f_{1}+(18-1) \times-3$ $=2 \times 75+17 \times-3$ $=150-51=99\, Hz$.