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JEE Mains · Physics · STD 11 - Waves and sound
  • JEE Mains 2022
  • Easy
  • Numerical

A set of \(20\) tuning forks is arranged in a series of increasing frequencies. If each fork gives \(4\) beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is \(\dots \; Hz\).

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Answer & Solution

Correct Answer

(B) \(152\)

Step-by-step Solution

Detailed explanation

\(f_{1}= f\)

\(f _{2}= f +4\)

\(f _{3}= f +2 \times 4\)

\(f _{4}= f +3 \times 4\)

\(f _{20}= f +19 \times 4\)

\(f +(19 \times 4)=2 \times f\)

\(f =76 \; Hz\)

Frequency of last tuning forks \(=2 \; f\)

\(=152 \; Hz\)

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