MHT CET · Physics · Waves and Sound
A tuning fork ' \(A\) ' produces 5 beats per second with a tuning fork of frequency 480
\(\mathrm{Hz}\). When a little wax is stuck to a prong of fork \(\mathrm{A}\), the number of beats heard per
second becomes \(2 .\) What is the frequency of tuning fork \(\mathrm{A}\) before the wax is stuck
to it?
- A \(485 \mathrm{~Hz}\)
- B \(478 \mathrm{~Hz}\)
- C \(475 \mathrm{~Hz}\)
- D \(482 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(A) \(485 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
(D)
Frequency of tuning fork \(\mathrm{B}=480 \mathrm{~Hz}\). It produces 5 beats with tuning fork \(\mathrm{A}\). Therefore frequency of tuning fork \(\mathrm{A}\) can be \(485 \mathrm{~Hz}\) or \(475 \mathrm{~Hz}\). On loading the tuning fork \(\mathrm{A}\) with wax its frequency will decrease. Since the number of beats decreases, it means its frequency is higher than that of B. Hence its frequency is \(485 \mathrm{~Hz}\).
#
Frequency of tuning fork \(\mathrm{B}=480 \mathrm{~Hz}\). It produces 5 beats with tuning fork \(\mathrm{A}\). Therefore frequency of tuning fork \(\mathrm{A}\) can be \(485 \mathrm{~Hz}\) or \(475 \mathrm{~Hz}\). On loading the tuning fork \(\mathrm{A}\) with wax its frequency will decrease. Since the number of beats decreases, it means its frequency is higher than that of B. Hence its frequency is \(485 \mathrm{~Hz}\).
#
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