Physics, asked by amitsainiamit4219, 9 months ago

When a tuning fork of frequency 341 is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first fork, the number of beats is two per second. The natural frequency of the second tuning fork is

Answers

Answered by Anonymous
6

Given:

Frequency of the fork = 341 = nA

Beats per second = 6 per sec = x

Number of beats in second fork = 2 per sec

To Find:

Natural frequency of the second tuning fork = nB

Answer:

As per the question x is decreasing (i.e, x↓) after loading (from 6 to 1) Therefore, the unknown tuning fork is also loaded, hence nB↓

= nB↓− nA=x↓

nB = nA + x

= 341 + 6

= 347

Answer: The frequency of second tuning fork is 347Hz.

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