    Question

# When two tuning forks, fork A and fork B are sounded together, 4 beats per second are heard. When the tuning fork B is loaded with wax then 6 beats per second are heard. If the frequency of fork A is 200 Hz then what was the original frequency of fork B?

A
209 Hz
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B
196 Hz
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C
202 Hz
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D
200 Hz
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Solution

## The correct option is B 196 HzGiven, fb=4 Hz We know, beat frequency fb=|f1−f2| ⇒f2=f1±fb Frequency of tuning fork A, f1=200 Hz ⇒f2=200±4 Hz So, frequency of tuning fork B can be, f2=204 Hz or 196 Hz When tuning fork B is loaded with wax then, fb=6 Hz ⇒f′2=200±6 Hz So, frequency of tuning fork B can be, f′2=206 Hz or 194 Hz As the tuning fork B is loaded with wax, therefore its time period should increase and frequency should decrease. So, by intuition, the original frequency should be 196 Hz which will decrease to 194 Hz when loaded with wax. OR Given, fb=4 Hz We know, beat frequency fb=|fA−fB|=4 Hz On waxing B, its frequency reduces and beat requency increases. Hence, fA−fB=4 Hz 200−fB=4 Hz fB=196 Hz  Suggest Corrections  0      Similar questions  Explore more