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Grade:
        
Good morning sir.
Beats application sir .
Explain fastly sir.
2 years ago

Answers : (1)

Arun
24740 Points
							
Dear student
 
frequency of unknown tuning fork will be = 288 +4 = 292 Hz   or   288 – 4 = 284 Hz
 
on placing a little wax on unknown tuning fork, its frequency decreases but now the number of beats produced is 2 i.e. the frequency difference now decreases. It is possible when frequency of unknown tuning fork  = 292 Hz
 
 
2 years ago
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