question_answer

When two tuning forks A and B are sounded together, 4 beats per second are heard. The frequency of the fork B is 384 Hz. When one of the prongs of the fork A is filed and sounded with B, the beat frequency increases, then the frequency of the fork A is

A)  \[388Hz\]         

B)  \[389Hz\]

C)  \[380Hz\]        

D)  \[379Hz\]

Correct Answer: A

Solution :

: Let the frequency of the fork A be \[{{\upsilon }_{A}}\]. As it produces 4 beats per second with the fork B of frequency \[({{\upsilon }_{B}}=)384Hz,\] \[\therefore \]   \[{{\upsilon }_{A}}={{\upsilon }_{B}}\pm A=(384\pm 4)Hz\] \[=388Hz\] or  \[380Hz\] When one of the prongs of A is filed, its frequency increases. If, \[{{\upsilon }_{A}}=380Hz\]further increase in \[{{\upsilon }_{A}}\] will result in decrease in the beat frequency when sounded with B. If, \[{{\upsilon }_{A}}=388Hz\] further increase in \[{{\upsilon }_{A}}\] will result in increase in the beat frequency when sounded with B. Thus, the frequency of the fork A is 388 Hz.