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.You need to login to perform this action.
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