Question

tuning forkA of frequency 258hz gives 8 beats with a tuning forkB.when the tuning fork A is filled again A and B are sounded the no of beats heard decreases.the frequency of B is
A.250hz
B.266hz
C.258hz
D.242hz​

Answer 1

Answer:

250hz

Explanation:∣F  

A

​  

−F  

B

​  

∣=8

|258 - F_B| = 8∣258−F  

B

​  

∣=8

F_B = 250 / F_B = 266F  

B

​  

=250/F  

B

​  

=266

When BB prongs cut f \pifπ equation increases so that means initially 2+2+ was = 250250

So later become 266266

So initially |258 - 250| = 8∣258−250∣=8

so later |258 - 266| = 8∣258−266∣=8