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

Question

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) 380 Hz (B) 388 Hz (C) 379 Hz (D) 389 Hz

Tardigrade Admin · Accepted Answer

Given, frequency of fork B, vB=384 Hz We know that, vA-vB= 4 vA-384=4 vA=384+4 Frequency of fork A, v A =388 Hz

Q. 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

380 Hz

388 Hz

379 Hz

389 Hz

Given, frequency of fork $B$ ,
$v_{B} = 384 Hz$
We know that,
$v_{A} - v_{B} = 4$
$v_{A} - 384 = 4$
$v_{A} = 384 + 4$
Frequency of fork $A$ ,
$v_{A} = 388 Hz$

Q. When two tuning forks A and B are sounded together. 4 beats per second are heard. The frequency of the fork B is 384Hz. 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