Q.
Two tuning forks A and B produce 6 beats per second. If the frequency of A is 512 Hz. The fork B is loaded with wax and they produce 4 beats per second- The frequency of B is:
Number of beats per second n1−n2=6 Given: frequency of tuning fork A = 512 Hz Hence, frequency of B is 512±6 Hence, frequency of B is either =506Hz or 518Hz After loading the fork 5, the frequency decreases Hence, frequency of fork B become 512±B=4B=516 or 508 Therefore, the frequency of fork before loading must be 518 Hz.