Two violin players tuning their instruments together hear 8 beats in 2 s. What is the frequency difference between the two violins

The beat frequency 8/2 = 4 Hz, equals the difference in frequencies.

To find the frequency difference between the two violins, we need to calculate the number of beats per second (BPS). The number of beats per second is equal to the number of beats divided by the time taken.

Given that the two players hear 8 beats in 2 seconds, we can calculate the BPS as follows:

BPS = 8 beats / 2 seconds = 4 beats per second

Since each beat corresponds to a frequency difference of one cycle, the frequency difference between the two violins is 4 cycles per second (4 Hz).

To find the frequency difference between the two violin players, we need to calculate the number of beats per second and then convert it into frequency.

Given that the two violin players hear 8 beats in 2 seconds, we can determine the number of beats per second by dividing the total number of beats (8) by the time taken (2 seconds):

Beats per second = 8 beats / 2 seconds = 4 beats/second

Now, to find the frequency difference, we need to consider that each beat is produced when there is a constructive or destructive interference between the two violins' frequencies.

When two frequencies interfere constructively, a beat is heard. In this scenario, the actual frequency difference will be equal to the beat frequency. Hence, the frequency difference between the two violins is 4 beats/second.

Therefore, the two violin players have a frequency difference of 4 beats/second.