# Two identical guitar strings under 108 N of tension sound tones with frequencies of 528 Hz. The peg of one string slips slightly, and the tension in it drops to 91.0 N. How many beats per second are heard?

## v= sqrt(T/mu)= freq*wavelength

so wavelength remais the same, so

(108/mu / 91/mu)=(528/f)^2

f^2=528^2/(108/91)

f= 528 (sqrt(91/108) about 485 (work it out)

beats= 528-485

## To find the number of beats per second heard, we need to calculate the difference in frequency between the two guitar strings.

1. Start by finding the frequency of the guitar string under 108 N tension. Since the two guitar strings are identical, the frequency of the first string is also 528 Hz.

2. Next, calculate the frequency of the guitar string under 91.0 N tension using the formula for the frequency of a vibrating string:

frequency = (1/2L) * sqrt(Tension/mass)

In this case, the tension is 91.0 N, and the other variables (2L and mass) are constant since the two strings are identical.

3. Now subtract the frequency of the second string from the frequency of the first string to find the difference in frequency.

frequency difference = 528 Hz - frequency of second string

4. The absolute value of the frequency difference gives the number of beats per second heard.

beats per second = |frequency difference|

By following these steps, you can find the number of beats per second heard in the given scenario.