A guitar string with a tension of 150 N is played together with a tuning fork whose frequency is 256 Hz, and beats of 2 Hz are heard. Assuming the guitar string is higher in frequency than the tuning fork, to what new tension should the guitar string be adjusted for the string and tuning fork to have the same frequency?

To solve this problem, we need to understand the relationship between the tension in the guitar string, the frequency of the sound it produces, and the frequency of the tuning fork.

The frequency of a vibrating string is given by the formula:

f = (1/2L) * sqrt(T/μ)

where:

• f is the frequency,
• L is the length of the string,
• T is the tension in the string, and
• μ is the linear density of the string.

Since we want the guitar string and the tuning fork to have the same frequency, we can set their frequencies equal to each other and solve for the new tension:

f_tuning_fork = f_string 256 Hz = (1/2L) * sqrt(T_new/μ)

We know that the original frequency of the string was 2 Hz higher than the tuning fork, so:

f_string_original = 256 Hz + 2 Hz = 258 Hz

Substituting this into the formula gives us:

258 Hz = (1/2L) * sqrt(T_original/μ) 256 Hz = (1/2L) * sqrt(T_new/μ)

We can then divide these two equations to get:

258/256 = sqrt(T_original/T_new)

Squaring both sides gives:

(258/256)^2 = T_original/T_new

We know that the original tension was 150 N, so we can solve for the new tension:

T_new = T_original / (258/256)^2 T_new = 150 N / (1.015625) T_new = 147.71 N

So, the guitar string should be adjusted to a tension of approximately 147.71 N for the string and tuning fork to have the same frequency.