Two strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times greater than in the other. Waves on the string with the lower tension propagate at 35.2 m/s. The fundamental frequency of that string is 258 Hz. What is the beat frequency when each string is vibrating at its fundamental frequency?
A) 11.0 Hz
B) 9.00 Hz
C) 13.7 Hz
D) 12.6 Hz
E) 17.1 Hz

Question

Quizplus · Accepted Answer

Let's call the tension in the first string T and the tension in the second string 1.10T (since it is 1.10 greater).

We know that waves on the second (higher tension) string will travel faster than waves on the first (lower tension) string. We can use the equation for wave speed to find the speed of waves on the second string:

v = sqrt(T/μL)

where v is the wave speed, T is the tension, μ is the linear mass density (mass per unit length), and L is the length of the string.

Since the strings have the same mass and length, their linear mass densities will be the same. This means that we can calculate the wave speed on the second string by using 1.10T as the tension:

35.2 m/s = sqrt(1.10T/μL)

Squaring both sides, we get:

(35.2 m/s)^2 = 1.10T/μL

Rearranging, we get:

T = (35.2 m/s)^2 * μL / 1.10 = 1.25 μL

Now we can use the equation for the frequency of a vibrating string:

f = (1/2L) * sqrt(T/μ)

For the first string (lower tension), we can plug in the given values:

258 Hz = (1/2L) * sqrt(T/μ)

Squaring both sides, we get:

(258 Hz)^2 = (1/2L)^2 * T/μ

Substituting in our expression for T in terms of μ and L, we get:

(258 Hz)^2 = (1/2L)^2 * 1.25 μ

Rearranging, we get:

μ = (258 Hz)^2 * 4L / 5

Now we can use this value of μ to find the frequency of the second string (higher tension):

f = (1/2L) * sqrt(1.10T/μ)

Plugging in the values we know, we get:

f = (1/2L) * sqrt(1.10 * 1.25 μ / μ)

Simplifying, we get:

f = 1.05 * (1/2L) * sqrt(μ)

Now we can find the beat frequency by taking the difference between these two frequencies:

beat frequency = | 258 Hz - 1.05*258 Hz | = 12.6 Hz

Two Strings of Identical Mass and Length Are Stretched with Their