Question

A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequencies is 195 Hz. The next higher resonance frequency is 260 Hz.

Required:
a. What is the fundamental frequency of this string?
b. Which harmonics have the given frequencies?
c. What is the length of the string?

Answer 1

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Step-by-step explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

`195 = (n)/(2l) \sqrt{(T)/(\mu) } ---(1)\\\\260 = (n+1)/(2l) \sqrt{(T)/(\mu) } ---(2)\\\\divide \ (2) \ by (1)\\\\(260)/(195) = (n+1 )/(n) \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = (195)/(65) \\\\n = 3`

(c) From any of the equations, solve for Length of the string (L);

`195 = (n)/(2l) \sqrt{(T)/(\mu) } \\\\195 = (3)/(2l)\sqrt{(380)/(3* 10^(-3)) } \\\\l = (3)/(2* 195)\sqrt{(380)/(3* 10^(-3)) }\\\\l = 2.74 \ m`

(a) the fundamental frequency is calculated as;

`f_o = (1)/(2l) \sqrt{(T)/(\mu) } \\\\f_o = (1)/(2* 2.74) \sqrt{(380)/(3* 10^(-3) ) }\\\\f_o = 65 \ Hz`

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.