Question

A string with a mass density of 4.5 ✕ 10^-3 kg/m is under a tension of 400 N and is fixed at both ends. One of its resonance frequencies is 195 Hz. The next higher resonance frequency is 260 Hz.

(a) What is the fundamental frequency of this string?
Hz

(b) Which harmonics have the given frequencies? (Enter 1 for the first harmonic, 2 for the second harmonic, etc.)

195 Hz
260 Hz

(c) What is the length of the string?
m

Answer 1

$\mu$ = mass density = 4.5 x 10^-3 kg/m

T = tension force = 400 N

n^th harmonic frequency

f_n = (n/(2L)) sqrt(T/$\mu$)

let f_n = 195 Hz and f_n+1 = 260 Hz

(n/(2L)) sqrt(T/$\mu$) = 195 Eq-1

((n + 1)/(2L)) sqrt(T/$\mu$) = 260 Eq-2

Dividing Eq-1 by Eq-2

n/(n + 1) = 195/260

n = 3

f = fundamental frequency = f_n/n = 195/3 = 65 Hz

b)

195 Hz : Third harmonic

260 Hz : Fourth harmonic

c)

fundamental frequency is given as

f =  (1/(2L)) sqrt(T/$\mu$)

65 =  (1/(2L)) sqrt(400/(4.5 x 10^-3))

L = 2.3 m