Question

A nylon guitar string has a linear density of 7.20 g/m and is under a tension of 150 N. The fixed supports are distance D = 90.0 cm apart. The string is oscillating in the standing wave pattern shown in figure. Calculate the (a) speed. (b) wavelength, and (c) frequency of the travelling waves whose superposition gives this standing wave.

Answer 1

Given

Linear density of nylon guitar string , $\mu =7.20g/m$

Tension $,T=150N$

Distance (D) = 90 cm

We know that for a transverse wave on a string the wave speed can be given by-

$v=\sqrt{T/\mu }$

$V=(150/0.0072{)}^{1/2}=144.33m/sec$

Wavelength =$2\times 90cm=180cm=1.8m$

We know that f = V/ λ where V is the speed, f is the frequency and λ is the wavelength.

By substitution the values

$f=144.33/1.8=80.1Hz$

$f=80.1Hz$