Question

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

Answer 1

Hint: You could simply substitute the given values of tension and linear density in expression for speed of a transverse wave which could be easily derived using dimensional analysis if you know the proportionality relation between the quantities. We can find Wavelength from the given wave pattern. Thus obtained values of speed and wavelength can be substituted to find the frequency.
Formula used:
Expression for speed of a transverse wave on a stretched string,
$v=\sqrt{\dfrac{T}{\mu }}$
Expression for frequency,
$f=\dfrac{v}{\lambda }$

Complete answer:
We are given a nylon guitar string with a linear density (μ) 7.20g/m and it is under a tension of 150N. Also, we have two fixed supports kept at a distance of 90.0cm apart from each other supporting the given guitar string. We are also given the standing wave pattern of the string’s oscillation. We are now supposed to find the speed, wavelength and frequency of the traveling waves.
Let us first find the speed of the traveling wave. For that let us discuss the speed of a transverse wave on a stretched string.
Keep this fact in mind that, restoring force along with inertial properties of the medium determines the speed of a mechanical wave. Speed is expected to be directly proportional to restoring force and inversely to inertial properties. Since, we are dealing with waves on a string, restoring force here is the tension T on the string and mass density μ is the inertial property. Let us derive a formula for the speed using dimensional analysis on the above information.
$\mu =\dfrac{m}{L}\Rightarrow \left[ M{{L}^{-1}} \right]$
$T\Rightarrow \left[ ML{{T}^{-2}} \right]$
$\dfrac{\left[ ML{{T}^{-2}} \right]}{\left[ M{{L}^{-1}} \right]}=\left[ {{L}^{2}}{{T}^{-2}} \right]$ ………………… (1)
Let us assume that T and μ are the only relevant physical quantities in the expression for speed, then, we could write (1) as,
$v=C\sqrt{\dfrac{T}{\mu }}$
C is the constant of dimensional analysis which is known to have the value of unity. That is, C=1.
So, speed of transverse wave on a stretched string is given by,
$v=\sqrt{\dfrac{T}{\mu }}$
Now let us substitute the given values from the question,
$v=\sqrt{\dfrac{150}{7.20\times {{10}^{-3}}}}=144.34m{{s}^{-1}}$ ………………… (2)
Now, let us find the wavelength from the given figure.

We define wavelength as the distance between the two consecutive crests or troughs in a wave, which clearly in the above figure is two among the three parts of the distance between two fixed supports D. So, wavelength here is,
$\lambda =\dfrac{2}{3}\left( 90.0cm \right)=60.0cm$ …………………. (3)
Now, recall the expression for frequency of wave in terms of speed and wavelength.
$f=\dfrac{v}{\lambda }$
Where, f is the frequency.
From (2) and (3) frequency f of the wave is,
$f=\dfrac{144.34m{{s}^{-1}}}{60.0\times {{10}^{-2}}m}=240.57{{s}^{-1}}=240.57Hz$
Hence, we get the values of velocity, wavelength and frequency of the given wave as the following,
$v=144.34m{{s}^{-1}}$
$\lambda = 60.0 cm = 0.6 m$
$f=240.57Hz$

Note:
In order to get the frequency in Hz, make sure that you substitute the values in their SI units. All the required commodities in order to solve this question can be acquired directly from the question and the given wave pattern, if you know how to look for them. Also, solve this question in a step by step manner, in case if you were to begin from finding the frequency, you wouldn’t end up anywhere.