16.6 Standing waves and resonance  (Page 4/17)

Standing waves on a string

Strategy

1. The velocity of the wave can be found using $v=\sqrt{\frac{F_T}{\mu }}.$ The tension is provided by the weight of the hanging mass.
2. The standing waves will depend on the boundary conditions. There must be a node at each end. The first mode will be one half of a wave. The second can be found by adding a half wavelength. That is the shortest length that will result in a node at the boundaries. For example, adding one quarter of a wavelength will result in an antinode at the boundary and is not a mode which would satisfy the boundary conditions. This is shown in [link] .
3. Since the wave speed velocity is the wavelength times the frequency, the frequency is wave speed divided by the wavelength.
   

   

Solution

1. Begin with the velocity of a wave on a string. The tension is equal to the weight of the hanging mass. The linear mass density and mass of the hanging mass are given:
   
   $v=\sqrt{\frac{F_T}{\mu }}=\sqrt{\frac{mg}{\mu }}=\sqrt{\frac{2\text{kg}\left(9.8\frac{\text{m}}{\text{s}}\right)}{0.006\frac{\text{kg}}{\text{m}}}}=57.15\text{m/s}.$
2. The first normal mode that has a node on each end is a half wavelength. The next two modes are found by adding a half of a wavelength.
   
3. The frequencies of the first three modes are found by using $f=\frac{v_w}{\lambda }.$
   
   $\begin{array}{}\\ f_1=\frac{v_w}{{\lambda }_1}=\frac{57.15\text{m/s}}{4.00\text{m}}=14.29\text{Hz}\hfill \\ f_2=\frac{v_w}{{\lambda }_2}=\frac{57.15\text{m/s}}{2.00\text{m}}=28.58\text{Hz}\hfill \\ f_3=\frac{v_w}{{\lambda }_3}=\frac{57.15\text{m/s}}{1.333\text{m}}=42.87\text{Hz}\hfill \end{array}$

Significance

Check Your Understanding The equations for the wavelengths and the frequencies of the modes of a wave produced on a string:

$\begin{array}{}\\ {\lambda }_n=\frac{2}{n}Ln=1,2,3,4,5\text{...}\text{and}\hfill \\ f_n=n\frac{v}{2L}=n{f}_{1}n=1,2,3,4,5\text{...}\hfill \end{array}$

were derived by considering a wave on a string where there were symmetric boundary conditions of a node at each end. These modes resulted from two sinusoidal waves with identical characteristics except they were moving in opposite directions, confined to a region L with nodes required at both ends. Will the same equations work if there were symmetric boundary conditions with antinodes at each end? What would the normal modes look like for a medium that was free to oscillate on each end? Don’t worry for now if you cannot imagine such a medium, just consider two sinusoidal wave functions in a region of length L , with antinodes on each end.

Yes, the equations would work equally well for symmetric boundary conditions of a medium free to oscillate on each end where there was an antinode on each end. The normal modes of the first three modes are shown below. The dotted line shows the equilibrium position of the medium.

Note that the first mode is two quarters, or one half, of a wavelength. The second mode is one quarter of a wavelength, followed by one half of a wavelength, followed by one quarter of a wavelength, or one full wavelength. The third mode is one and a half wavelengths. These are the same result as the string with a node on each end. The equations for symmetrical boundary conditions work equally well for fixed boundary conditions and free boundary conditions. These results will be revisited in the next chapter when discussing sound wave in an open tube.

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