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Standing waves on a string

Consider a string of L = 2.00 m . attached to an adjustable-frequency string vibrator as shown in [link] . The waves produced by the vibrator travel down the string and are reflected by the fixed boundary condition at the pulley. The string, which has a linear mass density of μ = 0.006 kg/m, is passed over a frictionless pulley of a negligible mass, and the tension is provided by a 2.00-kg hanging mass. (a) What is the velocity of the waves on the string? (b) Draw a sketch of the first three normal modes of the standing waves that can be produced on the string and label each with the wavelength. (c) List the frequencies that the string vibrator must be tuned to in order to produce the first three normal modes of the standing waves.

A string vibrator is shown on the left of the figure. A string is attached to its right. This goes over a pulley and down the side of the table. A hanging mass m = 2 kg is suspended from it. The pulley is frictionless. The distance between the pulley and the string vibrator is L = 2 m. It is labeled mu equal to delta m by delta x equal to 0.006 kg per m.
A string attached to an adjustable-frequency string vibrator.

Strategy

  1. The velocity of the wave can be found using v = F T μ . 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.
    Figure a shows a string attached at both ends. Two waves on the string form a node at either end and another one in the centre. This is labeled possible mode. Figure b shows a string attached at both ends. Two waves on the string form a node at one end of the string and an antinode at the other. This is labeled impossible mode.
    (a) The figure represents the second mode of the string that satisfies the boundary conditions of a node at each end of the string. (b)This figure could not possibly be a normal mode on the string because it does not satisfy the boundary conditions. There is a node on one end, but an antinode on the other.

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 = F T μ = m g μ = 2 kg ( 9.8 m s ) 0.006 kg m = 57.15 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.
    Three figures of a string of length L=2 m are shown. Each has two waves. The first one has 1 node. It is labeled half lambda 1 = L, lambda 1 = 2 by 1 times 2 m = 4 m. The second figure has 2 nodes. It is labeled lambda 2 = L, lambda 2 = 2 by 2 times 2 m = 2 m. The third figure has three nodes. It is labeled 3 by 2 times lambda 3 = L, lambda 3 = 2 by 3 times 2 m = 1.33 m.
  3. The frequencies of the first three modes are found by using f = v w λ .
    f 1 = v w λ 1 = 57.15 m/s 4.00 m = 14.29 Hz f 2 = v w λ 2 = 57.15 m/s 2.00 m = 28.58 Hz f 3 = v w λ 3 = 57.15 m/s 1.333 m = 42.87 Hz

Significance

The three standing modes in this example were produced by maintaining the tension in the string and adjusting the driving frequency. Keeping the tension in the string constant results in a constant velocity. The same modes could have been produced by keeping the frequency constant and adjusting the speed of the wave in the string (by changing the hanging mass.)

Visit this simulation to play with a 1D or 2D system of coupled mass-spring oscillators. Vary the number of masses, set the initial conditions, and watch the system evolve. See the spectrum of normal modes for arbitrary motion. See longitudinal or transverse modes in the 1D system.

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

λ n = 2 n L n = 1 , 2 , 3 , 4 , 5 ... and f n = n v 2 L = n f 1 n = 1 , 2 , 3 , 4 , 5 ...

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.

Three figures of a string of length L are shown. Each has two waves. The first one has 1 node. It is labeled lambda 1 = 2 by 1 times L, f1 = vw by lambda 1 = vw by 2L. The second figure has 2 nodes. It is labeled lambda 2 = 2 by 2 times L, f2 = vw by lambda 2 = vw by L. The third figure has three nodes. It is labeled lambda 3 = 2 by 3 times L, f3 = vw by lambda 3 equal to 3 times vw by 2L.

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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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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