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Lecture 7: Many Coupled Oscillators & Wave Equations

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Lecture Topics

A diagram showing a black line going downward diagonally, with three red circles on it.
  • Many Coupled Oscillators
  • Wave Equation
  • Transverse Traveling Pulses and Waves

Learning Objectives

By the end of this lecture, you should:

  • understand transverse and longitudinal modes of a linear system, and boundary conditions.
  • know how to analyze the motion of a particle on a string of many particles.
  • know how to use normal modes to solve the system of differential equations for particles on a string.
  • understand how a given motion of a set of particles may be made up of a combination of normal mode motions.
  • understand the transition to a continuous string, including the change in normal mode frequencies.
  • set up the equation of motion for a small piece of a string and obtain a wave equation.
  • show that any single-valued function f(x±ct) solves the wave equation.
  • understand the effect of boundary conditions on waves (reflections).

Lecture Activities

Check Yourself

  • The derivative of a quantity with respect to time is a limit of a change in that quantity, divided by a change in time, as the time change approaches zero. Thus its units are [1/s] times the units of the original quantity. Knowing and extending this, verify that \[\frac{1}{{{v^2}}}\frac{{{\partial ^2}y}}{{\partial {t^2}}} = \frac{{{\partial ^2}y}}{{\partial {x^2}}}\]is a dimensionally correct form of the wave equation, no matter what the physical quantity represented by y.

› View/hide answer

If [y] are the units of y, then the units of the LHS are 1/[m2/s2] [y]/[s2] = [y][s2]/([m2][s2]) = [y]/[m2]. The units of the RHS are [y]/[m2]. The equation is dimensionally correct.

 

  • A person pulling hard on a rope can likely generate about 100 N since that would be about equivalent to pulling up a 10 kg mass, which most people could do. A fairly strong nylon rope could have a linear mass density of about 100 g/m. If a person pulls on such a rope attached to a wall, and a second person "twangs" it perpendicularly, at what speed will waves travel on the rope?

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The speed is \(v = \sqrt {\frac{T}{\mu }} \)in SI units. Here then \[v = \sqrt {\frac{{100}}{{0.1}}} = \sqrt {1000} = 32m/s\] This is about the speed of an automobile on a highway.

 

  • By very cleverly moving the end of the nylon rope described, located at x = 0, a Gaussian pulse is created which at time t = 0 has the profile \(y = 0.05{e^{ - {x^2}}}\)(for x>0) and is moving toward the wall fixation point to the right. Assuming the end of the string is moved so as to generate the other half of the Gaussian pulse, what is the equation of this wave?

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A function of form \(f(x - vt)\) moves toward the right and v is known to be 32 m/s. The wave is thus described by \[y = 0.05{e^{ - {{(x - 32t)}^2}}}\] Although this strictly speaking has a non-zero value at all points on the rope, in practice it is highly localized within about a meter of the nominal position of the pulse.

 

  • The pulse referred to above arrives at the fixation point, which is 32 m away, and is reflected. What wave do forces at the fixation point produce, as necessitated by the condition that the end of the rope remains stationary at that point?

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The pulse center arrives at the fixation point after 1 s. A leftward moving negative pulse is generated at the fixation point, with 1 s delay from the original pulse. This means that its center is at x = 32 at t = 1 s, and it is a leftward traveling pulse. Its equation is thus \[y = - 0.05{e^{ - {{(x - 32 + 32(t - 1))}^2}}} = - 0.05{e^{ - {{(x - 64 + 32t)}^2}}}\]

 

  • Check that the end point where the pulse reflects actually has displacement zero at all times, AND show that when the reflected pulse comes to the origin, it is a "valley" of opposite sign to the original pulse. This verifies that a fixed end point makes a "mountain come back as a valley."

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The reflection point is at x = 32 m. Putting x = 32 into the solutions allows evaluating them at all times at that point. The original pulse is\[y = 0.05{e^{ - {{(x - 32t)}^2}}} = 0.05{e^{ - {{(32 - 32t)}^2}}}\] The reflected pulse is \[y = - 0.05{e^{ - {{(32 - 64 + 32t)}^2}}} = - 0.05{e^{ - {{( - 32 + 32t)}^2}}}\] Due to the squared argument, these are equivalent functions of opposite sign and their superposition as the total solution always gives zero at x = 32. The reflected pulse is back at the origin at time t = 2. At this time we can evaluate the function for all x and see what it looks like. \[y = - 0.05{e^{ - {{(x - 64 + 32(2))}^2}}} = - 0.05{e^{ - {x^2}}}\]represents the rope at this time if x>0. This is exactly the reversed profile of the initial function.

 

  • As a final useful minor exercise, show by explicit differentiation that a traveling Gaussian pulse \(y = {e^{ - {{(x - vt)}^2}}}\)satisfies the wave equation. Your steps may vary but one suggested way is shown as the answer.

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\[\begin{gathered} \frac{{\partial y}}{{\partial t}} = - 2(x - vt)( - v){e^{ - {{(x - vt)}^2}}} = (2vx - 2{v^2}t){e^{ - {{(x - vt)}^2}}}  \\ \frac{{{\partial ^2}y}}{{\partial {t^2}}} = - 2{v^2}{e^{ - {{(x - vt)}^2}}} + {(2vx - 2{v^2}t)^2}{e^{ - {{(x - vt)}^2}}} \\ \frac{{\partial y}}{{\partial x}} = - 2(x - vt){e^{ - {{(x - vt)}^2}}} = ( - 2x + 2vt){e^{ - {{(x - vt)}^2}}}  \\ \frac{{{\partial ^2}y}}{{\partial {x^2}}} = - 2{e^{ - {{(x - vt)}^2}}} + {( - 2x + 2vt)^2}{e^{ - {{(x - vt)}^2}}}  \\ \end{gathered}\] Comparison of the second and fourth lines shows that \[\frac{1}{{{v^2}}}\frac{{{\partial ^2}y}}{{\partial {t^2}}} = \frac{{{\partial ^2}y}}{{\partial {x^2}}}\]as the wave equation requires, so that \(y = {e^{ - {{(x - vt)}^2}}}\)is indeed a solution.

 

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