PES 2130 Fall 2014, Spendier Lecture 10/Page 1

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1 PES 2130 Fall 2014, Spendier Lecture 10/Page 1 Lecture today: Chapter 16 Waves-1 1) Wave on String and Wave Equation 2) Speed of a transverse wave on a stretched string 3) Energy and power of a wave traveling along a string 4) The Principle of Superposition for Waves 5) Interference of Waves Announcements: - Exam 1 graded - HW 4 given out Lecture before exam 1: Defined wave as a disturbance (or oscillation) that propagates in space and in time. - Oscillating wave (traveling wave pulses that repeat themselves) - Soliton wave (a solitary, traveling wave pulse) Equation to describe oscillatory waves (wave function): y(x,t) = displacement x = position t = time Amplitude (magnitude): y m Wavelength: λ [m] Angular wave number: k = 2π/ λ [rad/m] or [1/m] Period: T [s] (the wave repeats itself every time T) Frequency: f = 1/T = ω/2π [1/s] = [Hz] (How many wave cycles pass through a given point x 0 per second) Angular frequency (angular speed): ω = 2π/ T [rad/s] (one repletion / time it takes to repeat itself = 2 π/t) Phase constant: ϕ [rad] Phase of the wave: kx-ωt

2 PES 2130 Fall 2014, Spendier Lecture 10/Page 2 The speed of a traveling wave: v = dx/dt= ω/k = λ/t Direction of a traveling wave: How to describe a wave that is moving to the right/left? We have y = y(x,t) for a moving pattern If you use a reference frame (call it x ) that is moving with the pulse at speed v. The origin of this pulse (or origin of moving frame) is at: x f = vt Take x as a coordinate in the moving frame, then x = x f + x and x = x x f = x - vt If we take a curve y(x ) which is at rest in the moving frame and has a peak at x =0 Example: bell curve To describe this function in the rest frame substitute in x = x - vt We see that this curve must move into the positive x-direction: To stay at the same x, to stay at the same position on the curve, position x must increase as time increases. The same argument applies to any function: y = f(x-vt) y = f(x+vt) is a wave moving in the positive x direction at speed v is a wave moving in the negative x direction at speed v Let s apply this to our wave function for an oscillatory wave: Wave traveling in positive x direction: y(x, t) = y m sin(kx - ωt) Wave traveling in negative x direction: y(x, t) = y m sin(kx + ωt)

3 PES 2130 Fall 2014, Spendier Lecture 10/Page 3 Transverse velocity and wave equation: The speed of a wave v (or wave propagation speed, the speed at which a bump/crest travels), is v = dx/dt = ω/k However, each element of the string has also a velocity transverse to the string (up, down, up down). In other words: If you have a wave on a water surface, the 'regular speed' is how fast the pattern moves horizontally and it is a steady value (v). But the transverse speed is how fast a point on the surface is moving up and down; the velocity changes like the velocity of a mass on a spring, bouncing vertically. So how do we calculate this transverse velocity (v y ) of a particular point x on a string for example? We simply take the derivative of the wave function y(x,t) with respect to t, keeping x constant: Let s do this, given the following wave function: y(x, t) = Acos(kx - ωt) This transverse velocity varies with time, as expected for simple harmonic motion. The in this expression is a modified d, used to remind us that y(x,t) is a function of two variables and that we allowing only (t) to vary. The other (x) is a constant because we are looking at a particular point on the string. This derivative is called a partial derivative. If you haven t reached this point yet in calculus, don t worry it is a simple idea. So now, we can also calculate the acceleration of a particle in a transverse wave: ( ) The acceleration of a particle on a transverse wave equals which is a result you should recognize from Physics 1. times its displacement, We can also compute the partial derivative with respect to (x) holding (t) constant: Which is the slope of the string at point x

4 PES 2130 Fall 2014, Spendier Lecture 10/Page 4 The second partial derivative with respect to (x) holding (t) constant gives us the curvature of the string at point x: Now we can use the relation for the speed of the wave: v 2 = ω 2 /k 2 Rearrange: and we arrive at the wave equation. The wave equation is one of the most important equations in all of physics. A traveling wave in the positive and negative direction will satisfy this wave equation. Here we derived it for a wave traveling into the positive x direction. Let s show that a wave traveling into the negative x direction also satisfies the wave equation: Example: Show that y(x, t) = A sin(kx + ωt) is a solution to the wave equation

5 PES 2130 Fall 2014, Spendier Lecture 10/Page 5 Speed of a transverse wave on a stretched string So, we talked a lot about one key property of any wave which is the wave speed. Light waves in air have a much greater speed of propagation than do sound waves in air (3.00 x 10 8 m/s versus 344 m/s). Reason for why you see the flash from a lightning bolt before you hear the thunder It turns out that the speed of many kinds of mechanical waves turn out to have the same basic mathematical expressions as does the speed of waves on a string which we will discuss next. What determines the speed of a transverse wave on a string? - Tension τ in the string (increasing tensions will increase the restoring force, increase the speed) - Mass per unit length µ (increasing the mass will make motion more sluggish and decrease the speed The book shows two derivations following these one can show that the speed of a wave along a stretched ideal string depends only on the tension and linear density of the string and not on the frequency of the wave: [ ] Τ= Fτ = ma = [kg*m/s 2 ] µ = mass/length = kg/m What about the speed for other types of mechanical waves? They all have the same general form: Tension in the string tends to bring the string back to its undisturbed, equilibrium configuration. The linear mass density provides the inertia that prevents the string from returning instantaneously to equilibrium.

6 PES 2130 Fall 2014, Spendier Lecture 10/Page 6 Energy and power of a wave traveling along a string Every wave motion has energy associated with it. Energy we receive from sunlight (sun burn) Energy of ocean surf (beach erosion) Energy from earthquakes (very destructive) To produce any wave motion, we have to apply a force to a portion of the wave medium; the point where the force is applied moves, so we do work on the system. As the wave propagates, each portion of the medium exerts force and does work on the adjoining portion. In this way we can transport energy from one region of space to another.

7 PES 2130 Fall 2014, Spendier Lecture 10/Page 7 Application: Surface waves and the Swimming speed of ducks Example: string When we set up a wave on a stretched string, we provide energy for the motion of the string. As the wave moves away from us, it transports that energy as both kinetic energy and elastic potential energy Kinetic energy: associated with its transverse velocity (just like simple harmonic motion mass on a spring) When the string element of mass dm oscillates transversely in simple harmonic motion, it is rushing through its y = 0 position (element b in Fig. 16-9), its transverse velocity and thus its kinetic energy is a maximum. When the element is at its extreme position y = y m (as is element a), its transverse velocity and thus its kinetic energy is zero. Elastic potential energy: associated with stretchiness of string (just like simple harmonic motion mass on a spring, spring stiffness!) To send a sinusoidal wave along a previously straight string, the wave must necessarily stretch the string. As a string element of length dx oscillates transversely, its length must increase and decrease in a periodic way if the string element is to fit the sinusoidal wave form. Elastic potential energy is associated with these length changes, just as for a spring. When the string element is at its y = y m position (element a in Fig. 16-9), its length has its normal undisturbed value dx, so its elastic potential energy is zero. However, when the element is rushing through its y = 0 position, it has maximum stretch and thus maximum elastic potential energy.

8 PES 2130 Fall 2014, Spendier Lecture 10/Page 8 Energy transport: The oscillating string element thus has both its maximum kinetic energy and its maximum elastic potential energy at y = 0. In the snapshot of Fig. 16-9, the regions of the string at maximum displacement have no energy, and the regions at zero displacement have maximum energy. As the wave travels along the string, forces due to the tension in the string continuously do work to transfer energy from regions with energy to regions with no energy. One can derive an expression (see book) for the average power of a transverse wave (the average rate at which energy of both kinds is transmitted by the wave) µ Mass per unit length v wave speed ω angular frequency (angular speed) y m amplitude (magnitude) of wave