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Lecture13 - Wave Mo/on Waves When you setup an...

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Unformatted text preview: Wave Mo/on Waves When you setup an oscilla/on in one element of an extended object or medium, this element will exert forces on neighboring elements and will make them oscillate as well. Thus the oscilla/on will ‘travel’ along the extended object. A traveling oscilla/on is called a ‘wave’. Chapter 16 In a wave, individual elements oscillate back and forth, but they do not themselves travel along the direc/on of the wave. What travels is the mo/on or disturbance which is transmiGed through the medium as neighboring elements cause each other to oscillate. Transverse and Longitudinal Waves Depending on the direc/on in which individual elements of the medium move with respect to the direc/on in which the wave travels, we dis/nguish: 1. Transverse Waves: Elements of the medium oscillate perpendicular to the wave direc/on. 2. Longitudinal Waves: Elements of the medium oscillate parallel to the direc/on of the wave travel. Travelling wave model Recall: For an oscilla/on, the displacement at a /me t is given by: x (t ) = A cos(ω t + ϕ ) For a wave, each point oscillates according to the above formula, but different points along the wave have different phases. The oscillatory mo/on does not only repeat itself in /me (period, T), but also in space, i.e. two points separated by a certain distance, called the wavelength, undergo synchronous oscilla/ons. Just as the period T has a corresponding frequency ω, the wavelength λ has a corresponding ‘spa/al frequency’, k, called the wave number. While ω tells us how fast the phase angle changes in /me, k tells us how fast the phase angle changes with distance. Consider a wave that looks like this at /me t = 0: Oscilla/on of x = 0 element: y(t ) = A sin(ω t ) (Remember for simple harmonic mo/on, we can either use sine or cosine, the sine fits the situa/on shown in the picture) A Travelling wave model (contd…) Now include the spa/al dependence: y( x, t ) = A sin ( kx + ω t ) At t=0 and x=0, the phase is zero in the above picture. Where would this phase move to at some /me t>0, i.e. in which direc/on does the wave travel ? ω ⋅ 0 + k ⋅ 0 = ω t + kx ⇒ x < 0 if t > 0 i.e. the way we have it wriGen, the wave is traveling to the le\. So in order to have it travel to the right we need: y( x, t ) = A sin( kx − ω t ) Amplitude, A, for a wave is the same as for an oscilla/on: the maximum displacement of each element as they oscillate. The phase tells us, at a /me t and posi/on x, in which stage of its oscilla/on each element is. Each element will repeat its mo/on a\er one period T and elements separated by the wavelength λ, will be performing the same mo/on. A repeat of the mo/on will occur when the phase changes by 2π (since sin 2π = sin 0). k λ = 2π ⇒ k = 2π λ Also note that frequency is defined as before: A wave has an amplitude of 0.1m, a period of 1 second and a wavelength of 1.5m. At Bme t=0, the element at x=0 is at a displacement of y = 0.05 m. What is the displacement funcBon and what is the displacement of an element at x = 1.2 m at Bme t = 2 seconds ? We now include a phase constant, ϕ, since at t = 0 and x = 0, because y is not equal to zero: Example y( x, t ) = A sin( kx − ω t + ϕ ) What is the value of ϕ? Displacement ‘y’, at /me t = 2 s and posi/on x = 1.2 m: Speed of a wave When a wave moves, the ‘paGern’ of the wave is moving. The paGern is determined by where individual elements along the wave are, in terms of their oscilla/ons. This in turn is determined by the phase. Thus what really moves are points of constant phase: Taking the /me deriva/ve yields: From this we find: v= ωλ = = λf kT Note: The expression λ/T makes easy sense: All elements will go through one complete oscilla/ons during a period T. Since at this point the wave will look exactly the same again, the wave will have traveled one wavelength . Thus v = λ/T. Wave speed on a stretched string The speed of a wave depends on the medium through which it moves. Once the speed is known, it determines the rela/onship between wavelength and frequency via v = lf. Let’s use Newton’s second law to calculate the speed of a wave on string. Also assume that we are moving along with the traveling pulse, i.e. the string is moving past us. Restoring force: due to tension in the string. The lateral components of the tension cancel out, but the normal components remain. The restoring force on a small element Dl of the string is: Mass of element: (where m is the linear mass density m = m/L) τ θ Since we are moving along with the pulse, from our point ­of ­view the string element is moving past us with speed v, moving along a circle of radius R as shown. In this case we find for the accelera/on of the string element: Newton’s second law: Solve for v: v= τ µ Energy of a wave To start a wave mo/on, energy is required. As each element of the medium exerts a force on its neighboring elements, it also does work on the neighboring elements, i.e. it transfers energy. While in a typical wave, no maGer is transported, a wave will always transport energy. The total energy is a combina/on of kine/c energy due to the oscilla/on of all elements along the medium and the poten/al energy due to the restoring force in the medium (e.g. the elas/c poten/al energy in a string). For kine/c Energy: Need to know the speed of a mass element as it oscillates. Since it oscillates only in the y ­direc/on and does not actually moves in the x ­direc/on, we want to take the deriva/ve of y, while keeping x constant (this is called a ‘par/al’ deriva/ve): dy ∂y ∂ u= = = ( A sin( kx − ω t )) = −ω A cos ( kx − ω t ) 1 1 ( dm) u 2 = ( µ dx )ω 2 A2 cos 2 ( kx − ω t ) 2 2 Integra/ng over a wavelength, we get total kine/c energy in K λ = 1 µω 2 A2 λ 4 one wavelength (see page 463, textbook) dt x = const ∂t ∂t Kine/c energy of a mass element: dK = In the same way, we get the expression for poten/al energy in one wavelength: Thus the total energy in one wavelength: 1 µω 2 A2 λ 4 1 Eλ = µω 2 A2 λ 2 Kλ = Energy transmission and power The total energy transmiGed in one wavelength: Therefore the average power transmiGed is : 1 µω 2 A2 λ 2 ΔE Eλ 1 λ Pav = = = µω 2 A2 Δt T 2 T 1 ⇒ Pav = µω 2 A2 v 2 Eλ = Thus the rate of energy transfer (i.e., power) in any sinusoidal wave is propor9onal to the square of the angular frequency and to the square of the amplitude of the wave. A stretched string has linear density μ =525 g/m and is under tension τ = 45 N. We send a sinusoidal wave with frequency f = 120 Hz and amplitude A = 8.5 mm along the string. What is the average rate at which the wave transports energy ? Pav = 1 µω 2 A2 v 2 Example The wave equa/on We found some differen/al equa/ons to describe oscilla/ons (free, damped, or forced) by using Newton’s second law. Can we do the same for waves ? Note : Waves depend on x, y and t! Need par/al deriva/ves. Let’s think of a wave on a string. An an element of string that is /lted, but completely straight. We can see that for a straight piece of τ τ string under tension τ, the x ­ and y ­ component of the forces cancel out, so there is no net force on the string element. However, if the string is bent, the y ­components of the forces will be different, and there will be net restoring force on the string element. The ra/o of y ­ and x ­component of the force is just the local slope of the string: where we assumed that the slope is small and therefore Fx ≈ τ. So, the net force on the element is the difference between the y ­components of the tension in each end: The wave equa/on (contd…) Newton’s second law: Take the deriva/ve w.r.t. x: Now using and wri/ng deriva/ves as par/als: ∂2 y 1 ∂2 y =2 2 2 ∂x v ∂t Note: the first term indicates the curvature of the string. This equa/on turns out to hold for all types of waves! Ques<on: Is our wave model a solu/on to this equa/on ? y( x, t ) = A sin( kx − ω t ) ∂y = k A cos ( kx − ω t ) ∂x ∂2 y = − k 2 A sin ( kx − ω t ) 2 ∂x ⇒ − k 2 A sin ( kx − ω t ) = − With v = ω , our wave model sa/sfies the wave equa/on. k 12 ω A sin ( kx − ω t ) v2 ...
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