Waves Pulse Propagation The Wave Equation

Transcription

1 Waves Pulse Propagation The Wave Equation Lana Sheridan De Anza College May 16, 2018

2 Last time oscillations simple harmonic motion (SHM) spring systems energy in SHM introducing waves kinds of waves wave speed on a string

3 Warm Up Question If you stretch a rubber hose and pluck it, you can observe a pulse traveling up and down the hose. What happens to the speed of the pulse if you stretch the hose more tightly? (A) It increases. (B) It decreases. (C) It is constant. (D) It changes unpredictably. 1 Serway & Jewett, page 499, objective question 2.

4 Warm Up Question If you stretch a rubber hose and pluck it, you can observe a pulse traveling up and down the hose. What happens to the speed of the pulse if you stretch the hose more tightly? (A) It increases. (B) It decreases. (C) It is constant. (D) It changes unpredictably. 1 Serway & Jewett, page 499, objective question 2.

5 Warm Up Question If you stretch a rubber hose and pluck it, you can observe a pulse traveling up and down the hose. What happens to the speed if you fill the hose with water? (A) It increases. (B) It decreases. (C) It is constant. (D) It changes unpredictably. 1 Serway & Jewett, page 499, objective question 2.

6 Warm Up Question If you stretch a rubber hose and pluck it, you can observe a pulse traveling up and down the hose. What happens to the speed if you fill the hose with water? (A) It increases. (B) It decreases. (C) It is constant. (D) It changes unpredictably. 1 Serway & Jewett, page 499, objective question 2.

7 Overview pulse propagation the wave equation

8 e string passes over a pulley and sup- Example A uniform string has a mass of m s and a length of l. The string passes over a pulley and supports an block of mass m b. Find the speed of a pulse traveling along this string. (Assume the vertical piece of the rope is very short.) (Example ion T in the ined by the ect. The ave traveling is given by T 2.00 kg m block g 5 0

9 e string passes over a pulley and sup- Example A uniform string has a mass of m s and a length of l. The string passes over a pulley and supports an block of mass m b. Find the speed of a pulse traveling along this string. (Assume the vertical piece of the rope is very short.) (Example ion T in the ined by the ect. The ave traveling is given by T 2.00 kg m block g 5 0 v = mb gl m s

10 Pulse Propagation A wave pulse (in a plane) at a moment in time can be described in terms of x and y coordinates, giving y(x). We know that the pulse will move with speed v and be displaced, say in the positive x direction, while maintaining its shape. That means we can also give y as a function of time, y(x, t). Consider a moving reference frame, S, with the pulse at rest, y (x ) = f (x ), no time dependence. Galilean transformation: x = x vt

11 to earthquakes). The slower transverse r secondary, travel through the Earth at the Pulse time interval Propagation between the arrivals, the distance from the seismograph to mined. This distance is the radius of an raph. The origin of the waves is located spheres from three or more monitoring r intersect at one region of the Earth, x = x vt Then in the rest-frame of the string urred. n a long string as shown in Figure ition of the pulse at time t 5 0. At this y be, can be represented by some math- 0) 5 f(x). This function describes the string located at each value of x at time, the pulse has traveled to the right a ssume the shape of the pulse does not shape of the pulse is the same as it was tly, an element of the string at x at this located at x 2 vt had at time t 5 0: x 2 vt, 0) ansverse position y for all positions and the origin at O, as x 2 vt) (16.1) transverse positions of elements of the displacements. y(x, t) = f (x At t 0, the ) shape = of f the (x vt) a pulse is given by y f(x). O y y S v vt P P S v x x 1 vt) (16.2) ave function, depends on the two variten y(x, t), which is read y as a function O x

12 Pulse Propagation The shape of the pulse is given by f (x) and can be arbitrary. Whatever the form of f, if the pulse moves in the +x direction: y(x, t) = f (x vt) If the pulse moves in the x direction: y(x, t) = f (x + vt)

13 Wave Pulse Example 16.1 A pulse moving to the right along the x axis is represented by the wave function y(x, t) = 2 (x 3.0t) where x and y are measured in centimeters and t is measured in seconds. What is the wave speed? Find expressions for the wave function at t = 0, t = 1.0 s, and t = 2.0 s. 1 Serway & Jewett, page This function is an unnormalized Cauchy distribution, or as physicists say it has a Lorentzian profile.

14 Wave Pulse Example 16.1 A pulse moving to the right along the x axis is represented by the wave function y(x, t) = 2 (x 3.0t) where x and y are measured in centimeters and t is measured in seconds. What is the wave speed? 3.0 cm/s Find expressions for the wave function at t = 0, t = 1.0 s, and t = 2.0 s. 1 Serway & Jewett, page This function is an unnormalized Cauchy distribution, or as physicists say it has a Lorentzian profile.

15 2 Wave Pulse Example 16.1 ple 16.1 A Pulse Moving to the Right moving to the right along the x axis is represented by the wave n y1x, t x 2 3.0t t = 0, y(x, 0) = 2 x x and y are measured in centimeters and t is measured in secind expressions for the wave function at t 5 0, t s, and s. TION 2 t = 1, y(x, 1) = (x 3.0) tualize Figure 16.6a shows the pulse represented by this wave n at t 5 0. Imagine this pulse moving to the right at a speed m/s and maintaining its shape as suggested by Figures 16.6b.6c. rize We categorize this example as a relatively simple analysis m in which we interpret the mathematical representation of a 2 t = 2, y(x, 2) = (x Figure 6.0) e The wave function is of the form y 5 vt). Inspection of the expression for nd comparison to Equation 16.1 reveal e wave speed is v cm/s. Furthery letting x 2 3.0t 5 0, we find that the um value of y is given by A cm. their location, and the resultant pulse moves around the stadium. Is this wave (a) transverse or (b) longitudinal? (Example 16.1) Graphs of the function y(x, t) 5 2/[(x 23.0t) 2 1 1] at (a) t 5 0, (b) t s, and (c) t s. a b c y (cm) t cm/s y (x, 0) y (cm) y (cm) cm/s t 1.0 s 7 8 y (x, 1.0) cm/s t 2.0 s y (x, 2.0) x (cm) x (cm) x (cm)

16 The Wave Equation tan u Can to express we find a general equation describing the displacement (y) of our medium as a function of position (x) and time? (16.22) Start by considering a string carrying a disturbance. he right end of e force T S. This resented by the xis for this disinstant of time, gents in Equa- u A A x T S Figure An element of a B u B T S (16.23) string under tension T.

17 sin u < tan u to express The Wave Equation Consider a(16.22) small length of string x. d from the right end of enting the force T S. This n be represented by the the x axis for this disrticular instant of time, r the tangents in Equa- u A A x B u B T S T S (16.23) Figure An element of a As we did for oscillations, string under start tension from Newton s T. 2nd law. For small angles F y = ma y T sin θ B T sin θ A = (µ x) 2 y t 2 sin θ tan θ

18 The Wave Equation We can write tan θ as the slope of y(x): tan θ = y x Now Newton s second law becomes: ( y T x y ) x=b x = (µ x) 2 y x=a t 2 µ 2 y y x y T t 2 = x=b x x x=a

19 The Wave Equation We can write tan θ as the slope of y(x): tan θ = y x Now Newton s second law becomes: ( y T x y ) x=b x = (µ x) 2 y x=a t 2 µ 2 y y x y T t 2 = x=b x x x=a We need to take the limit of this expression as we consider an infinitesimal piece of string: x 0.

20 The Wave Equation We can write tan θ as the slope of y(x): tan θ = y x Now Newton s second law becomes: ( y T x y ) x=b x = (µ x) 2 y x=a t 2 µ 2 y y x y T t 2 = x=b x x x=a We need to take the limit of this expression as we consider an infinitesimal piece of string: x 0. µ 2 y T t 2 = 2 y x 2 where we use the definition of the partial derivative.

21 The Wave Equation µ 2 y T t 2 = 2 y x 2 Remember that the speed of a wave on a string is T v = µ The wave equation: 2 y x 2 = 1 2 y v 2 t 2 Even though we derived this for a string, it applies much more generally!

22 The Wave Equation We can model longitudinal waves like sound waves by a series of masses connected by springs, length h. u 1 u 2 u 3 u is a function that gives the displacement of the mass at each equilibrium position x, x + h, etc. For such a case, the propagation speed is KL v = µ where K is the spring constant of the entire spring chain, L is the length, and µ is the mass density. 1 Figure from Wikipedia, by Sebastian Henckel.

23 The Wave Equation u 1 u 2 u 3 u is a function that gives the displacement of the mass at each equilibrium position x, x + h, etc. Consider the mass, m, at equilibrium position x + h F = ma k(u 3 u 2 ) k(u 2 u 1 ) = m 2 u t 2 m k 2 u t 2 = u 3 2u 2 + u 1 1 Figure from Wikipedia, by Sebastian Henckel.

24 The Wave Equation m k 2 u t 2 = u 3 2u 2 + u 1 We can re-write m k in terms of quantities for the entire spring chain. Suppose there are N masses. m = µl N and k = NK and N = L h µ 2 u u(x + 2h) 2u(x + h) + u(x) = KL t2 h 2 Letting N and h 0, the RHS is the definition of the 2nd derivative. Same equation! 1 2 u v 2 t 2 = 2 u x 2

25 The Wave Equation The wave equation: 2 y x 2 = 1 2 y v 2 t 2 We derived this for a case of transverse waves (wave on a string) and a case of longitudinal waves (spring with mass). It applies generally!

26 Solutions to the Wave Equation Earlier we reasoned that a function of the form: y(x, t) = f (x ± vt) should describe a propagating wave pulse. Notice that f does not depend arbitrarily on x and t. It only depends on the two together by depending on u = x ± vt. Does it satisfy the wave equation? 2 y x 2 = 1 2 y v 2 t 2 (Next lecture...)

27 Summary wave speed on a string pulse propagation the wave equation Homework Serway & Jewett: Ch 16, onward from page 499. OQs: 5; Probs: 3, 23, 24, 29, 53, 59, 60