Physics 214 Spring 1997 PROBLEM SET 2. Solutions

Transcription

1 Physics 214 Spring 1997 PROBLEM SET 2 Solutions 1. Tipler, Chapter 13, p.434, Problem 6 The general expression for the displacement field associated with a traveling wave is y(x,t) = f(x vt) in which v is the wave speed, with v positive for waves propagating in the positive x- direction, and negative for waves propagating in the negative x-direction. (a). For y 2 (x,t) = A Cos[ k(x + 34t) ], by comparison with the general formula we see v = 34 m / sec ; the wave is propagating in the negative x-direction. (b). For y 3 (x,t) = A Exp[ k(x 20t) ], by comparison with the general formula we see v =+20 m / sec; the wave is propagating in the positive x-direction. B (c). For y 1 (x,t) =, by comparison with the general formula we see 2 C +(x 10t) v =+10 m / sec; the wave is propagating in the positive x-direction. 2. We are given a triangular shaped pulse of length L on a stretched string. When the wave pulse arrives at one end of the string, it is reflected. L End of string Suppose the end at which reflection occurs is fixed. Then the wave will be reflected with an inversion in sign. In the following figures, the light solid lines indicate the incident wave; the dotted lines represent the reflected wave; and the heavy solid lines represent the superposition of both, which is the shape of the total wave pulse.

2 (a). L/4 reflected: L/4 Fixed end of string (b). L/2 reflected: Fixed end of string L/2 (c). 3L/4 reflected: Fixed end of string 3L/4 (d). L reflected: L Fixed end of string =

3 Now suppose the end at which reflection occurs is free. Then the wave will be reflected without an inversion in sign. In the following figures, the light solid lines indicate the incident wave; the dotted lines represent the reflected wave; and the heavy solid lines represent the superposition of both, which is the shape of the total wave pulse. (a). L/4 reflected: (b). L/2 reflected: L/4 Free end of string Free end of string (c). 3L/4 reflected: L/2 (d). L reflected: 3L/4 Free end of string = Free end of string L

4 3. We are given the following form for a traveling wave pulse: b 3 y(x,t) = b 2 + (x vt) 2 with b=5 cm and v=2.5 cm/sec. The figure below shows the profile of this pulse at t=0 sec and t=0.2 sec. The solid line corresponds to t=0, the dashed line to t=0.2 sec. y (cm) x (cm) Displacement field y(x,t) at two different times: solid line: t=0 sec; dashed line: t=0.2 sec. The transverse velocity of a point located at x on the string, at time t, is y(x,t). This may be estimated numerically from the two curves given above by t subtraction: y(x,t) y(x,t + t) y(x,t t) y(x,0.2) y(x,0) = t 2 t 2(0.1) This estimate is plotted in the following figure: dy/dt (cm/sec) x (cm) String transverse velocity, calculated from the difference between y(x,t) at t=0 sec and y=0.2 sec

5 If the explicit analytical form of the wave pulse is known, as in this case, the transverse velocity may be calculated exactly: y(x,t) = b 3 t t b 2 + (x vt) 2 = 2b3 v(x vt) [ b 2 + (x vt) 2 ] 2 This expression is plotted in the following figure at t=0.1 sec: dy/dt (cm/sec) x (cm) String transverse velocity at t=0.1 sec, calculated analytically The numerical estimate is very close to the analytical result in this case. 4. Tipler, Chapter 13, p. 437, Problem 60. We are given two displacement fields y 1 (x,t) = y 2 (x,t) = 0.02 m 3 2 m 2 +(x 2t) m 3 2 m 2 +(x + 2t) 2 (a). The figure below graphs each of these functions at t=0. The solid curve gives y 1 (x,0); the dashed curve gives y 2 (x,0). Since y 1 (x,t)=f(x-2t), it corresponds to a traveling wave moving to the right with velocity 2 m/s. Similarly, y 2 (x,t)=f(x+2t) is a traveling wave moving to the left with velocity 2 m/sec.

6 0.01 y (m) x (m) Displacements y 1 (x,0) (solid) and y 2 (x,0) (dashed) vs. x (b). The resultant wave function is the sum of y 1 and y 2. Since, at t=0, y 1 =-y 2, their sum is zero, so the resultant wave function is zero for all x at t=0. (c). At t=1 sec, 0.02 y total (x,1) = y 1 (x,1)+ y 2 (x,1) = 2 +(x 2) m (x + 2) m x = 36 4x 2 + x m 4 (d). The figure below represents the resultant displacement field at t=1. y (m) x (m) Resultant displacement y total (x,1) = y 1 (x,1)+ y 2 (x,1)

7 5. The motion of the contact starts a wave, which propagates along the string. The speed of the wave is determined by the tension in the string and the mass per unit length of the string, according to τ v = µ in which τ= tension in the string, and µ=mass per unit length of the string. The mass per unit length is µ = 5 g 12.5 m = 4x10 4 kg / m The tension in the string is set by the force of gravity acting on the 10 kg mass: So the wave speed is τ = Mg = 10 kg x 10 m / sec 2 = 100 N τ v = µ = 100 m / sec = 500 m / sec. 4 4x10 (b). With this information, and given the measured displacement vs. x (from graph (b)), we can reconstruct the displacement vs. time, at x=0. This corresponds to the displacement of the contact. The leading edge of the wave, corresponding to the earliest motion of the contact, consists of an increase from zero to 5 mm over a distance of 0.4 m. Since the wave speed is 500 m/sec, this distance of 0.4 m corresponds to a time of 0.4 m/ (500 m/sec) = 0.8 msec (milliseconds). Then, the displacement is constant for another 0.4 m (another 0.8 msec); finally it goes back to zero (the contact opens) in 0.2 m/ (500 m/sec) = 0.4 msec. So, the graph of y vs. t at x=0 is y(mm) t (msec) Displacement of the magnetically operated contact vs. time. We take t=0 to be when the contact starts moving. msec. (a). From the figure above, the contact stays closed (constant at 5 mm) for 0.8

8 (c). From the figure above, the maximum speed of the contact occurs during its opening, and is equal to v contact = 5 mm = 12.5m / sec 0.4m sec (d). The leading edge of the wave pulse starts at t=0, x=0, and reaches x=6 m at the time of the photograph, moving at a speed of 500 m/sec. So the time of the photograph is t photo = 6 m = 12 m sec 500 m / sec