"Waves I "

44: A wave travels along a string at a speed of 280~m/s. What will be the speed if the string is replaced by one made of the same material and under the same tension but having twice the radius?

Twice the radius implies four times the cross-sectional area, so the linear density of the new string is four times that of the original string. Since the propagation speed of a wave in a stretched string is equal to the square root of the ratio of the tension to the linear density, the new propagation speed is one half of the old one.

54: A wave on a string is described by

D(x,t) = (2.00 cm)*sin[(12.57 rad/m)*x - (638 rad/s)*t],

(a) One can read directly off the formula for D(x,t) the amplitude D_max = 2.00 cm = 0.02 m, the wavenumber k = 12.57 rad/m, the angular frequency = 638 rad/s and the initial phase (zero). Since the propagation speed c = /k = 50.76 m/s is the square root of the ratio of the tension T to the linear density µ = 5.00 g/m = 0.005 kg/m, we have 50.76² = 2576 = T/0.005 or T = 12.88 N. (Note that we have to convert everything to SI units if we want to use the general formulae.)
(b) Given: D_max = 2.00 cm = 0.02 m.
(c) v = dD/dt = - D_max cos(kx - t), which has its maximum value when the cosine function equals -1: v_max = D_max = 638*0.02 = 12.76 m/s.

66: Lasers can be used to drill or cut material. One such laser generates a series of high-intensity pulses rather than a continuous beam of light. Each pulse contains 500 mJ or energy and lasts 10 ns. The laser fires 10 such pulses per second.
(a) What is the peak power of the laser light? The peak power is the power during one of the 10 ns pulses.
(b) What is the average power output of the laser? The average power is the total energy delivered per second.
(c) A lens focuses the laser beam to a 10-µm-diameter circle on the target. During the pulse, what is the light intensity on the target?
(d) The intensity of sunlight at midday is about 1100 W/m². What is the ratio of the laser intensity on the target to the intensity of the midday sun?

(a) 500 mJ / 10 ns = 0.5 J / 10^-9 s = 5*10⁸ W.
(b) The laser is "on" for only 100 ns altogether every second, which amounts to 10^-7 of the time. The average power is thus 10^-7 of the power when it's on, or 50 W.
(c) The spot has a radius of r = 5*10^-6 m so its area is r² = 0.785*10^-10 m². The intensity I is the power per unit area, so I = 5*10⁸/0.785*10^-10 = 6.37*10¹⁸ W/m².
(d) The question does not specify whether you should calculate the peak intensity I = 5.79*10¹⁵ I_sun or the average intensity, <I> = 5.79*10⁸ I_sun, so we'll do both. Either way, it's pretty hot!

72: A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it around her head in a horizontal circle at 100 rpm. What are the highest and lowest frequencies heard by a student in the classroom?

The frequency of sound from a moving source is shifted by a ratio 1/(1 - v/v_s) where v_s is the speed of sound (typically 343 m/s) and v is the speed of the source coming toward the observer. (This form also works when the source is moving away; then v -s negative.) So we need only calculate v = r where r = 1 m and = 2 = 2*100/60 = 10.5 s^-1. Thus v = 10.5 m/s, v/v_s = 0.0305 and f = (600 Hz)/(1 0.0305), giving a lowest frequency of 582 Hz and a highest frequency of 619 Hz.

76: Wavelengths of light from a distant galaxy are found to be 0.5% longer than the corresponding wavelengths measured in a terrestrial laboratory. Is the galaxy approaching or receding from the earth? At what speed?

The wavelength gets longer ("red shift") if the galaxy is receding from Earth, which is generally the case. The square of the ratio of the observed wavelength to the original wavelength of light emitted by a receding source is given by (1 + )/(1 - ) where is the ratio of the receding speed of the object to the speed of light. Thus (1.005)² = 1.010025 1.01 = (1 + )/(1 - ) or (1 + ) = 1.01 (1 - ) or 2.01 = 0.01 or = 0.01/2.1025 = 0.005 (note that we have kept extra significant figures during the calculation but dropped back to one significant figure at the end). Since the speed of light is (to this precision) 3*10⁸ m/s, we have v = 1.5*10⁶ m/s. A pretty good clip!