Home Work 16 – Chapter 16

Chapter 16, Concept Question 1

The following four waves are sent along strings with the same linear densities (x is in meters and t is in seconds).

1. y1=(3mm)sin(x-3t)

2. y2=(6mm)sin(2x-t)

3. y3=(1mm)sin(4x-t)

4. y4=(2mm)sin(x-2t)

In the following questions, you will need to rank the various waves. If multiple waves rank equally, use the same rank for each, then exclude the

intermediate ranking (i.e. if objects A, B, and C must be ranked, and A and B must both be ranked first, the ranking would be A:1, B:1, C:3). If all

waves rank equally, rank each as '1'.

Rank the waves according to their wave speed, greatest first.

Wave 1 1

Wave 2 3

Wave 3 4

Wave 4 2

1. Greatest

2. Second greatest

3. Third greatest

4. Fourth greatest

Rank the waves according to the tension in the strings along which they

travel, greatest first:

Wave 1 1

Wave 2 3

Wave 3 4

Wave 4 2

1. Greatest

2. Second greatest

3. Third greatest

4. Fourth greatest

Chapter 16, Concept Question 2

In Fig. 16-23, wave 1 consists of a rectangular peak of height 4 units and

width d, and a rectangular valley of depth 2 units and width d. The wave travels rightward along an x axis. Choices 2, 3, and 4 are similar waves, with

the same heights, depths, and widths, that will travel leftward along that axis and through wave 1. Right-going wave 1 and one of the left-going

waves will interfere as they pass through each other.

With which left-going wave will the interference give, for an instant, the

deepest valley?

(1) Wave 3

(2) Wave 2

(3) Wave 4

With which left-going wave will the interference give, for an instant, a flat

line?

(1) Wave 2

(2) Wave 3

(3) Wave 4

With which left-going wave will the interference give, for an instant, a flat

peak 2d wide?

(1) Wave 2

(2) Wave 3

(3) Wave 4

Chapter 16, Concept Question 5

If you start with two sinusoidal waves of the same amplitude traveling in phase on a string and then somehow phaseshift one of them by 5.4

wavelengths, what type of interference will occur on the string?

(1) Fully constructive

(2) Intermediate (closer to fully constructive)

(3) Intermediate (closer to fully destructive)

(4) Fully destructive

10. The equation of a transverse wave traveling along a very long string is y = 6.0 sin(0.020πx +

4.0πt), where x and y are expressed in centimeters and t is in seconds. Determine (a) the

amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation

of the wave, and (f) the maximum transverse speed of a particle in the string. (g) What is the

transverse displacement atx = 3.5 cm when t = 0.26 s?

Solutions

(a) The amplitude is ym = 6.0 cm.

(b) We find from 2/ = 0.020: = 1.0×102 cm.

(c) Solving 2f = = 4.0, we obtain f = 2.0 Hz.

(d) The wave speed is v = f = (100 cm) (2.0 Hz) = 2.0×102 cm/s.

(e) The wave propagates in the –x direction, since the argument of the trig function is kx + t

instead of kx – t (as in Eq. 16-2).

(f) The maximum transverse speed (found from the time derivative of y) is

1

max 2 4.0 s 6.0cm 75cm s.mu fy

(g) y(3.5 cm, 0.26 s) = (6.0 cm) sin[0.020(3.5) + 4.0(0.26)] = –2.0 cm.

17. The linear density of a string is 1.6 × 10-4

kg/m. A transverse wave on the string is described

by the equation

What are (a) the wave speed and (b) the tension in the string?

Solutions

(a) The wave speed is given by v = /T = /k, where is the wavelength, T is the period, is the

angular frequency (2/T), and k is the angular wave number (2/). The displacement has the

form y = ym sin(kx + t), so k = 2.0 m–1 and = 30 rad/s. Thus

v = (30 rad/s)/(2.0 m–1) = 15 m/s.

(b) Since the wave speed is given by v = , where is the tension in the string and is the

linear mass density of the string, the tension is

22 41.6 10 kg m 15m s 0.036 N.v

27. A sinusoidal wave is sent along a string with a linear density of 2.0 g/m. As it travels, the

kinetic energies of the mass elements along the string vary. Figure 16-36a gives the rate dK/dt at

which kinetic energy passes through the string elements at a particular instant, plotted as a

function of distance x along the string. Figure 16-36b is similar except that it gives the rate at

which kinetic energy passes through a particular mass element (at a particular location), plotted

as a function of time t. For both figures, the scale on the vertical (rate) axis is set by Rs = 10 W.

What is the amplitude of the wave?

Figure 16-36 Problem 27.

Solutions

We note from the graph (and from the fact that we are dealing with a cosine-squared, see Eq.

16-30) that the wave frequency is f = 1

2 ms = 500 Hz, and that the wavelength = 0.20 m. We

also note from the graph that the maximum value of dK/dt is 10 W. Setting this equal to the

maximum value of Eq. 16-29 (where we just set that cosine term equal to 1) we find

12 v 2 ym

2 = 10

with SI units understood. Substituting in 0.002 kg/m,= 2f and v = f , we solve for the

wave amplitude:

2 3

100.0032 m

2my

f .

28. Use the wave equation to find the speed of a wave given by

Solutions

Comparing 1 1( , ) (3.00 mm)sin[(4.00 m ) (7.00 s ) ]y x t x t to the general expression

( , ) sin( )my x t y kx t , we see that 14.00 mk and 7.00 rad/s . The speed of the wave is

1/ (7.00 rad/s)/(4.00 m ) 1.75 m/s.v k