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16.1 The Nature of Waves

16.2 Periodic Waves

16.3 The Speed of a Wave in a String

Conceptual Questions 1,2,3,7, 8, 11 – page 503

Problems 2, 4, 6, 12, 15, 16 page 501-502

Types of Waves

There are two main types of

disturbances in waves.

The wave created by Hand A is a

_________________ Wave.

The wave created by Hand B is a

_________________ Wave.

There is a third type? Can you guess what is?

There also two other classifications of waves:

Examples: Waves in a Spring, Sound, Water or Ocean Waves,

Example: Radio Waves, Microwaves, Infrared Waves, Visible, Ultraviolet, X Rays, Gamma Rayse

The Relationship between Wavelength, Frequency and Velocity

Wave Equations: 𝑓 =1

𝑇 Train speed = waves per period or 𝑣 =

𝜆

𝑇

What are units of T? v? f?

Which train has the greatest wavelength (λ ) ? A or B

Assuming the speed of the trains are the same, which train will

have more trains pass a stationary point in a given time? A or B

Time for one wave to pass is called a Period (T)

How often cars pass is called Frequency (f)

As wavelength increases Period will (increase or decrease)

Speed of a Wave – The material which a wave is

moving through is called the MEDIUM. (What is plural

of medium?) Example: Sound travels faster in higher

density substances. Speed of sound in air (which varies

with Temperature and Humidity) is about 340 m/s.

The motion of a wave in a string (such as a guitar string)

is well understood and accurately quantified with the

following Equation:

𝑣 = √𝐹

𝑚/𝐿 where F is the Tension in the String (N) , m is the Mass (kg) and L the unit Length (m) .

Notice that m/L is a Linear Density for the string.

As the Force increases the speed (increases / decreases).

As the Linear Density increases the speed (increases / decreases).

AT AN INTERFACE OR BOUNDARY

An interface (or boundary) is when a wave moves from one medium to another. Here are two examples:

Moving from low to high density string and moving from high to low density string.

What happens to the speed of the wave as it moves from low density to high density? (Increases or Decreases)

What happens to the speed of the wave as it moves from low density to high density? (Increases or Decreases)

The Frequency of the wave stays the same when passing from one medium to another.

What happens to the wavelength when a wave moves from low density to high density?

(Increases or Decreases)

What happens to the wavelength when a wave moves from high density to low density?

(Increases or Decreases)

Conceptual Questions 1,2,3,7, 8, 11 – page 503

1. Considering that nature of a water wave, describe the motion of a fishing float on the surface of a lake when

a wave passes beneath the float. Is it really correct to say that the float bobs straight up and down? Explain.

As Figure 16.4 shows, in a water wave, the wave motion of the water includes both transverse and longitudinal components. The water at the surface moves on nearly circular paths. When the wave passes beneath a fishing float, the float will simultaneously bob up and down, as well as move back and forth horizontally. Thus, the float will move in a nearly circular path in the vertical plane. It is not really correct, therefore, to say that the float bobs straight "up and down."

2. “Domino Toppling” is one entry in the Guinness Book of World Records. The event consists of lining up an

incredible number of dominoes and then letting them topple, one after another. Is the disturbance that

propagates along the line of the dominoes transverse, longitudinal, or partly both? Explain.

REASONING AND SOLUTION "Domino Toppling" is an event that consists of lining up an incredible number of dominoes

and then letting them topple, one after another. As the dominoes topple, their displacements contain both vertical and

horizontal components. Therefore, the disturbance that propagates along the line of dominoes has both longitudinal

(horizontal) and transverse (vertical) components.

3. Suppose that a longitudinal wave moves along a Slinky at a speed of 5 m/s. Does one coil of the Slinky move

through a distance of 5 m in one second? Justify your answer.

REASONING AND SOLUTION A longitudinal wave moves along a Slinky at a speed of 5 m/s. We cannot conclude that one coil of the Slinky moves through a distance of 5 m in one second. The quantity 5 m/s is the longitudinal wave speed, vspeed; it specifies how fast the disturbance travels along the spring. The wave speed depends on the

properties of the spring. Like the transverse wave speed, the longitudinal wave speed depends upon the tension F in the spring and its linear mass density m/L. As long as the tension and the linear mass density remain the same, the disturbance will travel along the spring at constant speed.

The particles in the Slinky oscillate longitudinally in simple harmonic motion with the same amplitude and frequency as the source. As with all particles in simple harmonic motion, the particle speed is not constant. The particle speed is a maximum as the particle passes through its equilibrium position and reaches zero when the particle has reached its maximum displacement from the equilibrium position. The particle speed depends upon the amplitude and frequency of the particle's motion. Thus, the particle speed, and therefore the longitudinal speed of a single coil, depends upon the properties of the source that causes the disturbance.

7. One end of each of two identical strings is attached to a wall. Each string is being pulled tight by someone at

the other end. A transverse pulse is sent traveling along one of the strings. A bit later an identical pulse is sent

traveling along the other string. What, if anything, can be done to make the second pulse catch up with and

pass the first pulse? Account for your answer. (A pulse is a single piece of a wave. What you do in a football

stadium is actually a “pulse” rather than a wave)

REASONING AND SOLUTION One end of each of two identical strings is attached to a wall. Each string is being pulled tightly by someone at the other end. A transverse pulse is sent traveling along one of the strings. A bit later, an identical pulse is sent traveling along the other string. In order for the second pulse to catch up with the first pulse, the speed of the pulse in the second string must be increased. The speed of the transverse pulse on the second

string is given by Equation 16.2: wave/( / )v F m L . This equation indicates that we can increase the speed of

the second pulse by increasing the tension F in the string. Thus, the second string must be pulled more tightly.

8. Would it take any time for a transverse wave to travel the length of the mass less rope? Justify your answer.

REASONING AND SOLUTION In Section 4.10 the concept of a "massless" rope is discussed. For a truly massless rope,

the linear density of the rope, m/L, is zero. From Equation 16.2, wave/( / )v F m L , the wave speed would be infinite

if m/L were zero. Therefore, if the rope were really massless, the speed of transverse waves on the rope would be

infinite, and a transverse wave would be instantaneously transmitted from one end of the rope to the other. It would

not take any time for a transverse wave to travel the length of a massless rope.

11. A loudspeaker produces a sound wave. Does the wavelength of the sound increase, decrease, or remain

the same, when the wave travels from air into water? Justify your answer.

A loudspeaker produces a sound wave. The sound wave travels from air into water. As indicated in Table 16.1, the

speed of sound in water is approximately four times greater than it is in air. We are told in the hint that the frequency

of the sound wave does not change as the sound enters the water. The relationship between the frequency f, the

wavelength , and the speed v of a wave is given by Equation 16.1: v f . Since the wave speed increases and the

frequency remains the same as the sound enters the water, the wavelength of the sound must increase.

Problems 2,4,6,12,15,16 page 501-502 2. A woman is standing in the ocean, and she notices that after a wave crest passes, five more crests pass in a

time of 50.0 s. The distance between two successive crests is 32 m. Determine the, if possible, the wave’s (a)

period, (b) frequency, (c) wavelength, (d) speed, and (e) amplitude. If it is not possible to determine any of

these quantities, then so state.

a. The period is the time required for one complete cycle of the wave to pass. The period is also the time for two successive crests to pass the person.

b. The frequency is the reciprocal of the period, according to Equation 10.5.

c. The wavelength is the horizontal length of one cycle of the wave, or the horizontal distance between two successive crests.

d. The speed of the wave is equal to its frequency times its wavelength (see Equation 16.1).

e. The amplitude A of a wave is the maximum excursion of a water particle from the particle’s undisturbed position.

SOLUTION

a. After the initial crest passes, 5 additional crests pass in a time of 50.0 s. The period T of the wave is

50.0 s10.0 s

5T

b. Since the frequency f and period T are related by f = 1/T (Equation 10.5), we have

1 10.100 Hz

10.0 sf

T

4. One tsunami, generated off the Aleutian islands in Alaska, had a wavelength of 750 km and traveled a

distance of 3700 km in 5.3 hours. (a) What was the speed (in m/s) of the wave? For reference, the speed of

the wave a 747 jetliner is about 250 m/s. Find the wave’s (b) frequency and (c) period.

The speed of a Tsunamis is equal to the distance x it travels divided by the time t it takes for the wave to travel that

distance. The frequency f of the wave is equal to its speed divided by the wavelength , f = v/ (Equation 16.1). The

period T of the wave is related to its frequency by Equation 10.5, T = 1/f.

SOLUTION

a. The speed of the wave is (in m/s)

33700 10 m 1 h190 m/s

5.3 h 3600 s

xv

t

b. The frequency of the wave is

4

3

190 m/s2.5 10 Hz

750 10 m

vf

(16.1)

c. The period of any wave is the reciprocal of its frequency:

3

4

1 14.0 10 s

2.5 10 HzT

f

(10.5)

6. A person lying on an air mattress in the ocean rises and falls through one complete cycle every five seconds.

The crests of the wave causing motion are 20.0 m apart. Determine (a) the frequency and (b) the speed of the

wave.

The period of the wave is the same as the period of the person, so T = 5.00 s.

a. f = 1/T = 0.200 Hz (10.5)

b. v = f = (20.0 m)(0.200 Hz) = 4.00 m/s (16.1)

12. A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far

apart. The tension in the wires is the same, and they have the same mass. A transverse travels on the shorter

wire with a speed of 240 m/s. What would be the speed of the wave on the longer wire?

The speed v of a transverse wave on a wire is given by / /v F m L (Equation 16.2), where F is the tension and m/L

is the mass per unit length (or linear density) of the wire. We are given that F and m are the same for the two wires, and that one is twice as long as the other. This information, along with knowledge of the wave speed on the shorter wire, will allow us to determine the speed of the wave on the longer wire.

SOLUTION The speeds on the longer and shorter wires are:

[Longer wire] longer

longer/

Fv

m L

[Shorter wire] shortershorter

/

Fv

m L

Dividing the expression for vlonger by that for vshorter gives

longerlonger longer

shorter shorter

shorter

/

/

F

m Lv L

v LF

m L

Noting that vshorter = 240 m/s and that Llonger = 2Lshorter, the speed of the wave on the longer wire is

longer shorterlonger shorter

shorter shorter

2240 m/s 240 m/s 2 340 m/s

L Lv v

L L

15. Two wires are parallel, and one is directly above the other. Each has a length of 50.0 m and a mass per unit

length of 0.020 kg/m. However, the same tension in a wire A is 6.00 × 102 𝑁, and the tension in wire B is

3.00 × 102 𝑁. Transverse waves pulses are generated simultaneously, one at the left end of the wire A and

one at the right end of wire B. The pulses travel towards each other. How much time does it take until the

pulses pass each other?

Each pulse travels a distance that is given by vt, where v is the wave speed and t is the travel time up to the point when

they pass each other. The sum of the distances traveled by each pulse must equal the 50.0-m length of the wire, since

each pulse starts out from opposite ends of the wires.

SOLUTION Using vA

and vB to denote the speeds on either wire, we have

vA t + v

B t = 50.0 m

Solving for the time t and using Equation 16.2 v F

m / L

, we find

t 50.0 m

vA v B

50.0 m

FA

m / L

FB

m / L

50.0 m

6.00 102 N

0.020 kg/m

3.00 10 2 N

0.020 kg/m

0.17 s

16. The drawing shows two transverse waves traveling on two strings. The linear density of each string is 0.065

kg/m, and the tension is provided by a 26-N block that is hanging from the string. Determine the speed of the

wave in part (a) and part (b) of the drawing.

The speed v of a transverse wave on a string is given by / /v F m L (Equation 16.2), where F is the tension and m/L

is the mass per unit length (or linear density) of the string. The strings are identical, so they have the same mass per unit length. However, the tensions are different. In part (a) of the text drawing, the string supports the entire weight of the 26-N block, so the tension in the string is 26 N. In part (b), the block is supported by the part of the string on the left side of the middle pulley and the part of the string on the right side. Each part supports one-half of the block’s weight, or 13 N. Thus, the tension in the string is 13 N.

SOLUTION

a. The speed of the transverse wave in part (a) of the text drawing is

126 N2.0 10 m/s

/ 0.065 kg/m

Fv

m L

b. The speed of the transverse wave in part (b) of the drawing is

113 N1.4 10 m/s

/ 0.065 kg/m

Fv

m L