WAVES, SOUND AND OPTICS LECTURER : MR. N MANDIWANA OFFICE : NSB 208

Chapter 16 Waves and Sound

Chapter 16 Waves and Sound 16.1 The nature of waves 16.2 Periodic waves 16.3 The Speed of a Wave on a String 16.5 The Nature of Sound Waves 16.6 The Speed of Sound Waves 16.7 Sound Intensity 16.8 Decibels 16.9 The Doppler Effect 16.10 Application of Sound in Medicine 16.11 Sensitivity of the human ear (Self Study)

16.1 The Nature of Waves When one end of a (stretched) string is given a single up-and-down jerk, a wave in the form a pulse travels along the string A wave is a traveling disturbance. A wave carries energy from place to place.

Transverse Wave A transverse wave is one in which the disturbance occurs perpendicular to the direction of travel of the wave Examples of transverse waves are radio waves, light waves and microwaves

Longitudinal Wave A longitudinal wave is one in which the disturbance occurs parallel to the line of travel of the wave A sound wave is a longitudinal wave

Some waves are neither transverse nor longitudinal Water waves are partially transverse and partially longitudinal.

16.2 Periodic Waves Periodic waves consist of cycles or patterns that are produced over and over again by the source. In the figures, the repetitive patterns occur as a result of simple harmonic motions.

𝑓= 1 𝑇 In the drawing, one cycle is shaded in color. The amplitude A is the maximum excursion of a particle of the medium from the particles undisturbed position. The wavelength is the horizontal length of one cycle of the wave. The period is the time required for one complete cycle. The frequency is related to the period and has units of Hz, or s-1. 𝑓= 1 𝑇

𝑣= 𝜆 𝑇 =𝑓𝜆

Example 1 The Wavelengths of Radio Waves AM and FM radio waves are transverse waves consisting of electric and magnetic field disturbances traveling at a speed of 3.00x108m/s. A station broadcasts AM radio waves whose frequency is 1230x103Hz and an FM radio wave whose frequency is 91.9x106Hz. Find the distance between adjacent crests in each wave. AM FM

16.3 The Speed of a Wave on a String The speed of a wave depends on the properties of the medium in which it travels. For a transvers wave (on a string) that has a tension 𝐹 and mass per unit length 𝑚 𝐿 , the speed is tension 𝑣= 𝐹 𝑚 𝐿 linear density

Example 2 Waves Traveling on Guitar Strings Transverse waves travel on each string of an electric guitar after the string is plucked. The length of each string between its two fixed ends is 0.628 m, and the mass is 0.208 g for the highest pitched E string and 3.32 g for the lowest pitched E string. Each string is under a tension of 226 N. Find the speeds of the waves on the two strings.

Conceptual Example 3 Wave Speed Versus Particle Speed Is the speed of a transverse wave on a string the same as the speed at which a particle on the string moves?

16.5 The Nature of Sound Waves LONGITUDINAL SOUND WAVES Sound is a longitudinal wave that can be created only in a medium, i.e. sound cannot exists in a vacuum The distance between adjacent condensations (or rarefaction) is equal to the wavelength of the sound wave.

Individual air molecules are not carried along with the wave. A given molecule vibrates back and forth about a fixed point

THE FREQUENCY OF A SOUND WAVE The frequency is the number of cycles per second. A sound with a single frequency is called a pure tone. The brain interprets the frequency in terms of the subjective quality called pitch. Human ear can detect sound of frequencies approximately from 20 Hz to 20 000 Hz (20 kHz) Sound waves with frequencies below 20 Hz are inaudible and are said to be infrasonic Frequencies above 20 kHz are ultrasonic

THE PRESSURE AMPLITUDE OF A SOUND WAVE Loudness is an attribute of a sound that depends primarily on the pressure amplitude of the wave.

16.6 The Speed of Sound Sound travels through gases, liquids, and solids at considerably different speeds.

In a gas, it is only when molecules collide that the condensations and rarefactions of a sound wave can move from place to place. For an ideal gas 𝑣 𝑟𝑚𝑠 = 3𝑘𝑇 𝑚 𝑘 𝑖𝑠 𝑡ℎ𝑒 𝐵𝑜𝑙𝑡𝑧𝑚𝑎𝑛 𝑛 ′ 𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡=1.38× 10 −23 𝐽/𝐾 The speed of sound in an ideal gas is given 𝑣= 𝛾𝑘𝑇 𝑚 𝛾= 5 3 𝑜𝑟 7 5

Conceptual Example 5 Lightning, Thunder, and a Rule of Thumb There is a rule of thumb for estimating how far away a thunderstorm is. After you see a flash of lighting, count off the seconds until the thunder is heard. Divide the number of seconds by five. The result gives the approximate distance (in miles) to the thunderstorm. Why does this rule work?

LIQUIDS SOLID BARS 𝑣= 𝑌 𝜌 𝑣= 𝐵 𝑎𝑑 𝜌

16.7 Sound Intensity 𝐼= 𝑃𝑜𝑤𝑒𝑟 (𝑃) 𝐴𝑟𝑒𝑎 (𝐴) Sound waves carry energy that can be used to do work. The amount of energy transported per second is called the power of the wave. The sound intensity is defined as the power that passes perpendicularly through a surface divided by the area of that surface. 𝐼= 𝑃𝑜𝑤𝑒𝑟 (𝑃) 𝐴𝑟𝑒𝑎 (𝐴)

Example 6 Sound Intensities In the figure below12x10-5W of sound power passed through the surfaces labeled 1 and 2. The areas of these surfaces are 4.0m2 and 12m2. Determine the sound intensity at each surface. 𝐼 1 = 𝑃 𝐴 1 = 12× 10 −5 𝑊 4 𝑚 2 =3.0 𝑊. 𝑚 −2 𝐼 2 = 𝑃 𝐴 2 = 12× 10 −5 𝑊 12 𝑚 2 =1.0× 10 −5 𝑊. 𝑚 −2

For a 1000 Hz tone, the smallest sound intensity that the human ear can detect is about 1× 10 −12 𝑊. 𝑚 −2 . This intensity is called the threshold of hearing. On the other extreme, continuous exposure to intensities greater than 1𝑊. 𝑚 −2 can be painful. If the source emits sound uniformly in all directions, the intensity depends on the distance from the source in a simple way. power of sound source 𝐼= 𝑃 4𝜋 𝑟 2 area of sphere

𝐼= 𝑃 4𝜋 𝑟 2 Conceptual Example 8 Reflected Sound and Sound Intensity Suppose the person singing in the shower produces a sound power P. Sound reflects from the surrounding shower stall. At a distance r in front of the person, does the equation for the intensity of sound emitted uniformly in all directions underestimate, overestimate, or give the correct sound intensity? 𝐼= 𝑃 4𝜋 𝑟 2

Ex 16-61 Deep ultrasonic heating is produced by applying ultrasound over the affected area of the body. The sound transducer is circular with a radius of 1.8 cm, and it produces a sound intensity of 𝟓.𝟗× 𝟏𝟎 𝟑 𝑾/ 𝒎 𝟐 . How much time is required for the transducer to emit 𝟔 𝟎𝟎𝟎 𝑱 of energy? Solution 𝐼=5.9× 10 3 𝑊 𝑚 2 𝑟=1.8× 10 −2 𝑚 𝐸=6000𝐽 𝐴 𝑐𝑖𝑟𝑐𝑙𝑒 =𝜋 𝑟 2 𝑃= 𝐸 𝑡 𝑃=𝐼𝐴 ∴𝐼𝐴= 𝐸 𝑡 𝑡= 𝐸 𝐼𝐴 𝑡= 𝐸 𝐼𝜋 𝑟 2 𝑡= 6000𝐽 5.9× 10 3 𝑊 𝑚 2 𝜋 1.8× 10 −2 𝑚 2 𝑡=999𝑠

16.8 Decibels 𝛽= 10𝑑𝐵 𝑙𝑜𝑔 𝐼 𝐼 0 𝐼 0 =1× 10 −12 𝑊. 𝑚 −2 The decibel (dB) is a measurement unit used when comparing two sound intensities. Because of the way in which the human hearing mechanism responds to intensity, it is appropriate to use a logarithmic scale called the intensity level: 𝛽= 10𝑑𝐵 𝑙𝑜𝑔 𝐼 𝐼 0 𝐼 0 =1× 10 −12 𝑊. 𝑚 −2 Note that log(1)=0, so when the intensity of the sound is equal to the threshold of hearing, the intensity level is zero.

Example 9 Comparing Sound Intensities Audio system 1 produces a sound intensity level of 90.0 dB, and system 2 produces an intensity level of 93.0 dB. Determine the ratio of intensities.

𝛽(𝑑𝐵)=10 log 𝐼 𝐼 0 𝛽 1 =10 log 𝐼 1 𝐼 0 𝛽 2 =10 log 𝐼 2 𝐼 0 𝛽 2 − 𝛽 1 =10 log 𝐼 2 𝐼 0 −10 log 𝐼 1 𝐼 0 𝛽 2 − 𝛽 1 =10 log 𝐼 2 𝐼 0 𝐼 1 𝐼 0 =10 log 𝐼 2 𝐼 1 3.0 𝑑𝐵=10 log 𝐼 2 𝐼 1 0.3= log 𝐼 2 𝐼 1 𝐼 2 𝐼 1 = 10 0.30 =2.0

16.9 The Doppler Effect The Doppler effect is the change in frequency or pitch of the sound detected by an observer because the sound source and the observer have different velocities with respect to the medium of sound propagation.

𝜆 ′ =𝜆− 𝑣 𝑠 𝑇 𝑓 0 = 𝑣 𝜆 ′ = 𝑣 𝜆− 𝑣 𝑠 𝑇 𝑓 0 = 𝑣 𝑣 𝑓 𝑠 − 𝑣 𝑠 𝑓 𝑠 MOVING SOURCE 𝜆 ′ =𝜆− 𝑣 𝑠 𝑇 𝑓 0 = 𝑣 𝜆 ′ = 𝑣 𝜆− 𝑣 𝑠 𝑇 𝑓 0 = 𝑣 𝑣 𝑓 𝑠 − 𝑣 𝑠 𝑓 𝑠 𝜈= 𝑓 𝑠 𝜆 𝑇= 1 𝑓 𝑠 𝑓 0 = 𝑓 𝑠 1 1− 𝑣 𝑠 𝑣 𝑓 𝑜 = 𝑓 𝑠 𝜈 𝜈− 𝑣 𝑠

𝑓 0 = 𝑓 𝑠 1 1− 𝑣 𝑠 𝑣 𝑓 𝑜 = 𝑓 𝑠 𝜈 𝜈− 𝑣 𝑠 𝑓 0 = 𝑓 𝑠 1 1+ 𝑣 𝑠 𝑣 source moving toward a stationary observer 𝑓 0 = 𝑓 𝑠 1 1− 𝑣 𝑠 𝑣 𝑓 𝑜 = 𝑓 𝑠 𝜈 𝜈− 𝑣 𝑠 𝑓 0 = 𝑓 𝑠 1 1+ 𝑣 𝑠 𝑣 source moving away from a stationary observer 𝑓 𝑜 = 𝑓 𝑠 𝜈 𝜈+ 𝑣 𝑠

Example 10 The Sound of a Passing Train A high-speed train is traveling at a speed of 44.7 m/s when the engineer sounds the 415-Hz warning horn. The speed of sound is 343 m/s. What are the frequency and wavelength of the sound, as perceived by a person standing at the crossing, when the train is (a) approaching and (b) leaving the crossing? 𝑣 𝑠 =44.7𝑚/𝑠 Solution 𝑓 𝑠 =415𝐻𝑧 𝜈=343𝑚/𝑠 𝑓 𝑜 = 𝑓 𝑠 1 1− 𝑣 𝑠 𝜈 𝑓 𝑜 =415𝐻𝑧 1 1− 44.7𝑚/𝑠 343𝑚/𝑠 𝑓 𝑜 =477𝐻𝑧 𝜆 ′ = 𝜈 𝑓 𝑜 = 343𝑚/𝑠 477𝐻𝑧 =0.719𝑚

𝑓 0 = 𝑓 𝑠 + 𝑣 0 𝜆 = 𝑓 𝑠 1+ 𝑣 0 𝑓 𝑠 𝜆 𝑓 0 = 𝑓 𝑠 1+ 𝑣 0 𝑣 MOVING OBSERVER 𝑓 0 = 𝑓 𝑠 + 𝑣 0 𝜆 = 𝑓 𝑠 1+ 𝑣 0 𝑓 𝑠 𝜆 𝑓 0 = 𝑓 𝑠 1+ 𝑣 0 𝑣 𝑓 𝑜 = 𝑓 𝑠 𝜈+ 𝑣 𝑜 𝜈

𝑓 0 = 𝑓 𝑠 1+ 𝑣 0 𝑣 𝑓 𝑜 = 𝑓 𝑠 𝜈+ 𝑣 𝑜 𝜈 𝑓 0 = 𝑓 𝑠 1− 𝑣 0 𝑣 𝑓 0 = 𝑓 𝑠 1+ 𝑣 0 𝑣 Observer moving towards stationary source 𝑓 𝑜 = 𝑓 𝑠 𝜈+ 𝑣 𝑜 𝜈 Observer moving away from stationary source 𝑓 0 = 𝑓 𝑠 1− 𝑣 0 𝑣 𝑓 𝑜 = 𝑓 𝑠 𝜈− 𝑣 𝑜 𝜈

𝑓 0 = 𝑓 𝑠 1± 𝑣 0 𝑣 1∓ 𝑣 𝑠 𝑣 GENERAL CASE Numerator: plus sign applies when observer moves towards the source 𝑓 0 = 𝑓 𝑠 1± 𝑣 0 𝑣 1∓ 𝑣 𝑠 𝑣 Denominator: minus sign applies when source moves towards the observer

Exercise 16-86 The siren on ambulance is emitting a sound whose frequency is 2450 Hz. The speed of sound is 343 m/s If the ambulance is stationary and you are sitting in a parked car, what are the wavelength and frequency of the sound you hear? Suppose the ambulance is moving toward you at the speed of 26.8 m/s, determine the wavelength and frequency of the sound you hear If the ambulance is moving toward you at a speed of 26.8m/s and you are moving toward it at a speed of 14m/s, find the wavelength and frequency of the sound you hear

If the ambulance is stationary and you are sitting in a parked car, what are the wavelength and frequency of the sound you hear? 𝑓 0 =𝑓=2450𝐻𝑧 𝜆= 𝑣 𝑓 𝑠 = 343𝑚/𝑠 2450𝐻𝑧 =0.140𝑚 Suppose the ambulance is moving toward you at the speed of 26.8 m/s, determine the wavelength and frequency of the sound you hear 𝜆 ′ =0.140− 26.8𝑚/𝑠 2450𝐻𝑧 =0.129𝑚 𝜆 ′ =𝜆− 𝑣 𝑠 𝑓 𝑠 𝜆 ′ =𝜆− 𝑣 𝑠 𝑇 𝑓 0 = 𝑓 𝑠 1 1− 𝑣 𝑠 𝑣 𝑓 0 =2450𝐻𝑧 1 1− 26.8𝑚/𝑠 343𝑚/𝑠 𝑓 0 ≈2660𝐻𝑧

𝑓 0 = 𝑓 𝑠 1± 𝑣 𝑜 𝑣 1∓ 𝑣 𝑠 𝑣 𝑓 𝑜 = 𝑓 𝑠 1+ 𝑣 𝑜 𝑣 1− 𝑣 𝑠 𝑣 If the ambulance is moving toward you at a speed of 26.8m/s and you are moving toward it at a speed of 14m/s, find the wavelength and frequency of the sound you hear 𝑣 𝑜 =14𝑚/𝑠 𝑓 0 = 𝑓 𝑠 1± 𝑣 𝑜 𝑣 1∓ 𝑣 𝑠 𝑣 𝑓 𝑜 = 𝑓 𝑠 1+ 𝑣 𝑜 𝑣 1− 𝑣 𝑠 𝑣 𝑣 𝑠 =343𝑚/𝑠 𝑓 𝑜 =2450 1+ 14𝑚/𝑠 343𝑚/𝑠 1− 26.8𝑚/𝑠 343𝑚/𝑠 𝑓 𝑜 ≈2770 𝐻𝑧

16.10 Applications of Sound in Medicine By scanning ultrasonic waves across the body and detecting the echoes from various locations, it is possible to obtain an image.

Cavitron Ultrasonic Surgical Aspirator (CUSA) Neurosurgeons use CUSA to remove brain tumors once thought to be inoperable Ultrasonic sound waves cause the tip of the probe to vibrate at 23 kHz and shatter sections of the tumor that it touches.

Doppler flowmeter The device measures the speed of blood flow, using a transmitting and receiving elements placed directly on the skin When the sound is reflected from the red blood cells, its frequency is changed in a kind of Doppler effect because the cells are moving. The Doppler flowmeter can be used to detect motion of a fetal heart as early as 8-10 weeks after conception

16.11 The Sensitivity of the Human Ear