Physics

PHYSICS

Bertujuan membantu calon STPM yang gagal atau berprestasi lulus sebahagian ke arah meningkatkan pencapaian agar mencapai tahap LULUS PENUH

Digubal oleh tenaga pengajar STPM berpengalaman di Johor Darul Takzim

PHYSICS STPM WAVE MOTION AND SOUND WAVE

PANEL PENGGUBAL:

1. EN. WONG KOK WAH

SMK TUN HABAB , KOTA TINGGI

2. EN. YUNG SEE BOON

SMK DATO’ BENTARA LUAR,BATU PAHAT

3. EN. TIOH KON POO

SMK YONG PENG, BATU PAHAT

4. EN. ABU BAKAR BIN IBRAHIM

SM SAINS SULTAN ISKANDAR,MERSING

5. PN. LIM SEANG KEE

SMK DATO’ JAAFAR, JOHOR BAHRU

2


Unit 1Objective: By the end of the lesson students should be able to:

State the definition of wave. Explain the propagation of sound waves.

A. Waves

1. A wave is a disturbance which moves with a fixed velocity through a medium. The wave form does not vary while in motion.

B. Propagation of Sound Waves

1. To produce sound waves, we must have (a) vibrating object and (b) a medium for the sound waves to propagate

2. Sound waves propagate pulses of pressure through the medium.3. Sound waves are mechanical longitudinal waves.

Self Test 1

1. What are the two different types of wave?

2. What is the difference between transverse and longitudinal wave motion?

3. The transverse wave in a string is represented by the equation

where x and y are measured in centimeters and t in seconds. The phase difference between two points 0.20 cm apart on the wave is A. 0.11 rad B. 0.13 radC. 0.20 rad D. 0.63 rad

4. The equation for a wave motion is represented by where x and y are in meter and t in seconds.

Calculate(a) amplitude, (b) wavelength,(c) speed, of the wave.

5. The equation of a transverse wave traveling along a string is ,where x and y are in cm and t in second.

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

3


Suggested answer for self test 1.

1. Longitudinal and transverse waves

2. Longitudinal wave : The direction of oscillation of the particles of the medium transmitting the wave is parallel to direction of wave propagation.Transverse wave : The direction of oscillation of the particles of the medium transmitting the wave is perpendicular to direction of wave propagation

3. D

4. Compare

to

(a) amplitude, a = 6 m

(b)

(c)

Speed,

5. Compare r

to

(a) amplitude, a = 6.0 cm

(b)

(c)

(d) Speed,

(e) Right to left

Unit 2

4


Objective: By the end of the lesson students should be able to: define displacement, amplitude, frequency, period, wavelength, and

wavefront interpret and use the progressive wave equation, or

sketch and interpret the displacement-time graph and the displacement-distance graph

use the formula

derive and use the relationship

A. Progressive Wave1. Progressive wave is waves propagated continuously outward from a

source of disturbance.

2. In Progressive wave, only the shape of the wave or the wave profile move forwards.

3. There are (a) progressive mechanical waves and (b) progressive electromagnetic waves.

4. The speed v of the progressive wave is given by

5. The progressive harmonic wave equation is given by

6. We can deduce the following data from equation (a) amplitude , A - maximum displacement for a particles from a equilibrium/mean position (b) speed/velocity of the wave using - distance traveled by a pulse of wave in a unit time in the direction of the wave propagation. (c) frequency , f - number of vibrations per unit time. (d) direction of propagation - - to the right, + to the left (e) wavelength , - distance between two successive particles moving in phase. (f) Period , T – time for a medium particle to make a complete vibration.

(g) is the phase difference

5

y

t/s0

y

x/m0

(a) y-t graph (b) y-x graph

T/2 T /2


6

0

4

-4

8 24

Displacement(mm)

Distance(cm)16


Example 1

Figure 1Figure 1 shows a displacement-time and a displacement –distance graph for

a progressive wave. Determine (a) the amplitude(b) the period of oscillation(c) the frequency of oscillation(d) the wavelength of the wave(e) the speed of the wave.

Solution:(a) amplitude,A=4.0mm (Refer both graphs)(b) T=20 ms=0.02s (Refer displacement-time graph)(c) f= 1/T=1/0.02=50Hz (d) =16cm(e) Speed ,v=f=50x 16=80cms -1=8ms-1

Example 2:

1. Diagram(a) below shows a graph of displacement y against distance x for a progressive wave at a certain time. At time 0.4 s later, the profile of the wave is shown in diagram (b).

7

25

Displacement(mm)

time(ms)

4

4

0 1015

y

x/m0

0.02

y

x/m0.10

0.12

(a) (b)


Self -Test 2

1.The figure above shows a displacement-distance graph of a ripple wave traveling 60cm in 2.0 seconds. Calculate a)the velocity of the wave ,b) the wavelength of the ripple wave,c)the frequency of the waves,d) the period of oscillation of water surface.

2. Which line ,A to D ,in the table shows correct relationships for the respective wavelengths,L, S and frequencies,fL , fS , of light waves and sound waves ?

3. Two points on a progressive wave differ in phase by /4 . The distance between them is 0.5m, and the frequency of the oscillation is 10Hz. What is the minimum speed of the wave ?

4.(a) The displacement y at distance x and time t of a sound wave propagating in air can be represented by y = 7.5 x 10 -4 sin (315 t – 1.05 x ),Where x and y are in metres and t is in seconds.(i) Sketch, on the same axes, a graph of y against x at time t=0 and t=T/4,

where T is the period of the wave.(ii) Determine the velocity and the frequency of the wave.(iii) Calculate the phase difference of a point 2.0 m from the origin.

5. What moves in a progressive wave? 6. Define ‘frequency’.

7. Define ‘wavelength’.

8

y/mm

90 3 6 12

3

-3

x/cm


8. Define ‘amplitude’. 9. How is phase difference related to distance between two points along a progressive

wave?

10. A progressive wave is represented by the equation the speed of the progressive wave is A. ab B. ac

C. D.

Question 11, 12, and 13 all refer to the figure below.

Figure above represents the simple harmonic motion of a particle in a progressive wave traveling at a speed of 5.0kms-1.

11. The frequency of vibration is A. 2.5 kHz B. 5.0 kHzC. 25 kHz D. 50 kHz

12. The amplitude of vibration is A. B. 2 C. 2 D. 4

13. The wavelength isA. 10 mm B. 15 mmC. 20 mm D. 100 mm

14. Progressive waves of frequency 300 Hz are superimposed to produce a system of stationary waves in which adjacent nodes are 1.5 apart. What is the speed of the progressive waves?A. 100 ms-1 B. 200 ms-1

C. 450 ms-1 D. 900 ms-1

9

time/

+2

-2

00 10 20 30 40

Displacement/


15. The progressive equation is given by . Explain the

meaning of the symbols used in the equation.

16. A progressive wave is represented by the equation ,where x and y are in cm, and t in seconds. Find

(a) the wave amplitude,(b) the wave frequency,(c) the wave length, (d) the wave velocity.

17. A progressive wave moving in the negative x-axis direction has an amplitude of 0.50 m, frequency of 500 Hz and a speed of 300 ms-1.(a)Calculate wavelength of the progressive wave.(b)Write the equation for the progressive wave.

18. A form of progressive wave which has frequency of 800 Hz and a speed of 350ms-1 can be stated in the form of the following equation

(a)Calculate the value of (i)(ii) k(b)What is the phase difference between two points of distance 1.0 m apart along the progressive wave.

19. Transverse waves of frequency 200 Hz moves along stretched string with a speed of 5.0ms-1. Each particle of the string oscillates through a distance of 6.0 cm.(a) What is its wavelength?(b) Write an equation to represent the progressive wave along the string.

20. The water waves in Cendana lake is represented by the equation

with x and y in metres and t in seconds. (a) Calculate the velocity of the water waves.

(b) Write an equation which represents the water waves in Teratai lake which have the same

6.0

-6.0

50

y/cm

Distance/cm

10


frequency as the water waves in Cendana lake but have half the amplitude of that in Cendana lake and propagates with a speed twice that in Cendana lake in the opposite direction.

Suggested answer for self test 2

1. (a) Velocity,v= s/t = 60x 10-2/ 2.0 = 0.30ms-1

(b) From the graph, wavelength =6.0cm=6.0 x 10-2m(c) Using v=f

0.30= f x (6.0x10-2) f =5.0Hz(d) Period,T=1/f =1/5= 0.2 s.

2. A

3. phase difference= 2x / =/4 (given x=0.5m , f=10Hz ) =8 x 0.5 =4.0m Using , v= f = 10 x 4.0= 40ms-1

4. (i) t=0, y = 7.5 x 10 -4 sin (– 1.05 x ) = - 7.5 x 10 -4 sin (1.05 x )

(ii) From the equation: y = 7.5 x 10 -4 sin (315 t – 1.05 x )

And compare with

2f = 315, f = 315/2 = 50.1 Hz,2/ = 1.05, =2/1.05 =5.98 m.V = f = 50.1 x 5.98 = 300 m s-1.

(iii) m is equivalent to 2 radian, for 2.0 m is equivalent to =2.10 rad.

11

7.5 x 10-4

- 7.5 x 10-4

t=0 t=T/4


5. Energy6. f : number of oscillation per second7. : the distance between 2 adjacent positions on a progressive wave that are in

phase with other.8. a : The maximum displacement from the equilibrium position.

9. Phase difference ,x = distance between 2 points

= wavelength10. C11. D12. B13. D14. D

15.

y : displacement of particle in a medium from its equilibrium position a : amplitude of particle vibration/wave

: angular frequency= wavelength of the wave

x = distance of a particle from the source of the wave.

16. Compare to

(a) amplitude = 5 cm(b) = 50 rads-1, but

(c) cm

(d) velocity, cms-1

17. (a) m

(b) a=0.50m, rads-1

from

18. (a)(i) rads-1

(ii) ,

m-1

12


(b) phase difference, rad

19. (a)

(b) , a = 6.0 cm = 0.06 m

Equation,

20. Compare to

(a)

velocity,

(b) Same , ;

, direction = +

Equation :

Unit 3 Objective: By the end of the lesson students should be able to:

Use the principle of superposition to explain the formation of standing waves

define and interpret the standing wave equation describe quantitatively the formation of standing waves along stretched

strings and use the formula for the frequency of the sound waves produced

13


A :Stationary(Standing) Waves

1. The production of a resultant disturbance at a point due to the overlapping of similar waves crossing the point simultaneously is known as superposition.

2. In superposition, the net displacement of the particle is the algebraic sum of individual displacements produced by each pulse at that point.

3. Stationary waves are produced when two progressive waves of the same type with the same amplitude and frequency move towards each other(opposite directions) and undergo superposition in the same region.

4. The amplitude of the stationary wave is

5. A point in a stationary wave at which there is no oscillation is known as a node.

6. A point at which the amplitude is a maximum is known as an antinode.

(a) Distance between two adjacent nodes =

(b) Distance between two adjacent antinodes =

(c) Distance between a node and the nearest antinode =

B: Sources of Sound – Transverse stationary waves along stretched string

1. The speed v of waves traveling along a stretched string which vibrates transversely is given by

2.

3. When we pluck a stretched string, transverse waves are produced and they travel along the string in opposite directions. Stationary waves are set up along the string when the similar transverse waves traveling in opposite directions superimposed.

14

nodeAntinode

Amplitude


4. A stretched string vibrates in the fundamental mode when only one antinode is produced in between the two fixed ends of the string. The frequency of vibration of the string is the fundamental frequency f0.

5. The fundamental frequency f0 is given by

6. , o= 2L

7. L= length of the string, T = tension in the string, =mass per unit length8. The stretched string can vibrate in other modes besides the fundamental

mode.

Fundamental 1st overtone

2nd overtone

3rd overtone

4th overtone

stretchedString

f1 =2fo f2 =3fo f3 =4fo f4 =5fo

Example 1 A stationary wave in a stretched string is represented by the equation

y= 3 sin x/ 4 cos 20t, with x and y measured in centimeters and t in seconds .

a)What is the resultant amplitude ? b) what is the distance between two adjacent nodes of the stationary

wave ?Solution: a) Resultant amplitude =2Acos(2x /)

b)Given : y= 3 sin( x/ 4 )cos( 20t) General equation:y=[ 2Acos(2x/ )]sin2ft. Hence, 2x/ = x/4 =8 cm Therefore the distance between two adjacent nodes = /2 = 8/2 =4.0cm.

Example 2

A string is stretched by the weight of an object of mass 5.0kg, as shown in

15

P Qstring

object

Smooth pulley


figure above. It has the following properties: length PQ= 40 cm, diameter,d=0.50mm and density of string= 7.9 x 103kgm-3

Determine:a) the tension in the string.b) The mass per unit length of the stringc) The fundamental frequency if the string vibrates in the fundamental

mode.d) The frequency of the second overtone if the string can produce this

overtone.

Solution :a) Tension, T=mg= (5.0)(9,81)=49.05Nb)

d

A L If m represents the mass of PQ of length 40cm, then the mass per unit length is given by = m/L =V/L =AL/L= (d2/4) = (7.9 x 103)(0.5 x 10-3)2 /4=1.55 x 10-3kgm-1

c)

c) Frequency of the second overtone= frequency of third harmonic. =3f0

= 3 x 222 = 666 Hz

f1= v/= 3v/4l = 3(320)/4(0.64)= 375Hz.

Example 3The figure shows a standing wave on a stretched string. A, B, C, and D are points on the string.

16


Compare the oscillations of the points B, C, and D with that of the point A in terms of(i) amplitude and(ii) phase.

Solution:(i) Amplitude of the point B is smaller than that of A.Amplitude of the point C is zero.The points A and D have the same amplitude.(ii) The points A and B are in phase.The point C does not oscillate at all.The points A and D are in antiphase.

Self Test 3

1. Which of the following statements is not true about a stationary wave?A. The phases of oscillation of the medium are the same between consecutive nodes.B. The amplitudes of oscillation of the medium are different between consecutive nodes.C. Velocities of oscillation of the medium are the same between consecutive nodes.D. Frequencies of oscillations of medium are the same between consecutive nodes. (Answer:C)

2. A stationary wave is represented by the equation y=5sin (x/8)cos(50t),where x and y are in cm , and t in second. What is the distance between successive antinodes ?A. 4.0 cm B. 8.0 cm C. 12.0 cm D.25.0 cm

3. A stationary wave is set up on a stretched string XY as shown in the figure .Which points vibrates exactly in phase with P

17

PX

1 2 3Y


A. 1 and 2 B. 2 and 3 C. 1 only D. 3 only

4.The frequency of the fundamental note produced by a stretched string of length 1.0 m is 256Hz. When the string is shorterned to 0.4 m at the same tension, the fundamental frequency is A. 102Hz B. 312 Hz C. 416Hz D. 640Hz

5. A stretched wire of length 60.0cm and mass 10.0g vibrates transversely. Waves travel along the wire at a speed 210ms-1. Three antinodes can be found in the stationary waves formed in between the two ends of the wire. Determine a) the wavelength of the progressive waves which move along the wire.b)the frequency of vibration of the wire.c) the tension in the wire.

6. One end of a string of mass 1.45 g and length 50.0 cm is attached to a frequency generator and the other end to a weight holder which hangs over a fixed pulley. The part of the string between the generator and the pulley is horizontal. The string is set to vibrate. Several weights are added until a fundamental frequency of 120 Hz is achieved.(a) Describe the motion of the particles in the string.(b) Calculate the tension in the string when the fundamental frequency is achieved.

7. A stretched string between two fixed points is free to vibrate when plucked at the centre. (a) Sketch the fundamental mode of the vibrating string and state the equation relating the fundamental frequency f0, to the length of string, l and the speed of waves in the string, v.(b) Sketch the 2 lowest overtones, and state their respective frequencies in terms of f0.

8.a) State the principle of superposition of two waves.

b) Stationary waves are formed in a stretched string. Explain the meaning of the terms(i) node and antinode

(ii) fundamental frequency and overtones

9 A standing wave is represented by the equation y = 26cos 4x sin 5t where x and y are in m and t is s.(a) What is the frequency of the standing wave?(b) What is the distance between successive nodes?

18


I Calculate the wavelength.

10.Show in the table below the frequency, the wavelength, the number of antinodes (A), and nodes (N) for the various modes of vibration alonga stretched string. Sketch a diagram to show each mode of vibration.

11. The mass of a stretched string of length 2.4m is 60g. The tension in the string is 10N. Determine the frequency if the string vibrates

a) in three segmentsb)in the fundamental mode.

12.The frequency of the third overtone produced by a vibrating string of length 2.00 m is 1600 Hz.(a) Calculate the frequencies of(i) the fundamental note(ii) the first and second overtones(b) Calculate the velocity of the wave along the string.Solution:

Answers for self test 31. C 2. B 3. D 4. D

5.a)

The waveform of the stationary wave formed in the wire is as shown above. We have

b)

c)

19

½ ------- L ------------------

A A A


6.(a) The particles along the string vibrates vertically to set up a stationary

wave.(b)

where

7.

(a)

Fundamental mode of vibration,

(b)

8.a) Principle of superposition : The resultant displacement produced by 2 waves of a

point is the vector sum of the individual displacement of the 2 waves.

20


(b)(i) Node : are points in the medium that do not have any displacement or vibrate at all. Antinodes : Are points in the medium that vibrates with maximum amplitude. (ii) fundamental frequency : is the lowest frequency or mode of vibration possible by a vibrating string or air column.

Overtone are integral multiple of the fundamental frequency that can occur or produced by a vibrating string or an air column.

9. (a) 2π f = 5Frequency f = 0.76 Hz(b) At the nodes, amplitude = 0 cos 4x = 0

Distance between successive nodes = (4.71 – 1.57) m = 3.14 m(b) Wavelength = 2 x 3.14 m = 6.28 m

10.

11.a) Since the string vibrates in three segments,

therefore length of one segment= 1/3 (length of string) = 1/3 (240 cm) =80 cm

Distance between two consecutive nodes in one segment= /2=80cm = 160cm

Speed of the waves traveling along the string, v=f f= v/

21


But = m/L= 0.06kg/2.4m = 0.025kgm-1

Hence

f=12.5Hzb) f = 3 x fundamental frequency fundamental frequency,fo= 12.5/3 =4.2 Hz.

12.

Unit 4 Objective: By the end of the lesson students should be able to:

describe quantitatively the formation of standing waves in air columns and use the formula for frequency including the determination of end correction

A. Closed pipe

1. For the stationary wave formed in the air column in a closed pipe: (closed at one end and open at the other end )

, Hence,

22


2. There is always an anti node at the open end and a node at the closed end.

3. The air column in closed pipe can vibrate in other modes besides the fundamental mode.


2nd overtone 3rd overtone 4th overtone

ClosedPipe

f1 =3fo f2 =5fo f3 =7fo f4 =9fo

B. Open pipe

1. For the stationary wave produced in the air column in an open pipe:

, Hence,

2. There are antinodes at both the open ends.

3. The air column in an open pipe can vibrate in other modes besides the fundamental mode.

23

Air column vibrating with Fundamental frequency

Air column producing 1st overtone.

Air column vibrating with Fundamental frequency

Air column producing 1st overtone.



2nd overtone 3rd overtone 4th overtone

Openpipe

f1 =2fo f2 =3fo f3 =4fo f4 =5fo

Example 1An organ pipe which is closed at one end has an effective length of 0.64m. If the speed of sound is 320 m s-1 , Calculate the two lowest resonance frequencies.

Solution:

At the first resonance or the fundamental frequency:

= 125 Hz

At the second resonance or the first overtone: = 375 Hz.

Example 2A pipe, open at both sides ,has a fundamental frequency of 5.5 kHz. If the speed of sound in air is 330 ms-1, (a) calculate the effective length of the pipe .(b) calculate the frequencies of the first overtone and the fourth overtone.

Solution:

(a)

24

l =/4

l=31/4

L


(b) First overtone = f1=2fo = 2 x 5.5 x 103 = 1.1 x 104 Hz. Fourth overtone = f4 = 5fo = 5 x 5.5 x 103 = 2.75 x 103 Hz.

Self-test 4

1. How many nodes will be formed in an open pipe when the 3rd overtone sound wave is produced? A. 3 B 4 C 5 D 6

2. A student blows at the end of a drinking straw of length 10.0 cm. What is the lowest frequency of sound produced? [Speed of sound in air = 320 m s-1]

A. 800 Hz B. 1600 HzC. 3200 Hz D. 4800 Hz

3. A pipe is open at both ends. If the length of the pipe is l, the wavelength of the fundamental tone is slightly

A. less than l B. less than 2lC. more than l D. more than 2 l

4. A pipe which is opened at one end only has an effective length of 0.15 m. If the speed of sound in air is 330 ms-1, what is the fundamental frequency of the pipe.

A. 275 Hz B. 550 HzC. 1100 Hz D. 1650 Hz

5. A tube X closed at one end and an opened tube Y have the same length. If the end-correction effect can be neglected, what is the ratio of

when air is blown across the opened end of the tubes?

A. 1:4 B. 2:3C. 3:2 D. 3:4

6. An organ pipe is 0.33 m long, open at one end and closed at the other. The speed of sound is 330 m s-1. Assuming end corrections are negligible.

(a) Sketch the fundamental mode of vibration of the air column in the pipe

25


(b) Calculate the frequency of the fundamental mode.(c) Sketch the first and second overtone modes of vibration of the air column.(d) Calculate the frequencies of the first and second overtones.

7. An organ pipe is 0.33 m long and opened at both ends. The speed of sound is 330 ms-1. Assuming end corrections are negligible.

(a) Sketch the fundamental mode of vibration of the air column in the pipe(b) Calculate the frequency of the fundamental mode.(c) Sketch the first and second overtone modes of vibration of the air column.(d) Calculate the frequencies of the first and second overtones.

8. A well with vertical wall resonates with a note of 7.0 Hz and does not resonant with frequencies less than 7.0 Hz. Estimate the depth of the well if the speed of sound in air is 350 m s-1.

Answer: 1. B 2. B 3. D 4. B 5. D

26


6 (a)

fundamental mode

(b) fundamental frequency,

(c) 1st overtone

1st overtone,

2nd overtone

2nd overtone,

7. (a) fundamental mode.

(b) 1st. overtone

2nd overtone

8. Depth, L = m

27


UNIT 5Objective: At the end of the lesson, students should be able to:

understand and define sound intensity. relate sound intensity to distance from source. relate sound intensity to the amplitude of vibrations of air molecules. understand and define sound intensity level in Bel and decibel,dB. calculate the sound intensity level at different distances from source. relate sound intensity and sound intensity level.

A. Sound Intensity

1. The sound intensity I is the amount of energy transferred per second (or power) by the sound wave that passes through unit area of any plane surface normal to the direction of propagation of the waves.

2. Intensity, , where P is the power of the source and r is the

distance from the point source.

3. The unit of intensity is J s-1 m-2,or W m-2.4. The intensity at a point is inversely proportional to the square of the

distance r of the point from the energy source.

5.

6. The intensity at a point is directly proportional to the square of the amplitude A of the vibrating particle at that point.

7.

8. Therefore,

Example 1.A small source of sound radiates energy equally in all directions. The intensity of the sound 2.0 m from the source is 3.0 x 10-4 Wm-2. Assuming that the sound is propagated without energy loss, what will be the intensity of the sound at a distance of 3.0 m from the source.

Answer:

Using

28


Example 2.A point source of sound radiates energy uniformly in all directions. At a distance of 2.0 m from the source, the amplitude of vibration of air molecules is 5.0 x 10-3 m . Assuming that no sound energy is absorbed, calculate the amplitude of vibration of the air molecules at a distance of 5.0 m from the source.

Answer:

Using

B. Sound Intensity Level

1. Sound intensity level of a sound is the logarithm of the ratio of the sound intensity I to the threshold intensity of hearing Io.

2. =

3. or =10

4. change of sound intensity level:

5. The threshold of hearing is the minimum intensity of audible sound and the universally agreed standard intensity is 10-12 W m-2.

6. The intensity level of the threshold of hearing is 0 dB.

29


Example 1:The sound intensity level that is detected by an observer at a particular distance from a sound source is 105 dB. What is the intensity of the sound falling on the observer’s ear?

Answer:

Intensity level,

Example 2:The difference in intensity level between the first sound of intensity I1 and the

second sound of intensity I2 is 20 dB. What is the ratio

Answer:

,

Self-Test 5:1. A point source of sound emits energy equally in all directions at a constant

rate and a person 8 m from the source listens. After a while, the intensity of the source is halved. If the person wishes the sound to seem as loud as before, how far should he be from the source?A. 2 m B. C. 4 m D.

2. A sound wave of amplitude 0.20 mm has an intensity of 3.0 W m-2. What will be the intensity of a sound wave of the same frequency which has an amplitude of 0.40 mm?A. 4.2 W m-2 B. 6.0 W m-2

C. 9.0 W m-2 D. 12 W m-2

3. A point source produces a sound at a rate of 120 W uniformly in all directions.

30


What is the intensity level of the sound at a distance of 2.0 m from the source?[Take threshold of hearing = 10-12 W m-2]A. 30 dB B. 124 dBC. 130 dB D. 141 dB

4. If the level of intensity of a sound is raised by 10 dB, what is the ratio of the new sound intensity to the original sound intensity?A. 0.1 B. 1C. 10 D. 1010

5. The intensity of a sound save is 6 m W cm-2. If the intensity level is increased by 10 dB, what is the new intensity of the wave in m W cm-2?A. 0.06 B. 6.6C. 12 D. 60

6. A spherical sound wave is emitted by a point source. What is the intensity level of this sound at 5 m from the source compared to the intensity level at 1 m from the source?A. +7 dB B. -7 dBC. +14 dB D. -14 dB

7. A small source of sound radiates energy equally in all directions. At a particular frequency, the intensity of the sound 1.0 m from the source is

, corresponding to an amplitude of oscillation of the air molecules of 70 Assuming that the sound is propagated without energy loss, what will be (a) the intensity of the sound(b) the amplitude of oscillation of the air molecules, at a distance of 5.0m from the source?

8. Define intensity of sound at a point from a source. A point source of sound radiates energy uniform in all directions. At a distance

of 3.0 m from the source, the amplitude of vibration of air molecules is. Assuming that no sound energy is absorbed, calculate the

amplitude of vibration 5.0 m from the source.

9. The difference in intensity level between the first sound of intensity I1 and the

second sound of intensity I2 is 35 dB. What is the ratio ?

10. Define sound intensity level. The intensity level of a sound at a distance of 6.0 m from a source is 60 dB.

What is the intensity level of the sound at a distance of 12.0 m away from the source?

31


Suggested Answers to Self-Test 5:1.D 2. D 3. B 4. C 5. D 6.D

7.

(a)

(b)

(8)

r2= 5.0 m, r1=3.0m, a2 = ?,

32

S


(9) ,

10.

=10

Thus, 50=10 , ,

The resultant intensity 2 x

=10 = 53 dB

UNIT 6Objective: At the end of the lesson, students should be able to:

Understand the phenomena of beat. Calculate beat frequency.

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Understand the phenomena of Doppler Effect. Calculate the apparent change in frequency due to relative motion of

sound source and observer.

A. Beats

1. Beats: When two notes of slightly different frequencies but similar amplitude are sounded together, the loudness increases and decreases periodically and beats are said to be heard.

2. Beat frequency, f = f2 - f1

3. Frequency of resultant wave =

B. Doppler Effect

1. Doppler Effect is the apparent change in the frequency of a sound due to the relative motion between the observer and the source.

2. The motion of the source will cause the apparent change in wavelength.

3.

4. The motion of the observer will cause the apparent change in speed of sound.

5.

6. The apparent frequency, of the waves received by an observer who experience the Doppler effect is given by

7.

+vo Observer moving towards the source-vo Observer moving away from the

source+vs Source moving away from the

observer-vs Source moving towards the

observerExample 1:Two tuning fork of frequencies 350 Hz and 354 Hz are sounded together in an enclosed room. What is the beat frequency heard by an observer inside the room?

Answer:

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Beat frequency,

Example 2:A source producing sound waves of frequency 1500 Hz, is moving away from a stationary observer with a velocity of 45 ms-1. The velocity of sound =330 ms-1. What is the frequency of the sound heard by the observer?

Answer: source moving away, vs=+45 ms-1, observer stationary,vo=0

Using

Example 3:Velocity of sound = 330 ms-1. A source producing sound of frequency 1500 Hz is moving towards an observer at a speed of 30 ms-1, and the observer is moving towards the source at a speed of 20 ms-1. What is the frequency of the sound heard by the observer?

Answer:Source moving towards observer,vs=-30 ms-1, obserever moving towards source, v0=+20 ms-1.

Self-Test 6:1. A car sounds its horn while moving along a straight road. An observer who is

stationary by the side of the road finds that the frequency of the sound made by the horn is f1 when the car is approaching and f2 when the car is moving away from her after passing her. The speed of sound is 320 m s-1 and the speed

of the car is 40 m s-1. The ratio

A. 0.78 B. 1.29C. 0.89 D. 1.12

2. A car sounds its horn and moves towards a large building. The sound of the horn that is reflected from the building returns and undergoes superposition with the emitted sound. What is the beat frequency heard by the driver of the car? (Speed of the car = 10 m s-1; speed of sound = 330 m s-1; original frequency of the horn = 200 Hz)A. 6.06 Hz B. 6.25 HzC. 11.8 Hz D. 12.5 Hz

3. A tuning fork of unknown frequency makes 3 beats per second when sounded

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together with another standard tuning fork of frequency 384 Hz. The beat frequency becomes less if a little wax is fixed to the branches of the first tuning fork. What is the frequency of this tuning fork?

4. What is beat? Two sound waves with frequencies 450 Hz and 456 Hz undergo superposition.

What is the frequency of the resultant wave and what is the beat frequency?

5. What do you understand by Doppler’s Effect? A source moves towards a stationary observer with a speed one-quarter that of

sound. If the frequency from the source is 600 Hz, what is the apparent frequency heard by the observer?

6. A train moves away from an observer standing on the platform with a velocity of 20 ms-1. The siren emitted by the train has a frequency of 1 200 Hz. (a) What is the frequency of the sound heard by the observer?(b) What is the frequency of the sound heard by a passenger in the train?(Speed of sound = 330 ms-1)

7. A police patrol car moves with a peed of 30 ms-1 has its siren on. The sound emitted is at frequency 1 000 Hz. What is the apparent frequency of the sound heard by the driver of a car which is approaching the patrol car with a speed of 30 m s-1. (Speed of sound = 330 m s-1)

8. A stationary sound source emits sound of frequency 5000Hz. An observer moves towards this source with a speed 25 m s-1 in a straight line.(a) What is the frequency of the sound heard by the observer?(b) If a part of the sound is reflected by the observer and travels towards the source,what is the frequency of the reflected wave?

(speed of sound wave=300 m s-1)

9. Beats are produced at one point when two sound waves meet.(a) If the frequency of one of the waves is 600 Hz and the other is 596 Hz,what is the beat frequency?(b) If the intensity of each of the waves is 7.0 x 10-9 W m-2,what is the maximum intensity level heard at that point?

10. A whistle which emits sound of frequency 2 000 Hz moves away from a observer towards a cliff with a speed of 20 m s-1.(a) What is the frequency of the sound that the observer hears directly from the whistle?(b) What is the frequency of the sound reflected from the cliff and heard by the observer?(c) What is the frequency of the beats produced by the incident and reflected sound?(d) Canwe hear the beat?

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(Speed of sound = 330 m s-1)

Suggested answers to Self-Test 6:1. A 2. D3. or

when wax is fixed to f, and

(4) Beat frequency = 456 – 450 = 6 s-1

Frequency of resultant wave = Hz

(5) S V O

apparent wavelength ’

apparent frequency,

(6) S O

f

(a)

(b) frequency of sound heard by passenger in the train is the same as frequency of

siren emitted by train 1200 as there is no relative motion between the source and

observer.

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(7) S O

(8) (a) S O

(b) S O

f’

(9) (a) Beat frequency = 600-596 = 4 Hz(b) ,

maximum amplitude = a + a = 2amaximum intensity,

max. intensity level

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(10) S O

(a)

(b)

(d) Beat frequency = (e) Can, but the maxima are too close to hear clearly.

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