GUIDE PH1140E2009
STUDY - 3
Instructor : Rudra Kae
PHYSICS - 1140 : Oscillations and Waves, TERM - E , 2009 TEXT-BOOK : University Physics - Young and Freedman, 12th Edition Objectives 1. Describe the resonance in sound. 2. Describe interference of two sound waves. 3. Describe the formation of beats and calculate the beat frequency. 4. Explain why the pitch of sound rises and falls when there is a relative motion between a source and an observer (Doppler Eect) and calculate the Doppler shifted frequencies. 5. Describe the interference of light waves (Youngs double-slit experiment) and calculate the fringe spacings in the interference pattern. 6. Describe an interference in thin lms and calculate the thickness of thin lms using the idea of interference. OBJECTIVES - 1, 2, 3 (STUDY MATERIALS and DETAILS) : Resonance and Sound (16.5), Interference of Waves (16.6) and Beats (16.7) 1. Explain the resonance of sound waves. When the frequency of the driving force is equal to the natural frequency of an oscillating system, the amplitude of oscillation becomes a maximum. This phenomenon is called
resonance. For example, when the frequency of the sound from the speaker driven by an amplier becomes equal to one of the normal mode frequencies of the pipe, [ refer to Fig.(16.19a)], the amplitude of oscillating air becomes a maximum. Therefore, we get several amplitude peaks, [refer to Fig.(16.19b): Resonance Curve], if the frequency of the sound is varied by adjusting the amplier. These peaks are due to the matching of the normal mode frequencies of the pipe with the values of the driving frequency. 2. Explain the interference of two sound waves. When two sound waves of the same frequency overlap in some region of space, they tend to reinforce or cancel each other. This phenomenon is called interference, [refer to Fig.(16.21)]. If the overlapping sound waves of equal amplitudes are in phase, there is a reinforcement 1
PH1140E2009
STUDY GUIDE - 3
Instructor : Rudra Kae
due to the two waves and a maximum sound is detected at that point. This is called a constructive interference. Thus, a point of destructive interference occurs when the path dierence of the waves from two sources is an integral multiple of , i.e., path dierence, (x) = n, n = 0, 1, 2, ..., [refer to Fig.(16 .22a]. If the overlapping sound waves of equal amplitudes are out of phase, there is a complete cancelation and no sound is detected at that point. This is called a destructive interference. Thus, a point of destructive interference occurs when the path dierence of the waves from the two sources is an odd integral multiple of (x) = (2n + 1) , n = 0, 1, 2, ..., [ refer to Fig.(16.22b)]. 2 3. Explain the formation of beats and derive the beat frequency, fbeat = fa fb . When two sound waves of the same amplitude but of slightly dierent frequencies traveling in the same direction interfere, there is a rise and fall of the intensity of sound in regular intervals of time. This phenomenon is called beats,[ refer to Fig.(16.24)] . The period at which the loudness of the sound repeats is called the beat period and the frequency at which this happens is called the beat frequency. The beat frequency is equal to the dierence between the frequencies of the two sounds, i.e., fbeat = fa fb . Consider two traveling sound waves of the same amplitude and of slightly dierent frequencies, fa and fb in the form: ya = A sin a t and yb = A sin b t, where a = 2fa and b = 2fb . When these wave travel in the same direction and principle of superposition :
, 2
i.e., path dierence,
interfere, the resultant displacement is given by the y = ya + yb = A sin a t A sin b t. : sin A sin B = 2 cos 2A sin
a b 2 a b 2 A+B 2 a +b 2
By using trigonometric transformation formula , we get the resultant wave equation as,
2
sin t.
AB 2
y=
t cos
The intensity of the resultant sound varies as the
a b 2 a b 2
square of the amplitude, i.e., I sin sin t = 1 = sin (2n + 1) 2
t . For the maximum intensity ( a beat), t = (2n + 1) , n = 0, 1, 2, 3, ...,. After doing 2
some algebra, we get the beat frequency, fbeat = (fa fb ). Suggested problems Worked out problems : 16.13, 16.14 End-of-the-Chapter problems : 16.31, 16.33, 16.35, 16.37, 16.39,
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PH1140E2009
STUDY GUIDE - 3
Instructor : Rudra Kae
OBJECTIVES - 4 , 5 , 6 (STUDY MATERIALS and DETAILS) : Doppler Effect(16.8) and Doppler eect (Notes), Interference and Coherent Sources (35.1), Two-Source Interference of Light(35.2) and Interference in Thin Films (35.4) 1. Explain the Doppler Eect in sound and calculate the frequency of the sound heard by a listener in dierent cases: (i) the listener is moving and source is at rest, (ii) the source is moving and the listener is at rest, and (iii) both the source and the listener are moving. When there is a relative motion between a source of sound and a listener, the frequency heard by the listener is dierent from the frequency of sound emitted by the source. This phenomenon is called the Doppler Eect. For example, the pitch of the sound from the siren of an ambulance approaching a stationary listener is higher (than its pitch when the ambulance is at rest) and the pitch decreases when it passes and goes away from the listener. Setting up a proper coordinate system, the frequency of the sound heard by a listener in all cases is given by the formula: fL = v + vL v + vS fS ,
where fS is the frequency of the sound emitted by a source, fL is the frequency of the sound heard by the listener, vS is the velocity of the source, vL is the velocity of the listener and v is the speed of sound. The sign convention for the velocities are as follows: (i) The speed of sound is always positive. (ii) The direction from the listener to the source is positive and from the source to the listener is negative. If the listener is stationary, vL = 0 m/s and if the source is stationary, vS = 0 m/s. 2. Explain coherent sources, with examples. Two sources which emit monochromatic waves (waves of the same frequency) in phase (or waves with some denite, constant phase relationship) are called the coherent sources. For example, two identical slits separated by a small distance and illuminated by a monochromatic source of light can be considered to be coherent sources. When the waves emitted by coherent sources interfere, they form a regular pattern of intensity distribution by reinforcement or cancelation. If there is a reinforcement, the resultant intensity is a maximum and the interference is called a constructive interference. If there is a cancelation, the resultant intensity is a minimum and the interference is called a destructive interference. 3
PH1140E2009
STUDY GUIDE - 3
Instructor : Rudra Kae
D . d
3. Explain the double slit-experiment and derive an expression for the fringe-width, =
The double slit experiment was rst done by Thomas Young in 1800 which proved the wave nature of light. Consider two slits S1 and S2 separated by a small distance d and illuminated by a monochromatic light of wavelength , [refer (Fig.35.5)]. The light waves through the slits interfere constructively or destructively depending upon the path dierence and produce an interference pattern of bright and dark bands on the screen at a distance R. These bright and dark bands are called interference fringes, [refer (Fig.35.5a)]. The distance between the two consecutive bright or dark fringes is called a fringe width or fringe a spacing. The position of any point P on the screen, at an angle m with the horizontal line through the middle of the slits, (ym ) = R tan m , [refer (Fig.35.5b)]. The path dierence between the waves reaching the point P from the two slits, x = d sin m = m, m = 0, 1, 2, 3, ... , if P is the position of a bright fringe. Combining these relations for small angle, m , the position of the mth bright fringe, ym =
mR , d
m = 0, 1, 2, 3, .... The angular distance
ym R
of a bright fringe can be obtained from (ym ) = R tan m as tan m = the position of the (m + 1)th bright fringe, ym+1 = fringe spacing, ( ) = ym+1 ym =
R . d (2m+1)R , 2d (m+1)R . d
=
m . d
Similarly,
Therefore, the fringe width or
The position of a dark fringe is given by ym =
m = 0, 1, 2, 3, .... The anguym R
lar distance of a dark fringe can be obtained from (ym ) = R tan m as tan m =
=
(2m+1) . 2d
We will get the same fringe width by considering the dark fringes too. Hence the bright and dark fringes are equally spaced on the screen. The central fringe is the bright fringe and as we go outward, the dark and bright fringes are distributed alternately, [refer (Fig.35.6)]. 4. Explain the relation between the phase dierence () and the path dierence (x): () =
2 x.
5. Explain briey, about the interference in thin lms. When light traveling through air (refractive index = na ) falls on a transparent thin lm of refractive index (refractive index = nb ) perpendicularly, it is partially reected by the upper surface of the lm and is partially transmitted into the lm which gets reected from the lower surface. The light reected from the lower surface interferes with the light reected from the upper surface constructively or destructively depending upon the path (or phase) dierence. Because of this, a thin soap lm or an oil slick on water looks colorful. If the thickness of the lm is t, the path dierence = 2t. The conditions for constructive or destructive interferences are as follows: (i) If there is no relative phase shift, con4
PH1140E2009
STUDY GUIDE - 3
Instructor : Rudra Kae
structive interference takes place under the condition: 2t = m, m = 0, 1, 2, ...,. (ii) If there is no relative phase shift, destructive interference takes place under the condition: 2t = (2m + 1) , m = 0, 1, 2, ...,. (iii) If there is a relative phase shift of radians, 2 constructive interference takes place under the condition: 2t = (2m + 1) , m = 0, 1, 2, ...,. 2 (iv) If there is a relative phase shift of radians, destructive interference takes place under the condition: 2t = m, m = 0, 1, 2, ...,. Note: If light traveling through air hits water surface perpendicularly, there is a phase shift of a half cycle or radians on reection because the refractive index of air is less than that of water. On the other hand, if light traveling through water emerges into air perpendicularly, there is no phase shift on reection because the refractive index of water is less than that of air, [ refer (Fig. 35.13) and Equation (35.16)]. Suggested problems Worked out problems : 16.15-16.19, 35.1, 35.2, 35.4, 35.5, 35.6 End-of-the-Chapter problems : 16.41, 16.43, 16.44, 16.46, 16.47, 35.1, 35.2, 35.4, 35.8, 35.9, 35.11, 35.16, 35.27, 35.34, 35.36
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PH1140E2009 Home Assignments
STUDY GUIDE - 3
Instructor : Rudra Kae
Assignment-10 (DUE : Midnight, Friday, June 19, 2008) This is an online assignment, available on : http : //www.masteringphysics.com/site. Assignment-11 (DUE : Conference, Tuesday, June 23, 2008) Please read the Homework Rules and submit the solution accordingly. 1. Two organ pipes, both closed at one end and open at the other, have lengths of 0.98 m and 1.00 m. (a) Calculate the beat frequency when both pipes are played at their rst harmonic. (b) Calculate the beat frequency when both pipes are played at their third harmonic. 2. A train is traveling at 30 m/s in still air. The frequency of the note emitted by the train whistle is 250 Hz . A passenger is on another train moving in the opposite direction to the rst at 20 m/s. The speed of sound in still air is 344 m/s. (a) Calculate the frequency heard by this passenger as he approaches the rst train. (b) Determine the wavelength of the sound for this passenger. (c) Determine the frequency and the wavelength of the sound heard by the passenger after he has passed the train and is receding from it. 3. Two thin slits at a distance of 0.5 mm are illuminated by laser light with a wav...