Wave Reflected from a less Dense Medium - INTERFERENCE) OF WAVES AND STANDING WAVES

INTERFERENCE) OF WAVES AND STANDING WAVES

Case 2: Wave Reflected from a less Dense Medium

b) Calculate the frequencies of the first 3 harmonics.

Solution

331 m/s 273 K 331 293273 m/s 343 m/s v T

1 v1 243 Hz 172 Hz2 f O

2 1

₁

n n

f f

2 4 1 172 Hz 516 Hz; 2 6 1 172 Hz 860 Hz

f f

c) Calculate the wavelength and frequency of the 4^th harmonic Solution

11 ?; 11 ?

f O

11 2 11 1 1 21 172 Hz 3612 Hz

f f

11 11v 3612343 m 0.095 m O f

Requiring the x = 0 point to be a node will result in a node in one end and an antinode in the other end which has been discussed already. So here only the case where the x = 0 end is required to be an antinode will be discussed. That is, a case where both ends are antinodes.

An example is sound resonance formed in a pipe open at both ends. With a similar analysis, it can be shown that the equations for the case where both ends are antinodes are identical to the case where both ends are nodes: O_n ²_n^L, fⁿ n L2v ^, f¹ 2vL^, fⁿ nf¹, N = n.

Example: Consider sound resonance (standing waves) in a 2 m pipe open at both ends.

Assume temperature is 20^°C. Calculate the wavelength and frequency of the 8^th harmonic.

Solution:

8 8

2 m; 20 C 293 K; ?; ? L T q q O f

1 2

f v L

331 m/s 331 293 m/s 343 m/s

273 273

v § T ·

¨ ¸

1 343 Hz 85.7 Hz f 2 2

fn nf

8 8 1 8 85.7 Hz 426.7 Hz

f f

8 8

343 m 0.8 m 426.7

v O f A Beat

A beat is a wave with points of zero vibration which for a given location are separated by equal intervals of time; for example, for sound waves, at a given location, the event of no sound will be separated by equal intervals of time. The wave between two consecutive events of zero vibration is called a beat. The time taken for a beat is called the beat period and the number of beats per second is called the beat frequency. A beat is formed when two waves with close frequencies interfere. Consider two interfering waves ^y₁ ^A^cos

^k₁^x^Z₁^t

and ^y₂ ^A^cos

^k₂^x^Z₂^t

of frequencies Z₁ and Z₂ such that 1 2 1

Z 1 Z

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For simplicity it is assumed that both waves have the same frequency. Since the waves are travelling in the same medium they will have the same speed; that is, ¹ ²

1 2

k k v Z Z

. Thus,

1 1

2 2

cos cos

ynet A k xZt A k xZt . This expression can be expressed as a product of two cosines using the trigonometric identity cos cos 2cos cos

2 2

x y x y

x y §¨© ·¸¹ §¨© ·¸¹.

1 2 1 2 1 2 1 2

2 cos cos

2 2 2 2

net k k

y A ª«¬§¨©k ·¸¹x§¨©Z Z ·¸¹tº»¼ ª«¬§¨©k ·¸¹x§¨©Z Z ·¸¹tº»¼

As can be seen from this equation, the effect of the interference of two waves of close frequencies is the product of a high frequency wave ^§^¨_©^cos^ª^«_¬^§^¨_©^k¹₂^{k x}²^·^¸_¹ ^§^¨_©^{Z Z}¹₂ ²^·^¸_¹^t^º^»_¼^·^¸_¹ and a low frequency wave ^§^¨_©^cos^ª^«_¬^§^¨_©^k¹₂^{k x}²^·^¸_¹ ^§^¨_©^{Z Z}¹₂ ²^·^¸_¹^t^º^»_¼^·^¸_¹. The result is travelling wave packets (beats) where the amplitude of the high frequency vibrations are controlled by the low frequency wave.

One cycle of of the low frequency wave will result in two beats as a result of the product.

That is the beat frequency f_b is twice the frequency of the low frequency wave. Since the frequency of the low frequency wave is given by ¹ ²

2 f f

, it follows that

1 2

fb f f

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The beat frequency, which is the number of beats per a unit time, is equal to the difference of the frequencies of the interfering waves. And since the beat period (T_b), time taken for one beat, is equal to b ¹

T f ,

ܶ_௕ ൌ ͳ ȁ݂ଵെ݂_ଶȁ

Example: Consider the interference of the following two waves: y₁ 2cos 2

x200t

and

2 2cos 2.02 202

y x t .

a) Express the net wave (beat) as a product of two waves.

Solution:

1 2 1/m; 1 200 rad/s; 2 2.02 1/m; 2 202 rad/s; 2; _net ?

k Z k Z A y

1 2 1 2 1 2 1 2

2 cos cos

2 2 2 2

net k k

y A ª«¬§¨©k ·¸¹x§¨©Z Z ·¸¹tº»¼ ª«¬§¨©k ·¸¹x§¨©Z Z ·¸¹tº»¼

.02 2 4.02 402

2 2 cos cos

2 x 2 t 2 x 2 t

ª§¨ ·¸ § ·¨ ¸ º ª§¨ ·¸ §¨ ·¸ º

«© ¹ © ¹ » «© ¹ © ¹ »

¬ ¼ ¬ ¼

> @ > @

4cos .01x t cos 2.01 201x t

b) How long does one beat take?

Solution:

1 2

1 2 1 1 2 1 200 202 Hz 2 0.318 Hz

2 2 2 2 2

fb f f Z Z Z Z

S S S S S

1 11 s s

b b

T f S

Example: When a sound of frequency 1000Hz interferes with sound of unknown frequency 2 beats can be heard in 5 seconds.

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a) Calculate the two possible frequencies of the unknown wave.

Solution:

1 1000 Hz; 2 Hz 0.4 Hz; 2 ?

b 5

f f f

1 2

fb f f

1 2 b

f f rf

2 1 _b 1000 0.4 Hz

f rf f r

b) If both waves have amplitude of 10^-9 m and the temperature is 20°C, express the beat as a product of two waves (just for one of the unknown frequency).

Solution:

1 2 f1 2000 rad/s; 2 2 f2 2000.8 rad/s; A 10 m; 9 T 293 ; K y_net ?

Z S S Z S S q

331 m/s 331 293 m/s 343 m/s

273 K 273

v T

1 1 2000 1/m 18.318 1/m 343

k Zv S

2 2 2000.8 1/m 18.325 1/m 343

k Zv S

1 2 1 2 1 2 1 2

2 cos cos

2 2 2 2

net k k

2 10 cos 0.004 1.26 cos 18.329 6284

y etn u x t x t

Practice Quiz 14.2 Choose the best answer

1. Which of the following is not true about standing waves.

A. The speed of the wave in a string is proportional to the square root of the tension in the string.

B. For a standing wave of a given length, the greater the number of loops the greater the frequency of the wave.

C. A standing wave is a wave with equally spaced points of zero vibration.

D. The length of one loop of a standing wave is equal to half of the wavelength of the wave.

E. The points of zero vibration of a standing wave are called antinodes.

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2. Which of the following is correct about a beat?

A. The frequency of a beat is equal to the difference between the frequencies of the interfering waves.

B. A beat is formed by the interference of an oncoming and a reflected wave.

C. A beat is a wave with points of zero vibrations separated by equal distances.

D. The period of a beat is equal to the difference between the periods of the interfering waves.

E. A beat is formed by the interference of two waves of the same frequency.

3. A standing wave is formed when a wave of the form y = 0.63 cos (234t – 7.8x) m initiated at the origin interferes with its reflection reflected at x = 10 m 100% from a denser medium. Calculate the amplitude of the oscillation of a particle located at x = 9 m.

A. 1.258 m B. 1.636 m C. 0.755 m D. 1.132 m E. 1.51 m

4. Calculate the wave length of the 3^th harmonic of a standing wave formed in a 4.5 m string clamped at both ends.

A. 4.2 m B. 3 m C. 2.7 m D. 3.6 m E. 3.9 m

5. A string of mass 0.12 kg and length 1.2 m is clamped at both ends. It is subjected to a tension of 110 N. Calculate the frequency of the 6^th harmonic standing wave formed in the string.

A. 49.749 Hz B. 74.624 Hz C. 58.041 Hz D. 82.916 Hz E. 66.332 Hz

6. A sound wave resonance is formed in a pipe of length 5 m open at both ends.

Calculate the wavelength of the 11^th harmonic.

A. 0.636 m B. 0.818 m C. 0.909 m D. 1.091 m E. 0.545 m

7. A sound wave resonance is formed in a pipe of length 2 m open at both ends.

Calculate the frequency of the 7^th harmonic. (The temperature is 10°C.) A. 353.858 Hz

B. 766.693 Hz C. 648.74 Hz D. 589.764 Hz E. 707.716 Hz

8. A sound wave resonance is formed in a pipe of length 0.5 m open at one end closed at the other. Calculate the wavelength of the 12^th harmonic.

A. 0.07 m B. 0.087 m C. 0.122 m D. 0.096 m E. 0.061 m

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9. A sound wave resonance is formed in a pipe of length 5 m open at one end closed at the other. Calculate the frequency of the 6^th harmonic.

(The temperature is 35°C.) A. 193.368 Hz

B. 232.042 Hz C. 212.705 Hz D. 116.021 Hz E. 251.378 Hz

10. A beat is formed by the interference of a wave of frequency 104 Hz and a wave of frequency 109.5 Hz. Calculate the number of beats that can be heard per second.

A. 5.5 Hz B. 100 Hz C. 0.182 Hz D. 213.5 Hz E. 0.005 Hz

11. A beat is formed by the interference of a wave of frequency 1000 Hz and a wave of an unknown frequency. If a beat lasts 0.7 s, which of the following is a possible frequency for the unknown wave?

A. 600.857 Hz B. 1001.429 Hz C. 701 Hz D. 1101.571 Hz E. 1201.714 Hz

12. A beat is formed by the interference of the two sound waves ΔP₁ = 2 cos (30x – 10200t) Pa

ΔP₁ = 2 cos (30.294x – 10300t) Pa

Give a formula for the beat (net wave) as function of position and time.

A. ΔP₁ = 4 cos (56.23x – 37230t) cos (0.567x – 70t) Pa B. ΔP₁ = 4 cos (30.147x – 10250t) cos (0.567x – 70t) Pa C. ΔP₁ = 4 cos (30.147x – 10250t) cos (0.147x – 50t) Pa D. ΔP₁ = 4 cos (23.456x – 20346t) cos (0.567x – 70t) Pa E. ΔP₁ = 4 cos (23.456x – 20346t) cos (0.147x – 50t) Pa

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Dalam dokumen PDF Daniel Gebreselasie Sound and - Trường Đại Học Bách Khoa Hà Nội (Halaman 61-69)