Wave Reflected from a Denser Medium

INTERFERENCE) OF WAVES AND STANDING WAVES

Case 1: Wave Reflected from a Denser Medium

Let ^y_i and y_r be the incident and the reflected wave respectively. If

1cos

yi A Zt Kx I then ^y_r ^A₂^cos

^{Zt K L x 2 I S}

.And the net wave y_net is given by^y_net ^y_i ^y_r ^A₁^cos

^{Zt Kx I}

^A₂^cos

^{Zt K L x}

²

^{I S}

.

At a given location x, (x = constant) these two waves are harmonic

oscillators with phase angles E₁ KxM and E₂ K L x

2

M S. Therefore

1cos 1 2cos 2

ynet A KxE A Z Et which can be written as y_net Acos

Z Gt

with ^{A A}₁^2A₂^{22A A}_{1 2}^cos

^E₂^E₁

^{and 1 1 2 2}

1 1 2 2

1 sin sin

tan cos cos

A A

A A

E E

G E E

§ ·

¨ ¸

¨ ¸

© ¹

. Where

2 1 k L x2 kx 2 (k L x)

E E I S I S.

For simplicity, let’s assume that the wave is reflected 100% (even though some of it may be transmitted). Then A₁ = A₂. Therefore

1 2 2 1

1 2

2 2 2 cos

A A A A A E E ^2A₁^22A₁^{2cos 2}

^{k L x S}

^A₁ ^{2 1 cos 2}

^{k L x}

^{. And}

using the trigonometric identity ^{1 cos2} 2 T sin²T, the following expression for the net amplitude as a function of distance is obtained.

2 1 sin

A A k L x

With this as an amplitude, the net wave is given by

2 1 sin cos

net A k L x t

y Z G

This shows that the amplitude of the harmonic oscillators is a function of position, x. And since the amplitude varies like a sine, there are going to be points with zero amplitude or no vibration. These are the points called the nodes of the waves. And of course the points where sine is a maximum are the antinode points.

The nodes are the values of x for which the amplitude is zero. Let xm be the ^mth node with x0 L being the reflection point. Then ^sin

^{k L x}

_m

⁰ which implies ^{k L x}

_m

^{mS or}

12 xm L mO

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The distance between consecutive node equal to ^x_m1^x_m ^§¨©^Lm₂^{1O· §}¸ ¨¹ © ^Lm₂^O·¸¹ ^O₂^.. That is the size of one loop of the standing wave is half of the wavelength of the wave.

length of one loop x^m 1 x^m O2

At the point of reflection x = L. Therefore 2 ₁ sin

0

Ax L A K L L . That is, when a wave is reflected from a denser medium, the reflection point is a node.

Boundary conditions restrict the wavelengths of waves that can exist as standing wave in a medium. Two very common boundary conditions will be considered.

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a) The other end (x = 0) also required to be a node: That is, the standing wave will have nodes at both ends. An example of this is a standing wave in a string where both ends are fixed. The amplitude at arbitrary value of x is given b ^{A 2A}₁ ^sin

^{k L x}

(assuming 100%

reflection). The condition A|_{x 0} 0 implies that sin kL sin ²S L 0 O

§ ·

¨ ¸

© ¹ or ²

n L n S S

O where n is a positive integer. Therefore if both ends are nodes only waves with wavelengths

n 2L O n

Where n is a positive integer can form standing waves. In other words the only waves that can form standing waves in a standing wave of length L ₁ 2 ,1 ₂ ²2 , ₃ ²3 , ₄ ²4 2,

L L L L

L L

O O O O }

and so on. Since the speed of a wave depends on the properties of the medium only, the frequencies that can exist as a standing wave are also restricted and the frequencies of the standing waves are given as _n

f vn

O where v is the speed of the wave. But O_n ²_n^L. Therefore the only frequencies that can form standing waves are given by

n 2 f n

L v

§ · ¨ ¸

© ¹

Where n is a positive integer. The standing wave whose frequency is f_n called the n^th harmonic of the standing wave. The first harmonic is also called the fundamental harmonic. With

1, 1 2V

n f L. Therefore the frequency of the harmonic frequency may be expressed in terms of the fundamental frequency as

1 fn nf

This shows that the frequencies of all the harmonics of the standing wave are integral multiples of the fundamental frequency f₁. The part of the standing wave between two consecutive nodes is called a loop. The number of loops for the harmonic (N) can be obtained by dividing the length of the standing wave (L) by the size of one loop _§O₂ⁿ _·

¨ ¸

© ¹. Using the expression for On in terms of L, it follows that N = n, that is, the n^th harmonic has n loops.

Example: A wave of the form ݕ ൌ ͷ …‘•ሺͶߨݔ െ ʹͲߨݐሻ is reflected from a boundary (obstacle) 2m away from the point where it is initiated. If the wave is reflected 100%.

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a) Calculate the amplitude of the harmonic oscillation of a particle located at an antinode.

Solution:

At an antinode the amplitude is maximum: A₁ = 5

2 1 sin

A A k L x

The maximum value of A occurs when •‹ሾܭሺܮ െ ݔሻሿ ൌ ͳ because maximum of a sine function is 1. Therefore

2 1 2 5 10 m Amax A

b) Determine the size of one loop of the standing wave (that is distance between consecutive nodes).

Solution

4 1/m; _m 1 _m ?

k S x x

2 4 1/m

k S S

O 0.5 m O

1 2 0.5 0.25 m2

m m

x x O

c) Determine location of the nodes.

Solution:

0 1

2 m; 0.5 m; , ,...?

L O x x

12 xm L mO

1 2 3 4

2 m; 1.75 m; 1.5 m; 1.25 m; 1 m;

xo x x x x

5 0.75 m; 6 0.5 m; 7 0.25 m; 8 0

x x x x

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d) Determine the number of loops.

Solution:

0.5 m; N ? O

Since the size of one loop is λ/2

2 2 2 8

2 0.5

L L

N O O

e) Calculate the amplitude of the harmonic oscillation of a particle located at x=0.4 m.

Solution:

0.4 m; _x 0.4 ?

x A

2 1 sin

A A K L x

0.4 2 5 m sin 4 2 0.4 9.5 m

A x S

Example: Consider a standing wave formed in a string of length 4m fixed at both of its ends. The mass of the string is 0.02 kg. There is a tension of 100 N in the string.

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a) Determine the wavelengths of the first 3 harmonics.

Solution:

2 m; _n 2L ?

L O n

1

2 4 m 8 m O 1

2

2 4 m 4 m O 2

3

2 4 m 8 m

3 3

O

b) Determine the speed of the wave in the string.

Solution:

100 N; 4 m; 0.02 kg; ?

T l m v

v TP

0.02 kg/m 0.005 kg/m 4

m P l

1000.005 m/s 141 m/s v

c) Calculate the wavelength and frequency of the 15^th harmonic.

Solution:

15 15

15; ?; ?

n f O

1 v1 141 Hz 17.6 Hz8 f O

15 15 1 15 17.6 Hz 264 Hz

f f

15 15

141 m 0.534 m 264

v O f

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b) The x = 0 end required to be an Antinode: Remember the other end (the reflection end) is a node. Therefore this is a standing wave with a node on one end and an antinode on the other end. An example of this is sound resonance in a tube closed on one end and open on the other end. It will have an antinode on the open end and a node on the closed end.

Since ^{A 2A}₁^sin

^{k L x}

, for an antinode, the value of the sine should be one because the maximum of sine is one. Therefore

^{k L x}

_{x 0} ¹ which implies that k L_n ^{2 1n}₂ S where n is a positive integer. And with _n ²

n

k S

O , the following expression for the wavelength of the n^th harmonic can be obtained.

2 14

n L

O n

This implies that the only waves that can exist in a standing wave of length L with a node on one end and an antinode on the other end are waves with wavelengths.

1 2 1 14L 4 ,L 2 2 2 14L 43L, 3 2 3 14L 45L, and so on.

O O O

and so

If the speed of the wave is v (the speed depends only on the properties of the medium), then the frequency of the n^th harmonic is given by

⁴ (2 1

n n

v v

f L

O n

. Or 2 1₄

n n vL

f

Since, ^f_{1 4}^v_L, the frequency of the n^th harmonic may also be written in terms of the fundamental frequency ^P0 as

2 1

₁

n n

f f

The number of loops in a standing wave may be obtained as the ratio of the length of the standing wave to the size of one loop (half of the wavelength).That is, _{/ 2 1 4} ^{2 1}₂

2 2 1

n

L L n

N L

O n

.

The n^th harmonic has ^{2 1}

n2 loops when one end is a node and the other antinode.

Example: Consider sound resonance (standing wave) formed in a pipe open at one end closed at the other end. Its length is 0.5m. Assume the temperature is 20^°C.

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a) Calculate the wave lengths of the first 3 harmonics.

Solution: The sound resonance will have a node at the closed end and antinode at the open end.

0.5 m; 20 C 293 K; _n ? L T q q O

4 0.5

4 m 2 m

2 1 2 1 2 1

n L

n n n

O

1 2 1 12 2 ;m 2 2 2 12 m 23 m; 3 2 3 12 m 25 m

O O O

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

q

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b) Calculate the frequencies of the first 3 harmonics.

Solution

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

q

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