ch10 mechanical waves

7/30/2019 Ch10 Mechanical Waves

1/5

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x/ cm

y/ mm

0.2 0.6 0.8 1.0 1.2 1.40.4

10

10

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Q

R

S T

CHAPTER 10 Mechanical Waves

Question 10.1

The graph above shows the displacement of the particles in a transverse progressive waveagainst the distance from the source at a particular instant.(a) Write down the points where the speed of the particles is zero.

(b) Write down all the points where the acceleration of the particles is maximum.(c) Write down the points where the velocity of the particles is opposite to each other.(d) Write down the points where the particles are in phase.

(e) State the amplitude and wave length of the wave.(f) If the time required to complete one cycle is 0.25 s, what is the speed of the wave?

Solution 10.1

(a) Pand R

(b) Pand R(c) Q and S or S and T(d) Q and T(e) Amplitude = 10 mm, wave length = 0.8 cm(f) V = f

1=

T1

V = ()(0.8)0.25

= 3.2 m s1

Question 10.2

A wave is moving through a medium with a speed of 250 m s1. The frequency of thewave is 1 250 Hz.

(a) What is the wavelength of the wave?(b) Find the distance between two points which are out of phase by 120 in the medium.

Solution 10.2

(a) V = f250 = 1 250

= 0.20 m


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2(b) = X

360

120 = X0.20

X = 0.067 m

Question 10.3

A plane progressive wave in a medium is represented by the equationy = 0.025 sin (120t 20x)

where yand xare measured in metres (m)and t is in seconds.

Calculate(a) the frequency,(b) the wavelength,(c) the speed of the wave and(d) the maximum speed of a particle in the medium.

Solution 10.3

Compare y = 0.025 sin (120t 20x)

2with y = y

0sin (t x)

(a) = 120but = 2f

120 = 2(3.142)ff = 19.1 Hz

2(b) 20 =

= 0.314 m or 31.4 cm

(c) V = f= (19.1)(0.314) m s1

= 6.0 m s1

(d) For the vibrating particle

V = Y0

= (120)(0.025)= 3.0 m s1


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Question 10.4

A transverse wave in a medium has a frequency of 1 250 Hz and velocity of 300 m s1.Calculate

(a) the angular frequency, of the particles,(b) the value of the wave number and(c) the separation between the two points which are out of phase by 60.

Solution 10.4

(a) = 2f

= 2(1 250)= 7.855 rad s1

2(b) K =

Using v = f 300 = (1 250) = 0.24 m

2

Hence K = 0.24

K = 26.2 m1

2(c) Using = X

60 = 3

2 = X

3 0.24

X = = m6 6

= 0.04 m

Question 10.5

Two waves Y1

= Y0

sin (t KX )Y

2= Y

0sin (t KX)

Superimpose to produce a resultant wave represented by the equationY = 0.4 sin (20t 5.0X 0.8)

where Yand Xare in cm and t in seconds.Calculate(a) the wave length,

(b) the speed,(c) the phase difference between the two waves, and(d) the amplitude of the two waves Y

0.

Solution 10.5

Y = Y1

+ Y2

= Y0

sin (t KX ) + Y0

sin (t KX)

1 1= Y

0[2 sin (t KX + t KX) cos (t KX t+ KX)]

2 2

= 2Y0[sin (t KX ) cos ( )]

2 2

Y = 2Y0

cos sin (t KX ) 12 2


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Compare equation 1 with Y= 0.4 sin (20t 5.0X 0.8)

(a) K = 5.0

2but K = cm1

2

5.0 =

= 1.26 cm

(b) = 20but = 2f

20 = 2ff = 3.18 s1

v = f = (3.18)(1.26) cm s1

= 4.01 cm s1

(c) = 0.8

2

= 1.6 rad

(d) 2Y

0cos = 0.4 cm

2

Y0

= 0.29 cm

Question 10.6

Two transverse progressive waves with equal amplitude, wavelength and frequency butmoving in opposite directions superimpose to produce a resultant wave. The phasedifference between the two waves is .(a) By using the general symbols for wavelength, frequency, amplitude and other

common symbols for the common physical quantities, write an equation for theresultant wave.

(b) If the amplitude and the wavelength of the individual waves are 2.5 mm and 5.0 cmrespectively, determine the maximum amplitude of the resultant wave.

Solution 10.6

(a) Let Y1

= A sin (t KX+ )

Y2

= A sin (t+ KX) Y = Y

1+ Y

2

Y = A sin (t KX+ ) + A sin (t+ KX)

1 1= A[2 sin (t KX+ + t+ KX) cos (t KX+ t KX)]

2 2

= 2A cos ( KX) sin (t+ )2 2

(b) The amplitude of the resultant wave is given by 2A cos ( KX). The amplitude has

2

maximum value when cos ( KX) = 12

Maximum amplitude = 2A= 2(2.5 mm)= 5.0 mm


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1.6 m

0.4 0.4 0.40.1

0.20.1

Question 10.7

Two loud speakers face each other and are separated by a distance of 1.6 m apart.A stationary sound wave can be set up in the region between the two speakers when they

emit sound waves of frequency 400 Hz. The velocity of sound in air is 320 m s 1.(a) Calculate the distance between two adjacent nodes.(b) Sketch a graph to illustrate the stationary wave formed between the two speakers.

(c) What are the changes to the stationary wave formed if the frequency of the soundwave is increased to 800 Hz?

Solution 10.7

(a) V = f320 = (400)

= 0.80 m

1Distance betwen two adjacent nodes =

21

= (0.8) m2

= 0.4 m

(b) At the speaker, antinode must be formed.

(c) When f = 800 Hz

320 = (800) = 0.40 m

Distance between two adjacent nodes is reduced to 0.20 m and a total of seven loopscan be formed.

N N N

0.40.80 m

1.6 m

0.2 0.2

ch10 mechanical waves

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