physics waves sums

7/28/2019 physics waves sums

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PRACTICE PROBLEMS

BY NALLAMOTHU Page 33

Solution : Let frequency of the tuning fork = ft = 256 Hz.Fb = | ft - fpipe | =

10

23

when the tunning fork is loaded, frequency is decreased. As beat

frequency is decreases, ft > fp

ft = fpipe + 2.3 putting ft = 256 Hz we obtain fpipe = 253.7 Hz

Let at the temperature rise , the frequency of the pipe increases from253.7 Hz to 256 Hz

576

17.253

256

7.253

3.21

7.253

256

576

C2.5)576(2537

23 0

Problem 3 : A source emitting sound of frequency 180 Hz is placed infront of a wallat a distance of 2m from it. A detector is also placed in front of the wallat the same distance from it. Find the minimum distance between thesource and the detector for which the detector detects a maximum ofsound. Speed of sound air = 360 m/s.

Solution : The situation is shown in Figure. Suppose the detector is placed at adistance of x meter from the source. The direct wave received from the

source travels a distance of x-meter.

The wave reaching the detector

after reflection from the wall has

traveled a distance of2 [(2)2+x2/4]1/2 meter. The path

difference between the two waves

is

S

X

D

= {2[(2)2+x2/4]1/2-x} meter

Constructive interference will take place when = , 2 , Theminimum distance x for a maximum corresponds to


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PRACTICE PROBLEMS


= (1)

The wavelength is = m2s180

s/m360

f

v1

Thus by (1), 2 [(2)2+x2/4]1/2-x = 2

or (4+x2/4)1/2 = 1+x/2

or 4+ x4

x1

4

x 22 or 3 = x

Thus the detector be placed at a distance of 3 m from the source. Note

that there is no abrupt phase change as we are considering the

pressure wave.

Problem 4 : A wave travels out in all directions from a point source. Justify theexpression y=(a0/r)sin k (r-vt), at a distance r from the source. Find thespeed, periodicity and intensity of the wave.

Solution : If P be the power of the source then

Intensity I = 2r4

P

or I 2r

1But I a2

So a r

1or a =

r

a0 Here a0 is constant.

The equation in standard form is y = a sin k (r-vt)

Therefore above equation is written as y =ra0 sin k (r-vt)

Now comparing this equation with y = a sin (kr-t)

we have = kv, or n =2

kvand k =

2


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PRACTICE PROBLEMS


or =k

2; speed c = n =

2

kvx

k

2

c = v and T =kv

2

n

1

Thus intensity is given by

I =2

1 a22c I =2

1 2

20

r

ak2v2.v

I =2

1

2

3220

r

vka.

Problem 5 : A wave is propagating on a long stretched string along its length takenas positive x-axis the wave equation is given by

2x

T

t

0eyy

where y0 = 2mm, T = 1.0 sec and = 6 cm find.(a) The velocity of the wave(b) Find the function f(t) giving the displacement of the particle at x = 0.(c) Find the function g(x) giving the shape of string at t = 0(d) Plot the shape g(x) of the string at t = 0.(e) Plot the shape of the string at t = 6 sec.

Solution : The wave equation may be written as

Y = Y0

2

2 T/xt

T1

e

(a) Comparing it with the general equation

y = f

v

xt , we get, v =

0.1

cm6

T

= 6 cm/sec

(b) Putting x = 0 in the given equation


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PRACTICE PROBLEMS


f(t) = y02)T/t(e

(c) Putting t = 0 in the given equation

g(x) = y02)/x(e

x = 0

(d)

x = 0 x = 6

(e)

Problem 6 : A string of length L and Mass M hangs freely from a fixed point.Calculate(a) The velocity of the transverse wave along the string at any

position.(b) Time taken by a transverse pulse to traverse the string.

Solution : (a) Mass per unit length of the string=

L

M (i)

Let there be a point on the string at a distance x from the free end.

Tension at the point

T = (mass of the x part of the spring)

T =L

Mx.g

Hence, velocity of transverse wave along the string,


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PRACTICE PROBLEMS


v =m

T v =

L/M

L/)Mgx(

v = gx (ii)

(b) From equation (ii) we have

v = gx

dt

dx= gx or dt =

gx

dx

Required time t = L

0 gx

dx=

L

0

2/1 dxxg

1

t = 2

g

L.

Problem 7 : The vibrations of a string fixed at both ends are described by theequation y = (5.00 mm) sin [(1.57 cm-1)x] sin [(314 s-1)t](a) What is the maximum displacement of a particle at x = 5.66 cm?(b) What are the wavelengths and the wave speeds of the two

transverse waves that combine to give above vibration?(c) What is the velocity of the particle at x = 5.66 cm at time t = 2 s.(d) If the length of a string is 10.0 cm, locate the nodes and

antinodes. How many loops are formed in the vibration?

Solution : (a) The amplitude of vibration of a particle at position x isA = 5.00 mm sin [(1.57 cm-1)x] for x = 5.66 cm

A = (5.00 mm) sin [2x5.66]

= (5.00 mm) sin (2.5 +3

)

= (5.00 mm) cos3

= 2.50 mm.


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PRACTICE PROBLEMS


(b) From the given equation, the wave number R = 1.57 cm-1 and angularfrequency = 314 s-1, the wavelength is

=k2 =

57.114.32

= 4.00 cm and the frequency is

n =

2=

14.32

314

= 50 s

1

Now the wave speed is

v = n

v = (50 s-1) (4.00 cm)

v = 2.00 m/s

(c) The velocity of the particle at position x at the t is given byv =

dt

dy

= [(5.00 mm) sin (1.57 cm-1) x] 314 s-1 cos (314 s-1) t

= (157 cm/s) sin (1.57 cm-1 ) x cos(314 s-1) t

Putting x = 5.66 cm and t = 2.05, the velocity of this particle at the

given time is

= (157 cm/s) sin

32

5cos (200)

= (157 cm/s) cos3

x 1

= 78.5 cm/s

(d) The nodes occur where the amplitude is zero. i.e.sin (1.57 cm-1) x = 0


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PRACTICE PROBLEMS


1cm2

x = n, where n is integer

Thus x = 2 n cm

The nodes, occur at x = 0 cm 2 cm, 4 cm, 8 cm, and 10 cm, antinodesoccur in between them i.e. at x = 1 cm, 3 cm, 5 cm, 7 cm and 9 cm.

Hence it is clear that string vibrates in five loops.

Problem 8 : The loudest painless sound produces a pressure amplitude of 28 Nm-2.Calculate the intensity of this sound wave at S.T.P. Density of air at S.T.P.= 1.3 kg m-3, speed of sound at S.T.P. = 332 ms-1.

Solution : y = a sin (t-kx) is the equation of progressive wave and 1/2 a22v is itsintensity.

P =dx

Edy, v =

E

= v2(-ak) (cos t-kx)

= v2 ak sin (t-kx /2)

Pmax = v2ak

I =

v

kvkv

kv

P

2

1 22224

max2

=v

P

2

1 max2

=

3323.1

28

2

1 2

= 0.91 W-m-2.

Problem 9 : Taking the composition of air to be 75% of nitrogen and 25% of oxygenby weight, calculate the velocity of sound through air at STP.

Solution : The molecular weight of a mixture is given by

M

....mmm 321 = ......M

m

M

m

2

2

1

1


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PRACTICE PROBLEMS


M

2575 =

32

25

28

75

or, M = 28.9

of a mixture is given by

= ....1

n

1

n

1

nnn

2

2

1

1321

1

32

25

28

75

=

14.1

32

25

14.1

28

75

1

46.3

= 6.70 + 1.95

= 1.4

v =

P

By the perfect gas equation

P =M

RT

M

RTP

v =M

RT=

9.28

27310003.84.1

v = 331.3 ms-1

Problem 10 : The velocity of sound in hydrogen at 0C is 1200 m/s. When someamount of oxygen is mixed with hydrogen, the velocity decreases to500 m/s. Determine the ratio of H2 and O2 by volume in the mixture,given that the density of the oxygen is 16 times that of hydrogen.

Solution : Since we know thatV =

H

P

; 1200 =

H

P

(1)


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PRACTICE PROBLEMS


Let there be x volume of H2 and y volume of O2

Then (x+y) mix = x H + y0 = x H + 16 yH

mix = Hyx

)y16x(

; 500 = H)y16x(

)yx(P

(2)

Dividing (1) by (2),

yx

y16x

5

12

or

1

2.2

y

x .

Problem 11 : A steel wire of length 1 m, mass 0.1 kg and uniform cross sectional area10-6 m2 is rigidly fixed at both ends. The temperature of the wire islowered by 20C. If transverse waves are set up by plucking the string inthe middle, calculate the frequency of fundamental mode of vibration.Youngs modulus of steel = 2 x 1011 N/m2 and coefficient of linearexpansion of steel = 1.21 x 10 -5 C-1

Solution : Coefficient of linear expansion is given by

=

where = change in temperature

Hence internal strain = (.) stress = Y strain = Y

tension T = r2 ; stress = r2 Y

v =m

T

n = mT21 = mYr212

n =1.0

201021.110210

1x2

1 5116

n = 11 Hz.


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PRACTICE PROBLEMS


Problem 12 : At two points of position vector 21 randr , lying on a straight line througha point source of sound and on either side of the source, the amplitudesof waves are a1 and a2 respectively. Find the position vector of thesource in terms of 21 r,r , a1 and a2. Neglect absorption of waves by themedium.

Solution: Let S be the source whose position vector relative to 0 is .r I (intensitya2 )and also I 1/d2 where a = amplitude and d = distance.

a 1/d

a1d1 = a2d2 = k (a constant)

Let i be the unit vector along PQ Then r

= idr 11

id

r

d

r

1

1

1

and )i(drr 22

id

r

d

r

2

2

2

Adding2

2

1

1

21 d

r

d

r

d

1

d

1r

Substituting d1 = 21

2 da

a

21

2211

aa

rarar

.

Problem 13: A stationary observer received sound waves from two tuning forks, oneof which approaches and the other recedes with the same velocity. Asthis takes place, the observer hears beats of frequency = 2 HZ. Find thevelocity of each tuning fork if their oscillation frequency n = 680 and thevelocity of sound in air is vs = 340 ms.

Solution: n = nvc

vc

s

0

Here n = 680v340

340680

v340

0340


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PRACTICE PROBLEMS


n= 680v340

340680

v340

0340

It is given that n - n = 2

2v340 680340v340 680340

or, 340 x 680 2v340

v222

or 340 x 680 x v = 3402-v2=3402 ( v


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PRACTICE PROBLEMS


Here vw= velocity of water stream, vb = velocity of boat.

(b) Velocity of sound in still air (vsa) = (RTM)1/2 = 344 m/s.

f = f (vsa va) (vsa- va- vb) = 100698 Hz. Here va = velocity of air.

Objective

Problem 1: When the load on a wire is increasing slowly from 2 kg to 4 kg, theelongation increases from 0.6 mm to 1 mm. The work done during thisextension of the wire is (g = 10 m/s2)(A) 14 10-3 J (B) 0.4 10-3 J(C) 8 10-2 J (D) 10-3 J

Solution : W = )F2

1( 22 )F

2

1( 11 = 20 10

-3 - 6 10-3 = 14 10-3 J

Problem 2: If is the density of the material of a wire and is the breaking stress,the greatest length of the wire that can hang freely without breaking is(A) /g (B) /2g(C) 2/g (D) /2g

Solution : Stress = weight / area= Ag/A = g = g

= /g

Problem 3: The speed of sound through a gaseous medium bears a constant ratiowith the rms speed of its molecules. The constant ratio is


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PRACTICE PROBLEMS


(A)3

(B) - 1

(C)3

2 (D)

v

p

c

c

Solution : Vsound =M

RT

vrms =M

RT3

3v

V

rms

sound .

Problem 4: In the interference of waves from two sources of intensities Io and 4I0,the intensity at a point where the phase difference is is(A) I0 (B) 2I0(C) 3I0 (D) 4I0

Solution: I = I1 + I2 + 2 cosII 21

= I0 + 4I0 + 2 cos)I4I( 00 = I0.

Problem 5: An air column in a pipe, which is closed at one end, will be in resonancewith vibrating tuning fork of frequency 264 Hz, if the length of thecolumn is(A) 31.25 (B) 62.50(C) 93.75 (D) 125

Solution : f=L4

v

264 = L4

330

L =2644

330

= 31.25 cm

Problem 6: A policeman on duty detects a drop of 10% in the pitch of the horn of amoving car as it crosses him. If the velocity of sound is 330 m/s, thespeed of the car will be


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PRACTICE PROBLEMS


(A) 20 m/sec (B) 17.3 m/sec(C) 25 m/s (D) 27 m/sec

Solution : f1 / f2 =

vc

c

vc

c

vc

vc

f

f

2

1

vc

vc

90

100

100 c 100 v = 90 C + 90 v 10 c = 190 v 1/19 330 = 17.3 m/s.

Problem 7 : An accurate and reliable audio oscillator is used to standardize a tuningfork marked as 512 Hz. When the oscillaror reading is 514, two beatsare heard per second. When the oscillator reading is 510, the beatfrequency is 6 Hz. The frequency of the tuning fork is(A) 506 (B)510(C) 516 (D) 158

Solution : It should be remembered that the oscillaror reading is correct and thetuning fork frequency marked is wrong. When the oscillator reading is

514, two beats are heard. Hence the frequency f the tuning fork is 514 2 = 516 or 512. When the oscillator reading is 510, the frequency ofthe tuning fork is 510 6 = 516 or 504. The common value is 516.Hence the frequency is 516 Hz.

(C)

Problem 8: A sound wave of wavelength travels towards the right horizontallywith a velocity V. It strikes and reflects from a vertical plane surface,travelling at a speed v towards the left. The number of positive crestsstriking during a time interval of three seconds on the wall is(A) 3 (V + v) / (B)3(Vv)/(C) (V + v)/ 3 (D) (V v) /3


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PRACTICE PROBLEMS


Solution : The relative velocity of sound waves with respect to the walls is V + v.Hence the apparent frequency of the waves striking the surface of the

wall is (V + v)/ . The number of positive crests striking per second is thesame as frequency. In three seconds the number is [3 (V + v)]/ .

(A)

Problem 9 : Two waves represented by y1 = 10 sin (2000 t) andy2 = 10 sin (2000 t + /2) are superposed at any point at aparticular instant. The resultant amplitude is

(A) 10 units (B) 20 units(C) 14.1 units (D) zero

Solution : The resultant amplitude A of two waves of amplitudes a1 and a2 at aphase difference of is ( 2/121

22

21 )cosaa2aa( . Substituting

a1 = 10, a2 = 10 and = 900, we get A = 14.1.

(C)

Problem 10 : When pressure increases by 1 atmosphere and temperature increasesby 10C, the velocity of sound(A) increases by 0.61 ms1 (B) decreases by 0.61 ms1(C) increases by 61 ms1 (B) decreases by 61 ms1

Solution : The increase of pressure does not change the velocity of sound. Whenthe temperature increases by 10C, the velocity of sound increases roguly

by 0.6 ms1. (Vt = V0 + 0.6 t).

(A)

Problem 11 : When two simple harmonic motions of same periods, same amplitude,having phase difference of 3/2, and at right angles to each other aresuper imposed, the resultant wave form is a(A) circle (B) parabola(C) ellipse (D) figure of eight


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PRACTICE PROBLEMS


Solution: The SHM's are x = a sin t and y = a sin (t+ 3/2) = -a cos t x2 + y2 = a2.

(A)

Problem 12: A line source emits a cylindrical wave. If the medium absorbs noenergy, the amplitude will vary with distance r from the source asproportional to(A) r1 (B) r2(C) r1/2 (D) r1/2

Solution : The energy is inversely proportional to the square of distance. Hencethe amplitude is inversely proportional to the distance.

(A)

Problem 13 : A transverse wave is described by the equation y = y0 sin 2 (ft x/a).The maximum particle velocity is equal to four times the wave velocity ifa is equal to(A) y0 / 4 (B) y0/2(C) y0 (D) 2y0

Solution : The maximum particle velocity of a SHM of amplitude y0 and frequency fis 2fy0 . The wave velocity is f. For 2 fy0 to be equal to 4f, has tobe y0 / 2

(Here=a). (B)

Problem 14: Which of the following does not represent a travelling wave ?(A) y = f (x vt) (B) y = ym sin k (x + vt)(C) y = ym log (x vt) (D) y = f (x2 v2t2)


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PRACTICE PROBLEMS


Solution : (a) and (b) are travelling waves, while (d) is the superposition of twotravelling waves. However (c) is a logarithmic function which cannot

represent a wave motion.

(C)

Problem 15: The amplitude of a wave is given by A = c / (a + b - c). Resonance willoccurwhen(A) b = -c / 2 (B)b=-a/2(C) b = 0, a = c (D) none of these

Solution : Resonance occurs when the amplitude is maximum, i.e., when thedenominator of this equation is minimum.

(C)

Problem 16 : Inside a gas, sound transmission is possible for(A)longitudinalwaves only(B transverse waves only(C) neither longitudinal waves nor transverse waves(D) both longitudinal and transverse waves

Solution : Inside a gas, only the longitudinal mode of transmission is possible forsound waves.

(A)

Problem 17 : If spherical wavefronts are incident on a plane wall, then reflectedwavefront will be(A) spherical (B)plane(C) elliptical (D) none of the above

Solution : (A)


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PRACTICE PROBLEMS


Problem 18 : A flat horizontal platform moves up and down in S.H.M. with anamplitude of 1 cm. A small object is placed on the platform. What isthe maximum frequency the platform can have, if the object is not toseparate from it during any part of the motion ?(A) /2980 per second (B) 2/980 per second(C) 980 / 2 per second (D) 2 980 per second

Solution : The maximum restoring force of S.H.M. is the weight of the object in theplatform. If A is the amplitude, we have m2A = mg, where = 2 f.This solves to

f = g /2 = 980/2. (g in CGS system is 980 cm/s2).

(A)

Problem 19 : The amplitude of a wave disturbance propagating in the positive x-direction is given by y =1/ (1 + x2) at time t = 0 and by y = 1/[1 + (x 1)2at t = 2 seconds, where x and y are in metres. The shape of the wavedisturbance does not change during the propagation. The velocity ofthe wave is(A) 1 ms1 (B)0.5ms1(C) 1.5 ms1 (D) 2 ms1

Solution : Writing the general expression for y in terms of x asy =

2)vtx(1

1

. At t = 0, y = 1/ (1 + x2). At t = 2 s, y =

2)]2(vx[1

1

Comparing with the given equation we get 2v = 1 and v = 0.5 m/s.

(B)

Problem 20: The displacement y of a particle executing periodic motion is given by y= 4 cos2(1/2 t)sin (1000 t). This expression may be considered as a resultof the superposition of(A) two waves (B)threewaves(C) four waves (D) five waves


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PRACTICE PROBLEMS


Solution : If we have a term cos kx sin t, it is the superposition of two wavemotions. If we have an equation of the form cos2 at sin bt it can be

shown to be superposition of three sine waves.

Here, y = 4 cos2 (2

1t sin (1000 t)

= 2 [1 + cos t] sin 1000 t

= 2 [sin 1000 t +2

1(sin 100 t + sin 999 t)]

= this comprises three waves.

(B)

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