Suggested Answers of Exercise Wave Phenomena

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```					                           Suggested Answers of Exercise 15 (Acoustics)
----------------------------------------------
1. (a) Power emitted as sound = 50 x 0.05
= 2.5 W                          (d) (i) Interchange the wire
Hence sound intensity at 3 m                        connections to one of the loudspeakers.
(ii) The positions of maximum and
minimum intensity are interchanged.
P        2.5                                     ----------------------------------------------
                    0.0442 W m -2
2r 2
2 (3) 2                                  4. (a) The formation of heaps of powder is
due to the formation of longitudinal
stationary waves set up along the tube. The
(b) Intensity level
middle of each heap is a node position
0.0442
 10 log         106 dB                           where the air does vibrate whereas the air on
10 -2                                    both sides vibrates, causing the heap to
(c) The minimum intensity that the                      remain at the node position.
sound can be barely heard is 10-12 W m-2.                       (b) Longitudinal stationary waves are
2.5                                           formed in the tube only when the length of
 10 12
2r 2                                          the glass tube reaches some particular
values.
2.5                                        (c) v = f = 2 f L where L is the
r                  6.31 105 m
2  10  12
distance between successive heaps
(d) There is attenuation of sound. A                              = 2 x 500 x 0.33 = 330 m s-1
more realistic distance is a few kilometers.                    (d) Their velocities and wavelengths
----------------------------------------------             are different.
2. When they are sounded together, the                     ----------------------------------------------
intensity received is doubled, i.e. I’ = 2I.               5. (a)
Required intensity level

2I                      I
 10 log       10 log 2  10 log     3.01  70  73 dB
Io                     Io                          256 Hz
----------------------------------------------
3. (a) This is to ensure that the phase
difference between the two loudspeakers
stays constant with time. The frequencies of                                                     distance
two separate signal generators are not
exactly the same.                                              (b)
(b) (i) Sound of maximum intensity is                          v  vr 
heard. The intensity decreases along AB.                           vv 
fr  fs         
      s 
(ii) Maximum and minimum
intensity of sound is heard alternately.                               330         
270  256  330  v       v s  17.1 m s -1

Distance between successive maxima                                        s   
D       vD 340  20                                                                 
                           3.33 m                      (c) f r  256         330
a         fa   680  3                                                330  (17.1)   243 Hz

                
(c) There is a minimum intensity of                   ------------------------------------------------
sound at Z because a wave from L1 arrives                  6. (a) Microwaves
there exactly out of phase with a wave from
f 2u   1.03  103     2u
L2. Destructive interference occurs. When                                           
L2 is disconnected, no such cancellation                       (b) f   c    1  1010
3  108
occurs, so only one set of wave arrives, so                          u  15.4 m s -1
the intensity increases locally. At other
points where there was originally a
maximum intensity signal, there will be a
reduction in intensity when L2 is
disconnected. Overall energy is conserved.
= 10 log I / Io = 60 – 10 log 4 = 54 dB
(c) Beat frequency f = (2u/c) f                  ----------------------------------------------------
% error of beat frequency                    3. 50 = 10 log I / Io where
= % error of u                                    I = P / 4π(2)2
The lower limit of the car speed                  Now P  2P, 2 m  6 m, so
= 15.4 x 0.97 = 14.9 m s-1                        I  2P / 4π (6)2 = (2/9) I
Local speed limit = 50 x 1000/3600                So required intensity level
= 13.9 m s-1                                      = 10 log (2/9) I / Io
So the car speed still exceeds the                = 10 log 2/9 + 50 = 43.5 dB
speed limit.                                           ----------------------------------------------------
----------------------------------------------         4. Stationary waves are formed because
7. (a) The reflected sound can be regarded             the sound intensity varies. The formation of
as emitted from the image of the bat which             stationary wave is due to superposition of
is moving towards the wall at the same                 two waves, of same frequency, (one incident
speed 10.0 m s-1.                                      and one reflected) traveling in opposite
directions. So statement 1 is incorrect.
 v  vr             340  0                      Positions of soft sound corresponds to
fr  fs 
vv         60000
                  
      s            340  10                nodes.
 6.18 kHz
Wavelength = v/f = 340 / 1000 = 0.34 m
So distance between successive nodes is
(b) The bat itself is a moving receiver.               0.34 / 2 = 0.17 m. So statement 2 is
incorrect.
 v  vr             340  (10) 
fr  fs 
vv         60000
                                      Increasing frequency will lower the
      s            340  10                wavelength, so positions of soft sound
 63.6 kHz                                             (nodes) become closer.
----------------------------------------------------   ----------------------------------------------------
Multiple Choice                                        5. Statement 2 is correct because
ultrasonic has high frequency, so the
1-5       CBBBA                                        wavelength is short, as a result, diffraction
6-10 B D D C C                                         effect is less significant. (Longer
11-13 B C cancelled or A                               wavelength has more significant
14-15 B C                                              diffraction).
----------------------------------------------------   ----------------------------------------------------
1.                                                     6. In a stationary longitudinal wave,
 v  vr                                      displacement antinode = pressure node
vv 
fr  fs                                              ----------------------------------------------------
      s                                      7. 60 = 10 log (I60 / Io) => 106 = I60 / Io
336  0                                              40 = 10 log (I40 / Io) => 104 = I40 / Io
300              fs                                       => I60 / I40 = 106 / 104 = 100
336  v s                                        ----------------------------------------------------
336  0                                     8.
150                   fs
336  (v s )

On solving, we have vs = 112 m/s
----------------------------------------------------
2. Let I be the intensity due to one
loudspeaker. Since intensity is proportional
to (amplitude)2, so the resulting intensity at
a point of constructive interference will be
4I (not 2I).
4I
60  10 log       =10 log 4 + 10 log I / Io                Suppose observer O stands at O.
Io                                      Without loss of generality, suppose the boy
Required intensity level                               is moving in anticlockwise direction and is
at A at t = 0. Then at A, to observer O, he is
neither approaching nor receding, so the               15. R, S and T are positions of nodes.
frequency remains unchanged at fo.                           = 3 cm = 0.03 m
At B, the boy is moving away from O, and                    f = v /  = 3 x 108 / 0.03 = 1 x 1010 Hz
the drop in frequency must be the largest. At          ----------------------------------------------------
C, he is again neither approaching nor
receding, so the frequency is unchanged. At
D, the boy is approaching towards O, and
the rise in frequency would be the largest.
The time for travelling from B to D is
obviously longer that that from D to B, so
----------------------------------------------------
9. Wavelength is unchanged for stationary
source.  = v/f = 300 / 600 = 0.5 m
----------------------------------------------------
10. For a closed tube, fo = v / 4l, f1 = 3fo.
----------------------------------------------------
11. Δf / f = 2u / c
2uf 2u
f       
c      
2  100  103 / 3600
f                        1900Hz
0.03
----------------------------------------------------
12.
 v  vr 
vv 
fr  fs         
      s 

 300  0 
462  500  300  ( v ) 
            s 

vs = 24.7 m/s
----------------------------------------------------
13. The first resonance position corresponds
to the lowest frequency with simplest mode
of vibration.
 = 4 l => fo = v /  = v / 4l
For a closed tube, only odd harmonics are
present.
f1 = v /  = v / (4l / 3) = 3v / 4l = 3 fo
So if the length is unchanged, the next
higher frequency corresponds to 3fo.
Now if l is increased to 3l (by lowering the
water level), then f will be kept unchanged.
So the required length should be
5 cm x 3 = 15 cm
----------------------------------------------------
14. 70 = 10 log I / Io
Required intensity level
= 10 log 2I / Io
= 10 log 2 + 10 log I / Io
= 3.01 + 70 = 73.01 dB
----------------------------------------------------

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