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TOPICS
● BEATS
● STATIONARY
WAVES
Copyright © Mahnoor
PHYSICS PRESENTATION:
TOPIC:Beats And Stationary Waves
WHAT IS BEATS?
O
R
BEATS:
 Definition : The periodic alternation of sound between constructive and
destructive interference of sound are called beats.
 Example: Tuning fork.
 A= 32HZ ; B= 32HZ
 A= 32HZ ; B= 30HZ
 This fading in and out will occur repetitively.
 In the example above, we can see that the period for the
wave to go from loud to soft and back to loud is 1waves.
 This is true in general: two waves with frequencies f1 and f2,
when added together, will pulse or beat with a frequency
equal to the difference between the two original frequencies,
f1 and f2. Since this frequency should always be a positive
number.
 Beat caused due to superposition of two waves.
 Two waves of slightly different frequencies.
 One rise and one fall make one beat.
 At the start the two waves are in phase, At this point they add together
constructively.
 if we look at the figure above, this is not so for waves of different
frequencies: after about five oscillations, we see that the waves are definitely
out-of-phase: when one is a maximum the other is a minimum. So, at this
point, there will be destructive interference.
 If we listened to this wave, we would first hear the sound normally as it
started with constructive interference.
 The sound would then fade away as it went through destructive
interference, and then come back again.
 f beat = |f1 – f2|.
 For example, if we play two notes, one at 500 Hz and one at
502 Hz, they will beat together with a frequency of 2 Hz.
 1 beat = ¼ seconds.
 4 beats = 1 seconds.
 When the difference between the frequencies of two sounds is
more than 10 HZ, then it becomes difficult to recognize the
beats.
 No of beats = difference between frequencies.
 FORMULA: No. of beats = n= fA – fB
where fA and fB of tuning forks A and B and n is the number of
beats per second.
 RESULT: The number of beats per second is equal to
difference between the frequencies of the tuning forks.
2nd Topic: stationary waves
DEFINITION: The stationary waves are produced by the
superposition of two waves having same frequency and
travelling in the opposite direction.
Examples are:
Waves produce in Organ pipes
Waves produce in a string
Characteristics of Stationary Waves
The points of zero displacement in the stationary waves are
called nodes.
The points of maximum displacement in the stationary waves
are called antinodes.
No energy is transferred from particle to particle in stationary
waves.
Particles, except nodes, perform SHM with the same period as
the component waves.
Distance between the two consecutive nodes or anti-nodes is
equal to λ/2.
Distance between node and its neighboring anti-nodes is equal
to λ/4.
Types of Organ Pipe
There are two types of organ pipes:
(i) Closed Pipe (ii) Open Pipe
Closed Pipe
If one end of the organ pipe is closed, it is called closed
pipe.
Open Pipe
If both ends of the organ pipe are open, it is called open
pipe.
The phenomenon of stationary waves in air column.
Stationary waves can be set in air column, such as in
case of organ pipe. The relationship between
the incident wave and the reflected wave depends on
whether the reflecting end is open or close.
If the reflecting end is open, as in case of open
organ pipe, the air molecule has complete freedom of
motion and this behaves as an anti-node.
If the reflecting end is closed, as in case of close
organ pipe, the motion of the air molecules is
restricted and it behaves as a node.
Modes of vibrations in an Open Air Column
Let a vibrating tuning fork be held at
the mouth of an open pipe of length L.
If the pipe is open at both ends, then
its ends behaves as anti-nodes.
First Harmonic
If f1 and λ1 be the frequency and the
wavelength of the stationary wave for
the case of first harmonic, then from
figure:
L= λ1/2 λ1=2L
If v is the speed of the wave, then
v=f1λ1=f1(2L)
f1=v/2L
Second Harmonic
If f2 and λ2 be the frequency and
the wavelength of the stationary
wave for the case of second
harmonic, then from figure:
L=2(λ2 /2) λ2 =L
If v is the speed of the wave, then
v=f2λ2=f2(L)
f2=v/L=2(v/2L)=2f1
Third Harmonic
If f3 and λ3 be the frequency and
the wavelength of the stationary
wave for the case of third
harmonic, then from figure:
L=3(λ3/2) λ3=2L/3
If v is the speed of the wave,
then
v=f3λ3=f3(2L/3)
f3=3(v/2L)=3f1
Generalization
Similarly for the nth harmonic,
fn=nf1
Where n=1,2,3,4,…..
Hence, it is proved that all harmonics
are present in an open organ pipe.
Modes of vibrations in a Close Air Column
Let a vibrating tuning fork be held at the
mouth of an open pipe of length L. If the pipe
is close at one end and open at the other, the
close end acts as node while the open end
behaves as anti-nodes.
First Harmonic
If f1 and λ1 be the frequency and the
wavelength of the stationary wave for the
case of first
harmonic, then from figure:
L=λ1/4 λ1=4L
If v is the speed of the wave, then
v=f1λ1=f1(4L)
f1=v/4L
Second Harmonic
If f and λ2 be the frequency and the
wavelength of the stationary wave for
the case of second harmonic, then from
figure:
L=3(λ2/4) λ2=4L/3
If v is the speed of the wave, then
v=f2λ2=f2(4L/3)
f2= 3(v/4L)=3f1
Third Harmonic
If f3 and λ3 be the frequency and the wavelength
of the stationary wave for the case of third
harmonic, then from figure:
L=5(λ3/2) λ3=2L/5
If v is the speed of the wave, then
v=f3λ3=f3(2L/5)
f3=5(v/2L)=5f1
Generalization
Similarly for the nth harmonic,
fn= nf1
where n= 1,2,3,…..
Hence, it is proved that only the odd harmonics
are present in a close organ pipe.
 A closed organ pipe (closed at one end) is excited so as to support the third
overtone. It is then found that there are in the pipe
1) Three nodes and three antinodes 2) Three nodes and four
antinodes
3). Four nodes and three antinodes 4). Four nodes and four
antinodes
 The number of beats produced per second is equal to
1) the sum of the frequencies of two tuning forks 2) the
difference of the frequencies of two tuning forks 3)the ratio of
the frequencies of two tuning forks 4) the frequency of either
of the two tuning forks.
 Beats are the results of
1) diffraction of sound waves 2)constructive and destructive
interference 4) polarization
3)destructive interference.
 In open organ pipe
1)only even harmonics are present 2) only odd harmonics
are present 3) selected harmonics are present 4)both even
and odd harmonics are
Beats and stationary Waves
Beats and stationary Waves

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Beats and stationary Waves

  • 4. BEATS:  Definition : The periodic alternation of sound between constructive and destructive interference of sound are called beats.  Example: Tuning fork.  A= 32HZ ; B= 32HZ  A= 32HZ ; B= 30HZ
  • 5.  This fading in and out will occur repetitively.  In the example above, we can see that the period for the wave to go from loud to soft and back to loud is 1waves.  This is true in general: two waves with frequencies f1 and f2, when added together, will pulse or beat with a frequency equal to the difference between the two original frequencies, f1 and f2. Since this frequency should always be a positive number.
  • 6.  Beat caused due to superposition of two waves.  Two waves of slightly different frequencies.  One rise and one fall make one beat.  At the start the two waves are in phase, At this point they add together constructively.  if we look at the figure above, this is not so for waves of different frequencies: after about five oscillations, we see that the waves are definitely out-of-phase: when one is a maximum the other is a minimum. So, at this point, there will be destructive interference.  If we listened to this wave, we would first hear the sound normally as it started with constructive interference.  The sound would then fade away as it went through destructive interference, and then come back again.
  • 7.  f beat = |f1 – f2|.  For example, if we play two notes, one at 500 Hz and one at 502 Hz, they will beat together with a frequency of 2 Hz.  1 beat = ¼ seconds.  4 beats = 1 seconds.  When the difference between the frequencies of two sounds is more than 10 HZ, then it becomes difficult to recognize the beats.  No of beats = difference between frequencies.  FORMULA: No. of beats = n= fA – fB where fA and fB of tuning forks A and B and n is the number of beats per second.  RESULT: The number of beats per second is equal to difference between the frequencies of the tuning forks.
  • 9. DEFINITION: The stationary waves are produced by the superposition of two waves having same frequency and travelling in the opposite direction. Examples are: Waves produce in Organ pipes Waves produce in a string
  • 10. Characteristics of Stationary Waves The points of zero displacement in the stationary waves are called nodes. The points of maximum displacement in the stationary waves are called antinodes. No energy is transferred from particle to particle in stationary waves. Particles, except nodes, perform SHM with the same period as the component waves. Distance between the two consecutive nodes or anti-nodes is equal to λ/2. Distance between node and its neighboring anti-nodes is equal to λ/4.
  • 11. Types of Organ Pipe There are two types of organ pipes: (i) Closed Pipe (ii) Open Pipe Closed Pipe If one end of the organ pipe is closed, it is called closed pipe. Open Pipe If both ends of the organ pipe are open, it is called open pipe.
  • 12. The phenomenon of stationary waves in air column. Stationary waves can be set in air column, such as in case of organ pipe. The relationship between the incident wave and the reflected wave depends on whether the reflecting end is open or close. If the reflecting end is open, as in case of open organ pipe, the air molecule has complete freedom of motion and this behaves as an anti-node. If the reflecting end is closed, as in case of close organ pipe, the motion of the air molecules is restricted and it behaves as a node.
  • 13. Modes of vibrations in an Open Air Column Let a vibrating tuning fork be held at the mouth of an open pipe of length L. If the pipe is open at both ends, then its ends behaves as anti-nodes. First Harmonic If f1 and λ1 be the frequency and the wavelength of the stationary wave for the case of first harmonic, then from figure: L= λ1/2 λ1=2L If v is the speed of the wave, then v=f1λ1=f1(2L) f1=v/2L Second Harmonic If f2 and λ2 be the frequency and the wavelength of the stationary wave for the case of second harmonic, then from figure: L=2(λ2 /2) λ2 =L If v is the speed of the wave, then v=f2λ2=f2(L) f2=v/L=2(v/2L)=2f1
  • 14. Third Harmonic If f3 and λ3 be the frequency and the wavelength of the stationary wave for the case of third harmonic, then from figure: L=3(λ3/2) λ3=2L/3 If v is the speed of the wave, then v=f3λ3=f3(2L/3) f3=3(v/2L)=3f1 Generalization Similarly for the nth harmonic, fn=nf1 Where n=1,2,3,4,….. Hence, it is proved that all harmonics are present in an open organ pipe.
  • 15. Modes of vibrations in a Close Air Column Let a vibrating tuning fork be held at the mouth of an open pipe of length L. If the pipe is close at one end and open at the other, the close end acts as node while the open end behaves as anti-nodes. First Harmonic If f1 and λ1 be the frequency and the wavelength of the stationary wave for the case of first harmonic, then from figure: L=λ1/4 λ1=4L If v is the speed of the wave, then v=f1λ1=f1(4L) f1=v/4L Second Harmonic If f and λ2 be the frequency and the wavelength of the stationary wave for the case of second harmonic, then from figure: L=3(λ2/4) λ2=4L/3 If v is the speed of the wave, then v=f2λ2=f2(4L/3) f2= 3(v/4L)=3f1
  • 16. Third Harmonic If f3 and λ3 be the frequency and the wavelength of the stationary wave for the case of third harmonic, then from figure: L=5(λ3/2) λ3=2L/5 If v is the speed of the wave, then v=f3λ3=f3(2L/5) f3=5(v/2L)=5f1 Generalization Similarly for the nth harmonic, fn= nf1 where n= 1,2,3,….. Hence, it is proved that only the odd harmonics are present in a close organ pipe.
  • 17.
  • 18.  A closed organ pipe (closed at one end) is excited so as to support the third overtone. It is then found that there are in the pipe 1) Three nodes and three antinodes 2) Three nodes and four antinodes 3). Four nodes and three antinodes 4). Four nodes and four antinodes  The number of beats produced per second is equal to 1) the sum of the frequencies of two tuning forks 2) the difference of the frequencies of two tuning forks 3)the ratio of the frequencies of two tuning forks 4) the frequency of either of the two tuning forks.  Beats are the results of 1) diffraction of sound waves 2)constructive and destructive interference 4) polarization 3)destructive interference.  In open organ pipe 1)only even harmonics are present 2) only odd harmonics are present 3) selected harmonics are present 4)both even and odd harmonics are