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1. Waves-1
TOPICS: Wave motion, characteristics,
equation of plane progressive wave with
numerical problems
PREPARED BY,
NIRUPAMA,
P.G.T. PHYSICS,
KENDRIYA VIDYALAYA,
NEW TEHRI TOWN,
DEHRADUN REGION

2. WAVE MOTION
Wave motion is a kind of disturbance which travels through a
medium due to repeated vibrations of the particles of the of the
medium about their mean positions, the disturbance being handed
over from one particle to neighbouring particles continuously.
Ball dropped in in water
Ripples

3. displacement
Displacement
Characteristics of wave motion
1. Particles of medium vibrate at their own places. Matter
(particles ) are not transported from one point to other.
2. Energy and momenta are transported from one place to
other.
3. The wave velocity is constant in a medium while particle
velocity changes periodically.
Mean

4. TYPES OF WAVES
MECHANICAL WAVES: The waves which require material medium
for propagation are called mechanical waves.
Ex. Ripples on the surface of water, sound waves etc.
ELECTROMAGNETIC WAVES: The waves which do not require
material medium for propagation are called electromagnetic waves.
Ex. visible light, infrared light, microwave, X rays etc.
Essential properties of media for propagation of mechanical waves
1. ELASTICITY: The medium must be
elastic so that the particles return to their
mean positions after being disturbed.
2. INERTIA: The medium must have
inertia (mass) so that its particles
can store energy and have momenta.
3. LOW FRICTION: The medium must have low friction so that
the particles continue to vibrate for longer time.

5. TYPES OF MECHANICAL WAVES
LONGITUDINAL WAVES: in these
waves, particles of medium
oscillate along the direction of
propagation of wave.
Ex. Sound wave. They can travel
in all the media i.e. solids,
liquids and gases.
TRANSVERSE WAVES: in these
waves, particles of medium
oscillate perpendicular to the
direction of propagation of wave.
They can travel in solids and on
the surface of liquids. Ex. Ripples
on the surface of water, wave in
Sitar string or membrane of a
drum
Mean
displacement
rarefaction
compression
rarefaction
propagation
of wave
propagation
of wave

6. DEFINITIONS FOR TRANSVERSE WAVE
Amplitude (a): It is the
maximum displacement of
the particles of medium
from their mean positions.
The point of max upward
displacement is called crest
and the point of max
downward displacement is
called trough.
Time Period (T): It is the
time taken by the particles
of medium to complete on
vibration.
Frequency (𝝂): It is the
number of vibrations per
second done by the particles
of medium.
Wave Length (𝝀): It is the distance
covered by the wave in one time
period. OR
Distance between successive crests
or troughs.
OR
Distance after which wave pattern is
repeated.
Mean
displacement
propagation of wave
𝝀
𝝀
𝝀
a
a
crest
trough

7. DEFINITIONS FOR LONGITUDINAL WAVE
Compression & Rarefaction:
The region of maximum
density is called compression
and the region of minimum
density is called rarefaction.
Time Period (T): It is the time
taken by the particles of
medium to complete on
vibration.
Frequency (𝝂): It is the
number of vibrations per
second done by the particles
of medium.
Wave Length (𝝀): It is the distance between successive compressions
or rarefactions. OR
Distance after which wave pattern is repeated.
rarefaction
compression
rarefaction
Propagation
of wave
𝝀
𝝀

8. RELATION BETWEEN WAVE VELOCITY,
FREQUENCY AND WAVE LENGTH
𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 =
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒄𝒐𝒗𝒆𝒓𝒆𝒅
𝒕𝒊𝒎𝒆 𝒕𝒂𝒌𝒆𝒏
In one time period T, distance covered= 𝝀
𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, 𝒗 =
𝝀
𝑻
𝒗 = 𝝀
1
𝑻
= 𝝀𝝂
𝒗 = 𝝂𝝀

9. EQUATION OF A PLANE PROGRESSIVE WAVE
In plane progressive wave, the particles of medium execute
simple harmonic motion at their own places.
X
Y
O
𝒙
(𝒙, 𝒚)
𝒚
For a wave starting at the
origin, the oscillations of
successive particles will lag in
phase. Equation of SHM of
the particle at the origin,
𝒚 = 𝒂 sin 𝝎𝒕……..(1)
Eqn of a particle at distance 𝒙,
𝒚 = 𝒂 sin(𝝎𝒕 − 𝝓)…..(2)
𝝀
The phase diff between points 𝝀
distance apart is 𝟐𝝅.
So the phase diff for distance 𝒙,
𝝓 =
𝟐𝝅
𝝀
𝒙, so from eqn 2,
𝒚 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 −
𝟐𝝅
𝝀
𝒙),
𝟐𝝅
𝝀
= 𝒌(𝒂𝒏𝒈𝒖𝒍𝒂𝒓 𝒘𝒂𝒗𝒆 𝒏𝒐. )
𝒚 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 − 𝒌𝒙)……(3)
𝒚 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 − 𝒌𝒙)….(4)
𝒚 = 𝒂 𝒔𝒊𝒏(
𝟐𝝅
𝑻
𝒕 −
𝟐𝝅
𝝀
𝒙)
𝒚 = 𝒂 𝒔𝒊𝒏 𝟐𝝅(
𝒕
𝑻
−
𝒙
𝝀
)…..(5)
𝒚 = 𝒂 𝒔𝒊𝒏
𝟐𝝅
𝝀
(
𝝀𝒕
𝑻
− 𝒙)
𝒚 = 𝒂 𝒔𝒊𝒏
𝟐𝝅
𝝀
(𝒗𝒕 − 𝒙)…….(6)

10. EQUATION OF A PLANE PROGRESSIVE WAVE
Plane progressive wave travelling along +𝑿,
𝒚 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 − 𝒌𝒙)….(1)
𝒚 = 𝒂 𝒔𝒊𝒏 𝟐𝝅(
𝒕
𝑻
−
𝒙
𝝀
)…..(2)
𝒚 = 𝒂 𝒔𝒊𝒏
𝟐𝝅
𝝀
(𝒗𝒕 − 𝒙)…….(3)
𝑿
𝒀
𝑶
Plane progressive wave travelling along −𝑿,
−𝑿
𝒀
𝑶
𝒚 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 + 𝒌𝒙)….(1)
𝒚 = 𝒂 𝒔𝒊𝒏 𝟐𝝅(
𝒕
𝑻
+
𝒙
𝝀
)…..(2)
𝒚 = 𝒂 𝒔𝒊𝒏
𝟐𝝅
𝝀
(𝒗𝒕 + 𝒙)…….(3)
If the initial phase is 𝝓𝟎, general
equation of progressive wave,
𝒚 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 ± 𝒌𝒙 ± 𝝓𝟎)

11. PHASE OF A HARMONIC WAVE
Phase describes the state of vibration (position, velocity,
direction etc) of a particle of medium in wave motion. It is given
by the argument of sine or cosine function of equation of wave.
For, 𝒚 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 ± 𝒌𝒙 ± 𝝓𝟎), Phase 𝝓 = 𝝎𝒕 ± 𝒌𝒙 ± 𝝓𝟎
VELOCITY AND ACCELERATION OF PARTICLE IN WAVE
Velocity and acceleration execute SHM with the frequency
of the particle 𝐯 = 𝝎 𝒂𝟐 − 𝒚𝟐, Acceleration 𝑨 = −𝝎𝟐
𝒚
VELOCITY OF WAVE IS CONSTANT.
PHASE CHANGE WITH POSITION:
Phase change with distance 𝝀
is 𝟐𝝅, so with distance 𝒙,
phase diff ∆𝝓 =
𝟐𝝅
𝝀
𝒙
𝒀
𝑶
𝑿
𝝀
𝟐𝝅 𝟒𝝅
𝑻 𝟐𝑻
PHASE CHANGE WITH TIME: Phase change in time 𝑻 is 𝟐𝝅,
so in time 𝒕, phase diff ∆𝝓 =
𝟐𝝅
𝑻
𝒕

12. Q. A Simple harmonic wave is described by
𝒚 = 𝟕𝑿𝟏𝟎−𝟔
𝒔𝒊𝒏(𝟖𝟎𝟎𝝅𝒕 −
𝝅
𝟒𝟐.𝟓
𝒙 −
𝝅
𝟒
) where x and y are in cm
and t in second. Calculate (a) amplitude (b) frequency, time
period (c ) initial phase (d) wavelength (e) phase diff between
particles separated by 17.0 cm (f) phase diff of a particle after T/5.
(g) direction of motion of wave (h) wave velocity.
Comparing it with standard equation,𝒚 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 ± 𝒌𝒙 ± 𝝓𝟎)
(a) Amplitude 𝒂 = 𝟕𝑿𝟏𝟎−𝟔
𝒄𝒎
(b) 𝝎 = 𝟖𝟎𝟎𝝅 = 𝟐𝝅𝝂 , 𝝂 = 𝟒𝟎𝟎 𝑯𝒛, 𝐓 =
𝟏
𝝂
=
𝟏
𝟒𝟎𝟎
𝒔
(c )Initial phase 𝝓𝟎 = −
𝝅
𝟒
(d) 𝒌 =
𝝅
𝟒𝟐.𝟓
=
𝟐𝝅
𝝀
, 𝝀 = 𝟐𝑿𝟒𝟐. 𝟓 = 𝟖𝟓 𝒄𝒎
(e)phase diff ∆𝝓 =
𝟐𝝅
𝝀
𝒙 =
𝟐𝝅
𝟖𝟓
𝑿𝟏𝟕 =
𝟐𝝅
𝟓
𝒓𝒂𝒅
(f) phase diff ∆𝝓 =
𝟐𝝅
𝑻
𝒕 =
𝟐𝝅
𝑻
𝑻
𝟓
=
𝟐𝝅
𝟓
𝒓𝒂𝒅
(g) The wave is propagating along +X axis.
(h) Wave velocity 𝐯 = 𝝂𝝀 = 𝟒𝟎𝟎𝑿𝟖𝟓
𝒄𝒎
𝒔
= 𝟑𝟒𝟎 𝒎 𝒔−𝟏

13. Waves-2
TOPICS: Reflection of waves, Stationary
waves and its characteristics, standing waves
in a string with numerical problems
PREPARED BY,
NIRUPAMA,
P.G.T. PHYSICS,
KENDRIYA VIDYALAYA,
NEW TEHRI TOWN,
DEHRADUN REGION

14. REFLECTION OF HARMONIC WAVES
FROM FIXED END: When a pulse
reaches a rigid end, it applies force on
it in upward direction. According to
the law of action and reaction, the
fixed end applies force in opposite
direction. So the pulse returns in
opposite phase.
Reflected pulse
incident pulse
Fixed end
+𝑿
−𝑿
Equation of incident wave,
𝒚 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 − 𝒌𝒙)
Equation of reflected wave,
𝒚 = −𝒂 𝒔𝒊𝒏(𝝎𝒕 + 𝒌𝒙)

15. REFLECTION OF HARMONIC WAVES
FROM FREE END: When a pulse
reaches a free end, it applies force on
it in upward direction so the free
loop is also pulled upwards. So the
pulse returns in same phase.
Free end
+𝑿
−𝑿
Equation of incident wave,
𝒚 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 − 𝒌𝒙)
Equation of reflected wave,
𝒚 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 + 𝒌𝒙)
incident pulse
Reflected pulse

16. SUPERPOSITION PRINCIPLE
When two or more waves travel simultaneously in a medium,
the displacement of a particle of the medium is given by the
vector sum of its displacements due to each of the waves.
𝒚 = 𝒚𝟏 + 𝒚𝟐 + 𝒚𝟑 … … …
STATIONARY OR STANDING WAVES
When two identical waves travel along the same path in opposite
directions, they superpose to form a wave which does not propagate in
any direction. Such a wave is called stationary or standing wave.
N
N
N
N
A
A A
The points which do not suffer
displacement are called Nodes(N)
and the points which suffer
maximum displacement are called
Antinodes(A).

17. CHARACTERISTICS OF STATIONARY WAVES
1. All the particles execute SHM at
their own places with the frequency
of the wave(except at nodes).
2. All the particle pass through their
mean positions simultaneously.
3. KE energy of particles at nodes
remains zero.
4. KE of particles at antinodes is
maximum when they pass through
mean positions.
5. Every particle has a fixed amplitude
which is different from that of its
neighbouring particles.
6. Energy is not transported.
7. Pressure at the nodes is maximum,
and at antinodes minimum.
8. Distance between successive
nodes/antinodes is 𝝀/𝟐 .
9. All the particles in the same
segment are in same phase of
oscillation and in opposite
phase with that in neighbouring
segment.
N
N
N
N
A
A A
𝝀

18. N
N
N
N
A
A A
STANDING WAVE IN A STRING (Transverse)
Let the incident wave is traveling along +X
axis and getting reflected at the fixed end.
Equation of incident wave along +X axis,
𝒚𝟏 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 − 𝒌𝒙)
Equation of reflected wave along –X axis,
𝒚𝟐 = −𝒂 𝒔𝒊𝒏(𝝎𝒕 + 𝒌𝒙)(from rigid end)
Applying superposition principle, the
resulting wave, 𝒚 = 𝒚𝟏 + 𝒚𝟐
𝒚 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 − 𝒌𝒙) − 𝒂 𝒔𝒊𝒏(𝝎𝒕 + 𝒌𝒙)
𝒚 = 𝒂[𝒔𝒊𝒏 𝝎𝒕 cos 𝒌𝒙 − cos 𝝎𝒕 sin 𝒌𝒙 − 𝒔𝒊𝒏 𝝎𝒕 cos 𝒌𝒙 − cos 𝝎𝒕 sin 𝒌𝒙]
𝒚 = − 𝟐𝒂 cos 𝝎𝒕 sin 𝒌𝒙
𝒚 = (− 𝟐𝒂 sin 𝒌𝒙)cos 𝝎𝒕
𝒚 = 𝑨 cos 𝝎𝒕
As this equation does not contain terms like (𝝎𝒕 ± 𝒌𝒙) or (𝒗𝒕 ± 𝒙), it is
not representing a travelling wave.
Amplitude of standing wave is 𝐀 = −𝟐𝒂 sin 𝒌𝒙

19. N N
A
STANDING WAVE IN A STRING (Transverse)
Eq of standing wave,
𝒚 = (−𝟐𝒂 sin 𝒌𝒙)cos 𝝎𝒕
Amplitude of wave is 𝐀 = −𝟐𝒂 sin 𝒌𝒙,
Max amplitude 𝑨𝒎 = 𝟐𝒂
𝟐𝒂
𝒙 = 𝟎 𝒙 = 𝑳
𝑳 = 𝝀𝟏/𝟐
At 𝒙 = 𝟎, 𝐀 = −𝟐𝒂 sin 𝒌𝒙 = 𝟎 (node)
At 𝒙 = 𝑳, 𝐀 = −𝟐𝒂 sin 𝒌𝒙 = 𝟎 (node)
sin 𝒌𝒙 = 𝟎, 𝒌𝒙 = 𝒏𝝅 (𝒏 = 𝟏, 𝟐, 𝟑, … )
𝟐𝝅
𝝀
𝑳 = 𝒏𝝅, 𝝀𝒏 =
𝟐𝑳
𝒏
……..(1)
For Fundamental note or first harmonic:
𝒏 = 𝟏, 𝝀𝟏 = 𝟐𝑳, freq 𝝂𝟏 =
𝒗
𝝀𝟏
=
𝒗
𝟐𝑳
For 2nd, 3rd , harmonics,
𝝂𝟐 =
𝒗
𝝀𝟐
=
𝟐𝒗
𝟐𝑳
, 𝝂𝟑 =
𝒗
𝝀𝟑
=
𝟑𝒗
𝟐𝑳
𝝂𝟏: 𝝂𝟐: 𝝂𝟑 = 𝟏: 𝟐: 𝟑
Velocity of transverse wave in a string,
𝒗 =
𝑻
𝝁
where T is tension and 𝝁 is mass
per unit length of string.
𝑳 = 𝝀𝟐
𝑳 = 𝟑𝝀𝟑/𝟐
Fundamental note/ first harmonic
𝝂 =
𝟏
𝟐𝑳
𝑻
𝝁

20. Q. A wire emits a fundamental note of 256 Hz. Keeping
the stretching force constant and reducing the length of
wire by 10 cm, the frequency becomes 320 Hz. Calculate
the original length of the wire.
Fundamental note/ first harmonic
𝝂 =
𝟏
𝟐𝑳
𝑻
𝝁
Where T is tension in the string and 𝝁 is mass per unit
length of the wire.
𝟐𝟓𝟔 =
𝟏
𝟐𝑳
𝑻
𝝁
……(1), 𝟑𝟐𝟎 =
𝟏
𝟐(𝑳−𝟏𝟎)
𝑻
𝝁
………(2)
Divide Eq. (1) by (2).
𝟐𝟓𝟔
𝟑𝟐𝟎
=
𝟏/𝟐𝑳
𝟏/𝟐(𝑳 − 𝟏𝟎)
=
(𝑳 − 𝟏𝟎)
𝑳
(𝑳−𝟏𝟎)
𝑳
=
𝟒
𝟓
, 𝑳 = 𝟓𝟎 𝒄𝒎

21. Waves-3
TOPICS: Stationary waves in organ pipes
with numerical problems
PREPARED BY,
NIRUPAMA,
P.G.T. PHYSICS,
KENDRIYA VIDYALAYA,
NEW TEHRI TOWN,
DEHRADUN REGION

22. STANDING WAVES IN ORGAN PIPES (Longitudinal)
Y
O X
IN OPEN ORGAN PIPE:
Let the incident wave is travelling along
+X axis,𝒚𝟏 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 − 𝒌𝒙)……(1)
It is reflected from open end and returns
in same phase,
𝒚𝟐 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 + 𝒌𝒙)….(2)
Applying superposition principle, the
resulting wave, 𝒚 = 𝒚𝟏 + 𝒚𝟐
𝒚 = 𝒂 𝒔𝒊𝒏 𝝎𝒕 − 𝒌𝒙 + 𝒂 𝒔𝒊𝒏(𝝎𝒕 + 𝒌𝒙)
𝒚 = 𝒂[𝒔𝒊𝒏 𝝎𝒕 cos 𝒌𝒙 − cos 𝝎𝒕 sin 𝒌𝒙 + 𝒔𝒊𝒏 𝝎𝒕 cos 𝒌𝒙 + cos 𝝎𝒕 sin 𝒌𝒙]
𝒚 = 𝟐𝒂 sin 𝝎𝒕 cos 𝒌𝒙
𝒚 = (𝟐𝒂 cos 𝒌𝒙)sin 𝝎𝒕
𝒚 = 𝑨 sin 𝝎𝒕
As this equation does not contain terms like (𝝎𝒕 ± 𝒌𝒙) or (𝒗𝒕 ± 𝒙), it is
not representing a travelling wave.
Amplitude of standing wave is 𝐀 = 𝟐𝒂 cos 𝒌𝒙
L
𝒙 = 𝟎 𝒙 = 𝑳
Incident
wave
Reflected
wave

23. STANDING WAVES IN ORGAN PIPES (Longitudinal)
Y
O X
Eq of standing wave,
𝒚 = (𝟐𝒂 cos 𝒌𝒙)sin 𝝎𝒕
Amplitude of wave is 𝐀 = 𝟐𝒂 cos 𝒌𝒙,
Max amplitude 𝑨𝒎 = 𝟐𝒂
At 𝒙 = 𝟎, 𝐀 = 𝟐𝒂 (antinode)
At 𝒙 = 𝑳, 𝐀 = 𝟐𝒂 (antinode)
cos 𝒌𝒙 = ±𝟏, 𝒌𝒙 = 𝒏𝝅
(𝒏 = 𝟏, 𝟐, 𝟑, … )
𝟐𝝅
𝝀
𝑳 = 𝒏𝝅, 𝝀𝒏 =
𝟐𝑳
𝒏
……..(1)
For Fundamental note or first
harmonic:
𝒏 = 𝟏, 𝝀𝟏 = 𝟐𝑳, freq 𝝂𝟏 =
𝒗
𝝀𝟏
=
𝒗
𝟐𝑳
For 2nd, 3rd , harmonics,
𝝂𝟐 =
𝒗
𝝀𝟐
=
𝟐𝒗
𝟐𝑳
, 𝝂𝟑 =
𝒗
𝝀𝟑
=
𝟑𝒗
𝟐𝑳
𝝂𝟏: 𝝂𝟐: 𝝂𝟑 = 𝟏: 𝟐: 𝟑
Velocity of sound wave in a gas,
𝒗 =
𝜸𝑷
𝝆
where 𝑷 is pressure,
𝜸 =
𝑪𝑷
𝑪𝑽
and 𝝆 is density of gas.
Y
O X
𝑳 = 𝝀𝟏/𝟐
𝒙 = 𝟎 𝒙 = 𝑳
𝑳 = 𝝀𝟐
N
N N
A
A
A A

24. STANDING WAVES IN ORGAN PIPES (Longitudinal)
IN ORGAN PIPE CLOSED AT ONE END:
Let the incident wave is travelling along -X
axis,𝒚𝟏 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 + 𝒌𝒙)……(1)
It is reflected from closed end and returns
in opposite phase,
𝒚𝟐 = −𝒂 𝒔𝒊𝒏(𝝎𝒕 − 𝒌𝒙)….(2)
Applying superposition principle, the
resulting wave, 𝒚 = 𝒚𝟏 + 𝒚𝟐
Y
O X
𝒙 = 𝟎 𝒙 = 𝑳
N
A
𝟐𝒂
𝑳 = 𝝀/𝟒
𝒚 = 𝒂 𝒔𝒊𝒏(𝝎𝒕 + 𝒌𝒙) − 𝒂 𝒔𝒊𝒏(𝝎𝒕 − 𝒌𝒙)
𝒚 = 𝒂[𝒔𝒊𝒏 𝝎𝒕 cos 𝒌𝒙 + cos 𝝎𝒕 sin 𝒌𝒙 − 𝒔𝒊𝒏 𝝎𝒕 cos 𝒌𝒙 + cos 𝝎𝒕 sin 𝒌𝒙]
𝒚 = 𝟐𝒂 cos 𝝎𝒕 sin 𝒌𝒙
𝒚 = (𝟐𝒂 sin 𝒌𝒙)cos 𝝎𝒕
𝒚 = 𝑨 cos 𝝎𝒕
As this equation does not contain terms like (𝝎𝒕 ± 𝒌𝒙) or (𝒗𝒕 ± 𝒙), it is
not representing a travelling wave.
Amplitude of standing wave is 𝐀 = 𝟐𝒂 sin 𝒌𝒙
Incident
wave
Reflected
wave

25. Eq of standing wave,
𝒚 = (𝟐𝒂 sin 𝒌𝒙)cos 𝝎𝒕
Amplitude of wave is 𝐀 = 𝟐𝒂 sin 𝒌𝒙,
Max amplitude 𝑨𝒎 = 𝟐𝒂
At 𝒙 = 𝟎, 𝐀 = 𝟐𝒂 sin 𝒌𝒙 = 𝟎 (node)
At 𝒙 = 𝑳,
𝐀 = 𝟐𝒂 sin 𝒌𝒙 = 𝟐𝒂 (antinode)
sin 𝒌𝒙 = ±𝟏, 𝒌𝒙 =
𝟐𝒏−𝟏 𝝅
𝟐
(𝒏 =
𝟏, 𝟐, 𝟑, … )
𝟐𝝅
𝝀
𝑳 =
𝟐𝒏−𝟏 𝝅
𝟐
, 𝝀𝒏 =
𝟒𝑳
𝟐𝒏−𝟏
……..(1)
For Fundamental note or first harmonic:
𝒏 = 𝟏, 𝝀𝟏 = 𝟒𝑳, freq 𝝂𝟏 =
𝒗
𝝀𝟏
=
𝒗
𝟒𝑳
For 2nd, 3rd , harmonics,
𝝂𝟐 =
𝒗
𝝀𝟐
=
𝟑𝒗
𝟒𝑳
, 𝝂𝟑 =
𝒗
𝝀𝟑
=
𝟓𝒗
𝟒𝑳
𝝂𝟏: 𝝂𝟐: 𝝂𝟑 = 𝟏: 𝟑: 𝟓
STANDING WAVES IN ORGAN PIPE CLOSED AT ONE END
Y
O X
𝒙 = 𝟎 𝒙 = 𝑳
N
A
𝟐𝒂
𝑳 = 𝝀/𝟒
Y
O X
𝒙 = 𝟎 𝒙 = 𝑳
N
A
𝟐𝒂
𝑳 = 𝟑𝝀/𝟒

26. Q. Find the fundamental frequency of a 50 cm long organ
pipe,
(a) open at both ends (b) closed at one end
(speed of sound wave is 340 m/s)
O X
N
A
𝟐𝒂
𝑳 = 𝝀/𝟒
𝑳 = 𝝀/𝟐
(a) 𝑳 = 𝟓𝟎 𝒄𝒎 , 𝒗 = 𝟑𝟒𝟎 𝒎 𝒔−𝟏
𝑳 = 𝝀/𝟐
𝝀 = 𝟏𝟎𝟎 𝒄𝒎 = 𝟏 𝒎
𝝂 =
𝒗
𝝀
= 𝟑𝟒𝟎 𝑯𝒛
(b) 𝑳 = 𝟓𝟎 𝒄𝒎 , 𝒗 = 𝟑𝟒𝟎 𝒎 𝒔−𝟏
𝑳 = 𝝀/𝟒
𝝀 = 𝟐𝟎𝟎 𝒄𝒎 = 𝟐 𝒎
𝝂 =
𝒗
𝝀
= 𝟏𝟕𝟎 𝑯𝒛

27. Waves-4
TOPICS: Formation of Beats, beat frequency,
applications, velocity of sound wave with
numerical problems
PREPARED BY,
NIRUPAMA,
P.G.T. PHYSICS,
KENDRIYA VIDYALAYA,
NEW TEHRI TOWN,
DEHRADUN REGION

28. BEATS
When two waves with very small difference in frequencies
(𝚫𝝂 < 𝟏𝟎), travel in a medium simultaneously, the intensity at
a point in the medium becomes maximum and minimum
(waxes and wanes) alternately. Theses are called beats.
𝝂𝟏
𝝂𝟐

29. Let the displacement of a particle of medium due to the
waves are 𝒚𝟏 and 𝒚𝟐. So equations of SHM of the particle
due to the waves,
𝒚𝟏 = 𝒂 𝒔𝒊𝒏 𝝎𝟏𝒕 , 𝒚𝟐 = 𝒂 𝒔𝒊𝒏 𝝎𝟐𝒕, Using superposition
principle, the net displacement of the particle,
𝒚 = 𝒂 𝒔𝒊𝒏 𝝎𝟏𝒕 + 𝒂 𝒔𝒊𝒏 𝝎𝟐𝒕
𝒚 = 𝒂(𝟐 𝒔𝒊𝒏
(𝝎𝟏𝒕 + 𝝎𝟐𝒕)
𝟐
cos
(𝝎𝟏𝒕 − 𝝎𝟐𝒕)
𝟐
)
𝒚 = 𝟐𝒂 𝒔𝒊𝒏
𝟐𝝅(𝝂𝟏+𝝂𝟐)𝒕
𝟐
cos
𝟐𝝅(𝝂𝟏−𝝂𝟐)𝒕
𝟐
, ( 𝝎 = 𝟐𝝅𝝂 )
𝒚 = 𝟐𝒂 𝒔𝒊𝒏𝟐𝝅𝝂𝒕 cos 𝝅(𝝂𝟏−𝝂𝟐)𝒕 , ( av. Freq. 𝝂 =
𝝂𝟏+𝝂𝟐
𝟐
)
𝒚 = [𝟐𝒂 cos 𝝅(𝝂𝟏−𝝂𝟐)𝒕] 𝒔𝒊𝒏𝟐𝝅𝝂𝒕
𝒚 = 𝑨 𝒔𝒊𝒏𝟐𝝅𝝂𝒕 ……(1)
Amplitude of the particle 𝐀 = 𝟐𝒂 cos 𝝅(𝝂𝟏−𝝂𝟐)𝒕
BEATS

30. Amplitude of the particle 𝐀 = 𝟐𝒂 cos 𝝅(𝝂𝟏−𝝂𝟐)𝒕
For minima: cos 𝝅(𝝂𝟏−𝝂𝟐)𝒕 = 𝟎 = cos 2𝑛 − 1 𝝅/𝟐 (n=1,2,3,…..)
So at time, 𝒕 =
𝟐𝒏−𝟏
𝟐(𝝂𝟏−𝝂𝟐)
𝒕 =
𝟏
𝟐(𝝂𝟏−𝝂𝟐)
,
𝟑
𝟐(𝝂𝟏 − 𝝂𝟐)
,
𝟓
𝟐(𝝂𝟏 − 𝝂𝟐)
, … … … …
Time between successive minima 𝐓 =
𝟏
𝝂𝟏−𝝂𝟐
For maxima: cos 𝝅(𝝂𝟏−𝝂𝟐)𝒕 = ±𝟏 = cos 𝑛𝝅 (n=1,2,3,…..)
So at time, 𝒕 =
𝒏
𝝂𝟏−𝝂𝟐
𝒕 =
𝟏
𝝂𝟏 − 𝝂𝟐
,
𝟐
𝝂𝟏 − 𝝂𝟐
,
𝟑
𝝂𝟏 − 𝝂𝟐
, … … … …
Time between successive maxima 𝐓 =
𝟏
𝝂𝟏−𝝂𝟐
One maximum and one minimum make one beat.
So time period of beats, 𝐓 =
𝟏
𝝂𝟏−𝝂𝟐
, Beat frequency, 𝐟 = 𝝂𝟏 − 𝝂𝟐
The persistence of ear is
𝟏
𝟏𝟎
𝒔. So the beats can be heard
distinctly only when 𝝂𝟏 − 𝝂𝟐 < 𝟏𝟎.

31. Beat frequency =2
TUNING FORK: When lightly hit on rubber pad a tuning fork
vibrates with its fixed natural frequency.
Loading of tuning fork: when a little wax is attached to the prongs of
a tuning fork, its mass increases. So its frequency decreases.
Filing of tuning fork: when the prongs of a tuning fork are filed by
using a sand paper, its mass decreases. So its frequency increases.
Beat method is used for tuning of musical instruments.

32. VELOCITY OF SOUND WAVE
Velocity of sound wave in a gas,
𝒗 =
𝜸𝑷
𝝆
where 𝑷 is pressure, 𝜸 =
𝑪𝑷
𝑪𝑽
and 𝝆 is density of gas.
𝒗 =
𝜸𝑷
𝝆
=
𝜸𝑹𝑻/𝑽
𝑴/𝑽
=
𝜸𝑹𝑻
𝑴
1.No effect of pressure on velocity of sound in gas.
2. 𝒗 ∝ 𝑻, with increase in temp , velocity of sound increases.
3. 𝒗 ∝
𝟏
𝑴
, with increase in molecular mass of gas, velocity of
sound deceases.
4. 𝒗 ∝
𝟏
𝝆
, with increase in density of gas, velocity of sound
decreases.
5. With increase in humidity in air, velocity of sound increases.

33. Q. A two tuning fork arrangement gives 4 beats/second, with one of
them having frequency of 288 Hz. A little of wax is places on the
unknown fork which reduces the beats to 2. Find the frequency of the
unknown fork.
Ans. Beat frequency, 𝐟 = 𝝂𝟏 − 𝝂𝟐
Beat frequency, 𝐟 = 𝟒 ,
𝝂𝟏 = 𝟐𝟖𝟖 𝑯𝒛 ,
𝝂𝟐 = 𝝂𝟏 ± 𝟒 = 𝟐𝟗𝟐 𝑯𝒛 𝒐𝒓 𝟐𝟖𝟒 𝑯𝒛
𝟐𝟖𝟖 𝑯𝒛
𝟐𝟖𝟒 𝑯𝒛
𝟐𝟗𝟐 𝑯𝒛
4 beats
4 beats
2 beats
6 beats
On loading the fork, its frequency
decreases. In this case the beats are
decreasing. So the frequency of the
second fork is 𝟐𝟗𝟐 𝑯𝒛 .