physics. wave and sound - 3 session session objectives
TRANSCRIPT
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Physics
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Wave and Sound - 3
Session
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Session Objectives
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Session Objective
Intensity of Sound waves - Sound Level
Standing Waves (Wave Function)
Position of Nodes & antinodes in standing waves
Periodic Sound Waves
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Periodic Sound Waves
The general form of the wave function is
where n = Integer
y(x, t) f(x vt)
y (x n ), t y(x, t)
In a periodic wave, the displacement repeats
itself after a fixed periodic length , i.e.
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Intensity of Sound waves - Sound Level
2 2 2I 2 dvA
where v = Velocity of the wave in that medium
= Frequency of particle oscillation
A = Amplitude
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Intensity of Sound waves - Sound Level
Also the intensity level L of a sound of
intensity ‘I’ is represented as
100
L log
where I0 = Zero level of intensity or threshold intensity
12 20I 1 10 W m
Intensity level is generally expressed in ‘decibels’.
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Standing Waves (Wave Function)
Two equal sine waves travel along a stretched string(length l) from opposite directions.
1y A sin( t kx)
2y A sin( t kx) 1 2y y y (2A cos kx) sin t
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Position of Nodes in standing waves
For nodes, amplitude of the particles is zero.
2A coskx 0
1kx n
2 2
1or x n
2 2where n is an integer.
coskx 0
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Position of Antinodes in standing waves
For antinodes, amplitude of the particles
is maximum.
2A cosk x is max imum
nor x
2
coskx 1
kx n
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Class Test
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Class Exercise - 1
A sine wave is traveling in a medium. A
particular particle has zero
displacement at a certain instant. The
particle closest to it having zero
displacement is at a distance of
(a) (b)4 3
(c) (d)2
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Solution
Hence answer is (c).
Particles which are at distance
from each other, have same
displacement.
2
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Class Exercise - 2
A sonometer wire of length vibrates in
fundamental mode when excited by a
tuning fork of frequency 416 Hz. If the
length is doubled, keeping other things
same, the string will vibrate with a
frequency of
(a) 416 Hz (b) 208 Hz
(c) 832 Hz (d) Stop vibrating
' '
1 Tn
2 m
Solution :-
Hence answer is (a).
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Class Exercise - 3
A standing wave is produced on a
string clamped at one end and free
at the other. The length of the string
must be an integral multiple of
(a) (b)2
(c) (d)3 3
This is condition for standing waves.
Solution :-
Hence answer is (d).
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Class Exercise - 4
Two strings A and B, made of same
material, are stretched by same tension.
The radius of string A is double the radius
of B. The transverse wave travels on A
with speed VA and on B with speed
VB. The ratio is
(a) (b) 2
(c) (d) 4
A
B
V
V1
2
1
4
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Solution
Hence answer is (a).
1
Speed vr
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Class Exercise - 5
The amplitude of sound is doubled and
the frequency is reduced to one-fourth.
The intensity of sound at the same point
will be
(a) increased by a factor of 2
(b) increased by a factor of 4
(c) decreased by a factor of 2
(d) decreased by a factor of 4
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Solution
Hence answer is (d).
2 2I A
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Class Exercise - 6
A standing wave having three nodes
and two antinodes is formed between
two atoms having a distance of 1.21 Å
between them. The wavelength of the
standing wave is
(a) 1.21 Å (b) 2.42 Å
(c) 6.05 Å (d) 3.63 Å
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Solution
Hence answer is (a).
2 2
In this case
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Class Exercise - 7
In a sinusoidal wave, the time required
for a particular point to move from the
maximum displacement to zero
displacement is 0.170 s. The frequency
of the wave is
(a) 1.47 Hz (b) 0.36 Hz
(c) 0.73 Hz (d) 2.94 Hz
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Solution
Hence answer is (a).
T = 4 × 0.170 = 0.68 s
1 1
1.47 HzT 0.68
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Class Exercise - 8
Particle displacements (in centimeters)
in a standing wave are given by
.
The distance between a node and the
next antinode's is
(a) 2.5 cm (b) 5 cm
(c) 7.5 cm (d) 10 cm
y(x, t) 2sin(0.1 x)cos(100 t)
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Solution
Hence answer is (b).
Distance between a node and
an antinode is .4
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Class Exercise - 9
Transverse waves of same frequency are
generated in two steel wires A and B.
The diameter of A is twice that of B and
tension in A is half that in B. The ratio of
velocities of waves in A and B is
(a) 1 : 2 (b)
(c) (d) 1: 2
1: 2 2
3 : 2 2
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Solution
Hence answer is (c).
2
Tv
r
BA B A
Tr 2r ; T
2
2A A B
B B A
v T r
v T r
Putting the values, we get
A
B
v 1 1
v 8 2 2
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Class Exercise - 10
y acos(k x t)A wave represented by is
superimposed with another wave to form
a standing wave, such that point x = 0 is
a node. Equation of the other wave is
(a) (b)
(c) (d)
a sin(kx t) acos(k x t)
acos(k x t) a sin(k x t)
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Solution
Hence answer is (b).
For formation of stationary waves two
waves should travel in opposite
directions. Hence, the waves are (a)
and (c) but (a) is not fulfilling the given
condition completely.
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Thank you