Physics. Wave and Sound - 4 Session Session Objectives.
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Transcript of Physics. Wave and Sound - 4 Session Session Objectives.
![Page 1: Physics. Wave and Sound - 4 Session Session Objectives.](https://reader036.fdocuments.us/reader036/viewer/2022062519/5697bff01a28abf838cbaeb0/html5/thumbnails/1.jpg)
Physics
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Wave and Sound - 4
Session
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Session Objectives
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Session Objective
Beats
Conditions under which beats occur
Standing Waves in air columns
- All Conditions
Standing waves in a string
(with all conditions)
Fundamental Frequency
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Standing waves in a string (with all conditions)
Let the waves
1y A sin(kx t)
2y A sin k x t
with same amplitude, same frequency, having a
phase difference ‘’, superimpose to produce a
standing wave given by
y 2A sin k x cos t2 2
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String fixed at both ends
A
N N
Lx = Lx = O
Since x = 0 and x = L are nodes, at these points y = 0 for all ‘t’.
Hence, for x = 0,
y 2A sin cos t
2 2
Considering = 0,
y 2A sin(0)cos t y 0
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String fixed at both ends
Similarly, for x = L
2
sin L 0 sinn
where n = 0, 1, 2, …
2L2 L
n orn
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String fixed at both ends
If the length of the string is an integral
multiple of , standing waves are
produced.
2
Hence, the other natural frequencies with which
standing waves can be formed on string are
v nv n T
or2L 2L
v
Now
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String fixed at both ends
1 0
2 T2
2L
2 0
3 T3
2L
3 0
4 T4
2L
[First overtone or second harmonic]
[Second overtone or third harmonic]
[Third overtone or fourth harmonic]
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String fixed at both ends
N N
A AN
N N
A AN N Second overtone
Third harmonic
First overtone
Second harmonic
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String fixed at one end
N A
A
NAN
N
A AN N A
Fundamental
First overtone
Second overtone
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String fixed at one end
The boundary condition that x = 0
is automatically satisfied by
y = 2A sinkx cost.
For x = L to be an antinode,
sinkL 1
1kL n
2
2 L 1n
2
2L 1n
v 2
12
n1 v Tor n
2 2L 2L
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String fixed at one end
Putting n = 0,
0v
4L [Fundamental frequency]
Similarly,
1 03v
34L
2 05v
54L
[First overtone]
[Second overtone]
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Fundamental Frequency
The lowest frequency with which a
standing wave can be set up in a
string is called the fundamental
frequency.
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Standing Waves in Air Columns
v nvff or n 1,3,5,7,.....[closed organ pipe]
4L
L
L4
Fundamental Frequencyor
First Harmonic
L
3L
4
First Overtoneor
Third Harmonic
L Second Overtoneor
Fifth Harmonic
5L
4
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Standing Waves in Air Columns
L2
LFundamental Frequency
orFirst Harmonic
v nvff or n 1,2,3, 4.....[open organ pipe]
2L
2L
2
LFirst Overtone
orSecond Harmonic
3L
2
LSecond Overtone
orThird Harmonic
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Beats
The difference between two combining frequencies is called beats.
Beats are produced due to superposition of two or more waves having nearly equal frequencies.
[Beat frequency] beat 1 2f f f
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Class Test
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Class Exercise - 1
An open organ pipe of length L vibrates
in its fundamental mode. The pressure
variation is maximum
(a) at the two ends
(b) at the middle of the pipe
(c) at distance inside the ends
(d) at distance inside the ends
L
4
L
8
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Solution
Hence answer is (a).
Pressure variation is maximum
at the ends.
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Class Exercise - 2
An open pipe is suddenly closed at one
end with the result that the frequency of
third harmonic of the closed pipe is
found to be higher by 100 Hz than
fundamental frequency of the open
pipe. The fundamental frequency of the
open pipe is
(a) 200 Hz (b) 300 Hz
(c) 240 Hz (d) 480 Hz
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Solution
Hence answer is (a).
3v v v
100 Hz 200 H z4L 2L 2L
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Class Exercise - 3
A pipe, open at both ends, gives
frequencies which are
(a) only even multiple of fundamental
frequency
(b) only odd multiple of fundamental
frequency
(c) all integral multiple of fundamental
frequency
(d) all fractional multiple of fundamental
frequency
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Solution
Hence answer is (c).
Open pipe produces all integral multiple
of fundamental frequency.
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Class Exercise - 4
In a closed organ pipe, the fundamental
frequency is ‘’. What will be the ratio of
the frequencies of the next three
overtones?
(a) 2 : 3 : 4 (b) 3 : 4 : 5
(c) 3 : 7 : 11 (d) 3 : 5 : 7
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Solution
Hence answer is (d).
Closed pipe produces only odd
multiples of fundamental frequency.
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Class Exercise - 5
Two open organ pipes of length 25
cm and 25.5 cm produce 0.1 beats
per second. The velocity of sound is
(a) 350 m/s (b) 325.5 m/s (c)
303 m/s (d) 255 m/s
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Solution
Hence answer is (d).
v v
0.12(25) 2(25.5)
v = 255 m/s
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Class Exercise - 6
Velocity of sound in air is 320 m/s. A
pipe closed at one end has a length of 1
m. Neglecting end correction, the air
column in the pipe can resonate for
sound of frequency which is equal to
(More than one options may be correct)
(a) 80 Hz (b) 240 Hz
(c) 320 Hz (d) 400 Hz
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Solution
Hence answer is (a, b , d).
Fundamental frequency
0
v 32080 Hz
4L 4 1
Hence, possible frequencies are odd multiples of
fundamental frequency.
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Class Exercise - 7
Two waves of wavelengths 2 m and 2.02
m respectively, moving with the same
velocity superimpose to produce 2 beats
per second. The velocity of the wave is
equal to
(a) 400 m/s (b) 402 m/s
(c) 404 m/s (d) 406 m/s
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Solution
Hence answer is (c).
v v
22 2.02
v = 404 m/s
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Class Exercise - 8
A sonometer wire, 65 cm long, is in
resonance with a tuning fork of frequency
N. If the length of the wire is decreased by
1 cm and it is vibrated with the same
tuning fork, 8 beats are heard per second.
What is the value of N?
(a) 256 Hz (b) 384 Hz
(c) 480 Hz (d) 512 Hz
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Solution
Hence answer is (a).
1 TN'
2(65 1) m
1 T
N2 65 m
Given
Also or N' N 8 N' N 8
Hence, N is calculated.
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Class Exercise - 9
Beats are result of
(a) diffraction
(b) destructive interference
(c) constructive interference
(d) superposition of two waves of nearly
equal frequencies
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Solution
Hence answer is (d).
Definition of beats.
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Class Exercise - 10
An organ pipe vibrates in fundamental
resonance with the medium as air, nitrogen
and oxygen. Which is the correct statement?
(a) The wavelength changes with medium
change
(b) Both wavelength and frequency change
(c) Both and remain unaltered with the
medium change
(d) The frequency changes with the medium
change
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Solution
Hence answer is (a).
Only wavelength depends
on the medium.
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Thank you