Class 12 Physics Wave in pipes and strings Notes

Class 12 Physics Wave in pipes and strings Notes
UNIT 3
CLASS 12 PHYSICS • WAVE AND OPTICS

Wave in Pipes and Strings

Chapter 8
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Organ Pipe

A cylindrical hollow tube specially made for producing musical sound of different harmonics by the vibration of air molecules into it is called an organ pipe. For example: flute, whistle etc.

Types of Organ Pipe

  1. Closed organ pipe
  2. Open organ pipe

1. Closed Organ Pipe

An organ pipe whose one end is closed and another end is open is called closed organ pipe. For example: bottle, cap of pen etc.

When air is blown from open end it gets reflected from closed end, hence a stationary wave is produced. At the closed end node is formed and at the open end anti-node is formed.

Fundamental, second and third modes in a closed organ pipe Fundamental mode Node Antinode L=λ/4 Second mode / first overtone L=3λ/4 Third mode / second overtone L=5λ/4
Clean representation of the first three modes of a closed organ pipe

i) Fundamental Mode of Vibration

In the fundamental mode of vibration of a closed organ pipe, one node is formed at the closed end and one anti-node is formed at open end.

From figure:
L = λ/4
λ = 4L

If f0 be the frequency of fundamental mode:

f0 = v/λ
f0 = v/(4L)
First harmonic

ii) Second Mode of Vibration (First Overtone)

In the second mode of vibration two nodes and two anti-nodes are formed.

L = 3λ/4
λ = 4L/3
f1 = v/λ
f1 = 3v/(4L)
f1 = 3f0
Third harmonic

Hence, frequency of second mode of vibration is three times greater than frequency of fundamental mode of vibration in closed organ pipe.

iii) Third Mode of Vibration (Second Overtone)

In third mode of vibration three nodes and three anti-nodes are formed.

L = 5λ/4
λ = 4L/5
f2 = v/λ
f2 = 5v/(4L)
f2 = 5f0
Fifth harmonic

Hence, frequency of third mode of vibration is five times greater than frequency of fundamental mode of vibration in closed organ pipe.

The frequencies obtained in closed organ pipe are f0, 3f0, 5f0 … i.e. closed pipes give only odd harmonics.
In general:
fn−1 = (2n−1)v/(4L) = (2n−1)f0, where n = 1,2,3…

2. Open Organ Pipe

An organ pipe whose both ends are open is called open organ pipe. For example: flute.

When air is blown into the pipe through one end, a wave travels through the tube to the next end and from where it is reflected. Hence, a stationary wave is produced and at the open end anti-node is formed.

Fundamental, second and third modes in an open organ pipe Fundamental mode L=λ/2 Second mode L=λ Third mode / second overtone L=3λ/2
Clean representation of the first three modes of an open organ pipe

i) Fundamental Mode of Vibration

In fundamental mode of vibration of organ pipe, one node and two anti-nodes are formed.

L = λ/2
λ = 2L
f0 = v/λ
f0 = v/(2L)

ii) Second Mode of Vibration (First Overtone)

In second mode of vibration in open organ pipe, two nodes and two anti-nodes are formed.

L = λ
f1 = v/λ
f1 = v/L
f1 = 2f0

Hence, frequency of second mode of vibration is two times greater than frequency of fundamental mode of vibration in open organ pipe.

iii) Third Mode of Vibration (Second Overtone)

In third mode of vibration in open organ pipe, three nodes and four anti-nodes are formed.

L = 3λ/2
λ = 2L/3
f2 = v/λ
f2 = 3v/(2L)
f2 = 3f0

Hence, frequency of third mode of vibration is three times greater than frequency of fundamental mode of vibration in open organ pipe.

End Correction

In an organ pipe, the air molecules at the open end are free to vibrate so that the wave extends a little into air outside the pipe. The distance between the end point of pipe and position of antinode formed outside the free end of pipe is called end correction. It is denoted by ‘e’.
End correction of closed and open organ pipes Closed organ pipe L e Open organ pipe L e e
End correction in closed and open pipes

End Correction of Closed Pipe

L + e = λ/4
λ = 4(L + e)

End Correction of Open Organ Pipe

L + e + e = λ/2
L + 2e = λ/2
λ = 2(L + 2e)

It is found that the end correction ‘e’ and internal diameter ‘d’ of pipe are related as:

e = 0.3d

Thus, the value of end correction is greater for pipe of larger diameter.

Waves in Strings

Velocity of Wave in a Stretched String

Let us consider a transverse wave travelling along a string of length ‘l’ and mass ‘m’ with a velocity ‘v’ under tension ‘T’. Then it is found that velocity of wave depends upon:

  1. The tension ‘T’ acting on string: v ∝ Ta
  2. The mass ‘m’ of string: v ∝ mb
  3. The length ‘l’ of string: v ∝ lc
Combining all:
v ∝ Tamblc
v = kTamblc   — (i)

Dimensional form:

[M0LT−1] = k[MLT−2]a[M]b[L]c
= k[Ma+bLa+cT−2a]

Comparing:

a + b = 0   — (i)
a + c = 1   — (ii)
−2a = −1   — (iii)
a = 1/2
b = −1/2
c = 1/2
v = kT1/2m−1/2l1/2
v = k√(Tl/m)

Since μ = m/l:

v = √(T/μ)

Modes of Vibration in a Stretched String

First, second and third modes of vibration of a stretched string fixed at both ends First mode L=λ/2 Second mode L=λ Third mode L=3λ/2
First three normal modes of a fixed stretched string

1) First Mode of Vibration

In first mode of vibration two nodes and one anti-node are formed. If ‘L’ be length of string and ‘λ’ be wavelength:

L = λ/2
λ = 2L
f1 = v/λ
But, v = √(T/μ)
f1 = (1/2L)√(T/μ)

2) Second Mode of Vibration

In second mode of vibration three nodes and two anti-nodes are formed.

L = λ
f2 = v/λ = v/L
f2 = (2/2L)√(T/μ)
f2 = 2f1

3) Third Mode of Vibration

In third mode of vibration four nodes and three anti-nodes are formed.

L = 3λ/2
λ = 2L/3
f3 = v/λ
f3 = (3/2L)√(T/μ)
f3 = 3f1

Resonance

When a body capable of vibration is allowed to vibrate freely, it will vibrate with its own frequency known as natural frequency. If an external periodic force is applied to a body, then the body vibrates with the frequency of periodic force. When the frequency of periodic force is equal to natural frequency of the body, the amplitude of vibration is maximum. This phenomenon is called resonance.

Conditions for Resonance

  1. The frequency of applied force must be equal to the natural frequency of the system.
  2. The applied force must be in phase with the vibrating system.

Resonance Tube Experiment

Resonance tube apparatus with tuning fork, tube, reservoir, rubber tube and scale Tuning fork Tube T Reservoir R Scale S Rubber tube
Fig: Resonance tube apparatus

Resonance tube consists of tube ‘T’ and reservoir ‘R’ which contains liquid. The tube and reservoir are connected by rubber tube and scale ‘S’ is attached to measure the air column in the tube. The length of air column is adjusted by raising or lowering the level of liquid in the reservoir.

A vibrating tuning fork of known frequency is placed over the mouth of tube ‘T’. This vibrates the air column inside the tube.

First and second resonance positions in a resonance tube First resonance l₁ λ/4 Second resonance l₂ 3λ/4
First and second resonance of the air column

First Resonance

l1 + e = λ/4   — (i)

Second Resonance

l2 + e = 3λ/4   — (ii)

Subtracting equation (i) from equation (ii):

l2 − l1 = 3λ/4 − λ/4
l2 − l1 = λ/2
λ = 2(l2 − l1)   — (iii)

If f be frequency of tuning fork and v be velocity of sound:

v = λf
v = 2(l2 − l1)f   — (iv)

This gives the velocity of sound at room temperature.

Velocity of Sound at 0°C

Suppose v and v0 be velocity of sound at room temperature and 0°C.
v/v0 = √[(T + 273)/273]
v0 = v√[273/(T + 273)]

To Determine End Correction

From equation (i):
l1 + e = λ/4
e = λ/4 − l1   — (v)

Using λ = 2(l2 − l1):

e = [2(l2 − l1)]/4 − l1
e = (l2 − 3l1)/2

Laws of Transverse Vibration of a Fixed Stretched String

1. Law of Length

The frequency of transverse vibration of a stretched string is inversely proportional to its length ‘l’ when tension ‘T’ and mass per unit length ‘μ’ are constant.
f ∝ 1/l   for constant T and μ

2. Law of Tension

The frequency of transverse vibration of a stretched string is directly proportional to square root of tension ‘T’ when mass per unit length ‘μ’ and length ‘l’ are constant.
f ∝ √T   for constant l and μ

3. Law of Mass per Unit Length

The frequency of transverse vibration of a stretched string is inversely proportional to square root of mass per unit length ‘μ’ at constant tension ‘T’ and length ‘l’.
f ∝ 1/√μ   for constant T and l

Combined Relation

Combining all:
f ∝ (1/l)√(T/μ)
f = k(1/l)√(T/μ)
where k = 1/2
f = (1/2l)√(T/μ)

Verification of Laws of Vibration Using Sonometer

Sonometer apparatus with wooden box, bridges, stretched wire, pulley, weight and tuning fork Wooden box B₁ B₂ Tuning fork Weight
Fig: Sonometer

The laws of vibration of fixed stretched strings are verified by using a sonometer. It consists of a hollow wooden box with a wire fixed at one end and stretched with the help of a load at the other end.

A vibrating tuning fork is placed vertically on wooden box and two bridges B1 and B2 are adjusted for maximum vibration. The length between two bridges gives resonating length. The frequency of fundamental mode can be determined by:

f = (1/2l)√(T/μ)
Graphs verifying law of length, law of tension and law of mass per unit length f vs 1/l f vs √T f vs 1/√μ
Straight-line graphs used to verify the three laws

I) To Verify Law of Length: f ∝ 1/l

Take different tuning forks of known frequency and measure resonating length for each by keeping tension ‘T’ and mass per unit length ‘μ’ constant. A graph between f and 1/l is a straight line passing through origin.

II) To Verify Law of Tension: f ∝ √T

Observe resonance for different frequencies by varying tension on string for same resonating length of wire and same mass per unit length. A graph between f and √T is a straight line passing through origin.

III) To Verify Law of Mass per Unit Length: f ∝ 1/√μ

Take different wires of different mass per unit length and observe corresponding resonance by keeping length and tension constant. A graph between f and 1/√μ is a straight line passing through origin.

Solved Numericals

Q.1 — Change in Fundamental Frequency of Piano Wire

A pianoforte wire having a diameter of 0.99 mm is replaced by another wire of the same material but with diameter 0.93 mm. If the tension of the wire is as before, what is percentage change in the frequency of fundamental note?

f = (1/2l)√(T/μ)
μ = mass/length = ρA = ρπd2/4
Therefore, f = (1/ld)√(T/πρ)

For the two wires:

f1 = (1/ld1)√(T/πρ)
f2 = (1/ld2)√(T/πρ)
f1/f2 = d2/d1
= 0.93/0.99
f1/f2 = 0.94

Percentage change:

(1 − 0.94) × 100%
= 6.1%

Q.2 — Fundamental Frequency and Tension in Third Mode

A cord of length 1.5 m is fixed at both ends. Its mass per unit length is 1.2 g/m and the tension is 12 N. (a) What is the frequency of fundamental oscillation? (b) What tension is required if the n = 3 mode has frequency of 0.50 kHz?

l = 1.5 m
μ = 1.2 g/m = 1.2×10−3 kg/m
T = 12 N

(a)

f = (1/2l)√(T/μ)
= 1/(2×1.5) × √[12/(1.2×10−3)]
f = 33.33 Hz

(b)

f = 0.50 kHz = 500 Hz
For n = 3 mode:
f = (3/2l)√(T/μ)
500 = 3/(2×1.5) √[T/(1.2×10−3)]
250000 = T/0.0012
T = 300 N

Q.3 — End Correction of Open and Closed Pipes in Resonance

An open pipe 30 cm long and a closed pipe 23 cm long, both of the same diameter, each sound their first overtone. If they are in resonance find the end correction of these pipes.

Closed pipe, first overtone:

lc + e = 3λ/4
λ = 4(lc + e)/3

Open pipe, first overtone:

lo + 2e = λ

For resonance, wavelength is same:

lo + 2e = 4(lc + e)/3
3(0.30 + 2e) = 4(0.23 + e)
0.90 + 6e = 0.92 + 4e
2e = 0.02
e = 0.01 m

Q.4 — Resonance Air Column: Velocity and End Correction

In a resonance air column apparatus the first and second resonance positions are observed at 18 cm and 56 cm. The frequency of tuning fork used was 480 Hz. Calculate the velocity of sound in air and end correction of the tube.

The handwritten working labels the two lengths as l₁ = 56 cm and l₂ = 18 cm before substituting them. The calculation below preserves that source sequence.
l1 = 56 cm
l2 = 18 cm
f = 480 Hz
v = 2f(l1 − l2)
= 2 × 480 × (56 − 18)
v = 36480 cm/s

End correction:

e = (l1 − 3l2)/2
= (56 − 54)/2
e = 1 cm

Q.5 — Frequency of an Organ Pipe at 0°C

An organ pipe is tuned to a frequency of 440 Hz when the temperature is 27°C. Find its frequency when the temperature drops to 0°C. Assume both ends of the pipe open.

f1 = 440 Hz
T1 = 273 + 27 = 300 K
T2 = 273 K

Since frequency is proportional to velocity for fixed pipe length:

f2/f1 = v2/v1
v2/v1 = √(T2/T1)
f2/440 = √(273/300)
f2/440 = 0.95
f2 = 419.73 Hz

Q.6 — Closed Pipe Second Overtone and Open Pipe Third Harmonic

On a day when the speed of sound is 345 m/s, the fundamental frequency of a closed organ pipe is 220 Hz. The second overtone of this pipe has the same wavelength as the third harmonic of an open pipe. How long is the open pipe?

v = 345 m/s
f = 220 Hz
λ = v/f = 345/220
λ = 1.56 m

Length of closed pipe in fundamental mode:

lc = λ/4
= 1.56/4
lc = 0.39 m

For second overtone of closed pipe:

lc = 5λ′/4
λ′ = 4lc/5
= 4×0.39/5
λ′ = 0.312 m

For third harmonic of open pipe:

lo = 3λ′/2
= 3×0.312/2
lo = 0.47 m

Q.7 — Steel Wire in Unison with Open Pipe at 0°C

A steel wire of length 40 cm and diameter 0.25 mm vibrates in unison with a tube open at both ends and of effective length 60 cm. Find the tension in the wire. (Velocity at 0°C = 332 m/s, density of steel = 7800 kg/m³.)

Length of steel wire, ls = 40 cm = 0.4 m
Diameter, d = 0.25×10−3 m
Density of steel, ρ = 7800 kg/m3
Effective length of open pipe, lo = 0.6 m

For the open pipe fundamental:

lo = λ/2
λ = 2lo
f = v/λ = v/(2lo)
= 332/(2×0.6)
f = 276.16 Hz

For steel wire:

f = (1/2ls)√(T/μ)
μ = ρA = ρπd2/4
f = (1/lsd)√(T/πρ)

Using the values as in the source:

T = 18.77 N

Q.8 — Young’s Modulus of a Piano String

A piano string has a length of 2 m and a density of 8000 kg/m³. When the tension in the string produces a strain of 1%, the fundamental note obtained from the string in transverse vibration is 170 Hz. Calculate the Young’s modulus value for the material of string.

l = 2 m
Strain = 1% = 1/100
ρ = 8000 kg/m3
f = 170 Hz

We know:

Y = stress/strain = (T/A)/strain
T = YA × strain

Also:

f = (1/2l)√(T/μ)
μ = Aρ
f = (1/2l)√[(Y × strain)/ρ]
170 = 1/(2×2) √[Y/(8000×100)]
170×4 = √(Y/800000)
462400 = Y/800000
Y = 3.7×1011 N/m2

Q.9 — Steel Wire in Unison with Open Pipe at 27°C

A steel wire of length 40 cm and diameter 0.25 mm vibrates in unison with a tube open at both ends and of effective length 60 cm, when each is sounding its fundamental note. The air temperature is 27°C. Find the tension in the wire. (Velocity at 0°C = 332 m/s, density of steel = 7800 kg/m³.)

The handwritten source obtains v27 = 316.7 m/s from the temperature relation and continues the numerical with that value. The typed solution below preserves the source calculation rather than silently correcting it.
Velocity at 0°C, v0 = 332 m/s
T0 = 273 K
T27 = 300 K
v0/v27 = √(T0/T27)
332/v27 = √(273/300)
v27 = 316.7 m/s   [as written in source]

For open pipe:

λ = 2lo
f = v27/(2lo)
= 316.7/(2×0.6)
f = 263.92 Hz

For steel wire:

f = (1/lsd)√(T/πρ)
T = 17.06 N
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