Solution-Class-12-Physics-Chapter-6-Superposition of Waves-MSBSHSE

(D) 5

(D) 357m/s

(B) 1.12

(A) x = Asin (ky - ωt)

(A) must be an odd integral multiple of λ/4.

The wave is travelling along the negative x-direction.

• Nodes are the points where the vibrating string can be touched without disturbing its motion.
• When the string vibrates in its fundamental mode, the string vibrates in one loop. There are no nodes formed between the fixed ends. Hence, there are no point on the string which can be touched without disturbing its motion.
• When the string vibrates in its first overtone (second harmonic), there are two loops of the stationary wave on the string. Apart from the two nodes at the two ends, there is now a third node at its centre. Hence, the string can be touched at its centre without disturbing the stationary wave pattern.
• When the string vibrates in its second overtone (third harmonic), there are three loops of the stationary wave on the string. So, apart from the two end nodes, there are two additional nodes in between, at distances one-third of the string length from each end. Thus, now the string can be touched at these two nodes.

• A stationary wave is set up in a bounded medium in which the boundary could be a rigid support (i.e., a fixed end, as for instance a string stretched between two rigid supports) or a free end (as for instance an air column in a cylindrical tube with one or both ends open). The boundary conditions limit the possible stationary waves and only a discrete set of frequencies is allowed.
• The lowest allowed frequency n₁ is called the fundamental frequency of vibration. Integral multiples of the fundamental frequency are called the harmonics, the fundamental frequency being called the first harmonic. The second harmonic is twice the fundamental or 2n₁, the third harmonic is 3n₁, and so on.
• The higher allowed harmonics above the first harmonic or fundamental are called overtones. The first overtone is the higher allowed harmonic immediately above the first harmonic.
• Above the fundamental, the first allowed frequency is called the first overtone, the next higher frequency is the second over-tone, and so on. The relation between overtones and allowed harmonics depends on the system under consideration.

The fundamental is the first harmonic. Therefore, the ratio of the fundamental frequency (n) to the second harmonic (n₁) is 1 : 2

Given : y = 0.2 sin 4π\([\frac{t}{0.08}-\frac{x}{0.8}]\)  , = 0.2 sin 2π\([\frac{2t}{0.08}-\frac{2x}{0.8}]\)

= 0.2 sin 2π\([\frac{t}{0.04}-\frac{x}{0.4}]\)

Let us compare above equation with the equation of a simple harmonic progressive wave.

 y = A sin 2π\([\frac{t}{T}-\frac{x}{λ}]\) = 0.2 sin 2π\([\frac{t}{0.04}-\frac{x}{0.4}]\)

Comparing the quantities on both sides, we get,

A = 0.2 m, T = 0.04 s, λ = 0.4 m

∴ (a) Wavelength (λ) = 0.4 m

  (b) Frequency (n) = 1/T = 1/0.04  = 25 Hz

  (c) Amplitude (A) = 0.2 m

Characteristics of progressive waves:

1) Each particle in a medium executes the same type of vibration. Particles vibrate about their mean positions performing simple harmonic motion.

2) All vibrating particles of the medium have the same amplitude, period and frequency.

3) The phase, (i.e., state of vibration of a particle), changes from one particle to another.

4) No particle remains permanently at rest. Each particle comes to rest momentarily while at the extreme positions of vibration.

5) The particles attain maximum velocity when they pass through their mean positions.

6) During the propagation of wave, energy is transferred along the wave. There is no transfer of matter.

7) The wave propagates through the medium with a certain velocity. This velocity depends upon properties of the medium.

8) Progressive waves are of two types : transverse and longitudinal.

• In a transverse mechanical wave, the individual particles of the medium vibrate perpendicular to the direction of propagation of the wave. The progressively changing phase of the successive particles results in the formation of alternate crests and troughs that are periodic in space and time. In an electromagnetic wave, the electric and magnetic fields oscillate in mutually perpendicular directions, perpendicular to the direction of propagation.
• In a longitudinal mechanical wave, the individual particles of the medium vibrate along the line of propagation of the wave. The progressively changing phase of the successive particles results in the formation of typical alternate regions of compressions and rarefactions that are periodic in space and time. Periodic compressions and rarefactions result in periodic pressure and density variations in the medium. There are no longitudinal em wave.

9) A transverse wave can propagate only through solids, but not through liquids and gases while a longitudinal wave can propagate through any material medium.

Characteristics of stationary waves :

• Stationary waves are produced due to superposition of two identical waves (either transverse or longitudinal waves) traveling through a medium along the same path in opposite directions
• The overall appearance of a standing wave is of alternate intensity maximum (displacement antinode) and minimum (displacement node).
• The distance between adjacent nodes (or antinodes) is λ/2.
• The distance between successive node and antinode is λ/ 4.
• There is no progressive change of phase from particle to particle. All the particles in one loop, between two adjacent nodes, vibrate in the same phase, while the particles in adjacent loops are in opposite phase.
• A stationary wave does not propagate in any direction and hence does not transport energy through the medium.
• In a region where a stationary wave is formed, the particles of the medium (except at the nodes) perform SHM of the same period, but the amplitudes of the vibrations vary periodically in space from particle to particle.

When two progressive waves having the same amplitude, wavelength and speed propagate in opposite directions through the same region of a medium, their superposition under certain conditions creates a stationary interference pattern called a stationary wave.

Consider two simple harmonic progressive waves, of the same amplitude A, wavelength λ and frequency n = ω/2π, travelling on a string stretched along the x-axis in opposite directions. They may be represented by

y₁ = A sin (ωt − kx)   (along the + x-axis) and  ….(1)

y₂ = A sin (ωt + kx) (along the — x-axis)  …..(2)

where k = 2π/λ is the propagation constant.

By the superposition principle, the resultant displacement of the particle of the medium at the point at which the two Waves arrive simultaneously is the algebraic sum

y = y₁ + y₂ = A[sin (ωt − kx) + sin (ωt + kx)]

Using the trigonometrical identity,

Sin C + sin D = 2 sin\((\frac{C−D}{2})\)  cos\((\frac{C−D}{2})\)

y = 2A sin ωt cos (−kx)

= 2A sin ωt cos kx      ….[… cos (−kx) = cos (kx)]

= 2A cos kx sin ωt  ….(3)

 y = R sin ωt, ...(4)

where R = 2A cos kx.  ….(5)

Equation (4) is the equation of a stationary wave.

The amplitude of the resultant wave produced due to the interference of the two waves is

A = \(\sqrt{A_1^2+2A_1A_2 cos φ + A_2^2}\)

Laws of a Vibrating String :

The fundamental frequency of a vibrating string under tension is given as

n = \(\frac{1}{2L}\sqrt{\frac{T}{m}}\)

From this formula, three laws of vibrating string can be given as follows:

1) Law of length: The fundamental frequency of vibrations of a string is inversely proportional to the length of the vibrating string, if tension and mass per unit length are constant. if T and m are constant

n ∝ 1/L

2) Law of tension: The fundamental frequency of vibrations of a string is directly proportional to the square root of tension, if vibrating length and mass per unit length are constant. If L and m are constant.

n ∝ \(\sqrt{T}\)

3) Law of linear density: The fundamental frequency of vibrations of a string is inversely proportional to the square root of mass per unit length (linear density), if the tension and vibrating length of the string are constant. If T and L are constant

n ∝ \(\sqrt{\frac{1}{m}}\)

A sonometer is used to determine the frequency of a tuning fork and to verify the laws of vibrating strings.

1) Verification of first law (law of length) of a vibrating string:

According to this law, n ∝ 1/L, if T and m are constant. To verify this law, the sonometer wire of given linear density m is kept under constant tension T.

The length of the wire is adjusted for the wire to vibrate in unison with tuning forks of different frequencies n₁, n₂, n₃,... . Let L₁, L₂, L₃,….., be the corresponding resonating lengths of the wire. It is found that, within experimental errors,

n₁L₁ = n₂L₂ =n₃L₃ = …. . This implies that the product, nL = constant, which verifies the law of length.

2) Verification of second law (Law of tension) of a vibrating string:

According to this law, n ∝ , if L and m are constant. To verify this law, the vibrating length L of the sonometer wire of given linear density m is kept constant.

A set of tuning forks of different frequencies is used. The tension in the wire is adjusted for the wire to vibrate in unison with tuning forks of different frequencies

n₁, n₂, n₃,... ., T₁, T₂, T₃,…  be the corresponding tensions. It is found that, within experimental errors,

\(\frac{n_1}{\sqrt{T_1}}=\frac{n_2}{\sqrt{T_2}}=\frac{n_3}{\sqrt{T_3}}\)=......

This implies constant which verifies the law of tension.

3) Verification of third law (Law of linear density) of a vibrating string:

According to this law, n ∝ \(\sqrt{\frac{1}{m}}\)  , if T and L are constant. To verify this law, two wires having different linear densities m₁ and m₂ are kept under constant tension T.

A tuning fork of frequency n is used. The lengths of the wires are adjusted for the wires to vibrate in unison with the tuning fork.

Let L₁ and L₂ be the corresponding resonating lengths of the wires. It is found that, within experimental errors,

\(L_1\sqrt{m_1}=L_2\sqrt{m_2}\)   This implies \(L\sqrt{m}=\)  constant. According to the law of length of a vibrating string, n ∝ 1/L.

∴ n ∝ \(\sqrt{\frac{1}{m}}\) which verifies the law of linear density.

Consider a narrow cylindrical pipe of length l closed at one end.

When sound waves are sent down the air column in a cylindrical pipe closed at one end, they are reflected at the closed end with a phase reversal and at the open end without phase reversal.

Interference between the incident and reflected waves under appropriate conditions sets up stationary waves in the air column.

The stationary waves in the air column in this case are subject to two boundary conditions that there must be a node at the closed end and an antinode at the open end.

Taking into account the end correction e at the open end, the resonating length of the air column is

L = l + e.

Let v be the speed of sound in air. In the simplest mode of vibration [Fig.(a)] there is a node at the closed end and an antinode at the open end. The distance between a node and a consecutive antinode is λ/4 , where λ is the wavelength of sound. The corresponding wavelength λ and frequency n are

Λ = 4L and n = \(\frac{v}{λ} = \frac{v}{4L}=\frac{v}{4(l+e)}\)   …..(1)

This gives the fundamental frequency of vibration and the mode of vibration is called the fundamental mode or first harmonic.

In the next higher mode of vibration, the first overtone, two nodes and two antinodes are formed [Fig.(b) ]. The corresponding wavelength λ₁ and frequency n₁  are   and n₁ = v/λ₁ = 3v/4L = \(\frac{3v}{4(l+e)}\)  = 3n ……(2)

Therefore, the frequency in the first overtone is three times the fundamental frequency, i.e., the first overtone is the third harmonic.

In the second overtone, three nodes and three antinodes are formed [Fig (c)]. The corresponding wavelength λ₂ and frequency n₂ are

λ₂ = \(\frac{4L}{5}\)   and n₂ = v/λ₂ = 5v/4L = \(\frac{5v}{4(l+e)}\)  = 5n  ….(3)

which is the fifth harmonic,

Therefore, in general, the frequency of the p^th overtone (p = 1, 2, 3, ...) is

n_p= (2P+1)n   …..(4)

i.e., the p^th overtone is the (2p + 1)th harmonic.

Equations (1), (2) and (3) show that allowed frequencies in an air column in a pipe closed at one end are n, 3n, 5n,  ….. That is, only odd harmonics are present as overtones.

Consider a cylindrical pipe of length l open at both the ends.

When sound waves are sent down the air column in a cylindrical open pipe, they are reflected at the open ends without a change of phase.

Interference between the incident and reflected waves under appropriate conditions sets up stationary waves in the air column.

The stationary waves in the air column in this case are subject to the two boundary conditions that there must be an antinode at each open end.

Taking into account the end correction e at each of the open ends, the resonating length of the air column is L= l +2e.

Let v be the speed of sound in air. In the simplest mode of vibration, the fundamental mode or first harmonic, there is a node midway between the two antinodes at the open ends [Fig. (a)]. The distance between two consecutive antinodes is λ/2, where λ is the wavelength of sound. The corresponding wavelength λ and the fundamental frequency n are

λ = 2L and n = v/λ = v/2L = \(\frac{v}{2(l+2e)}\)    …..(1)

In the next higher mode, the first overtone, there are two nodes and three antinodes [Fig.(b)].

The corresponding wavelength λ₁ and frequency n₁

λ₁ = L and n₁ = v/λ₁ = v/L = \(\frac{v}{l+2e}\)  =2n    …..(2) i.e., twice the fundamental.

Therefore, the first overtone is the second harmonic.

In the second overtone, there are three nodes and four antinodes [Fig. (c)]. The corresponding wavelength λ₂ and frequency n₂ are

λ₂ = 2L/3 and n₂ = v/λ₂ = \(\frac{3v}{2L}\) = \(\frac{3v}{2(l+2e)}\)  = 3n   …..(3) or thrice the fundamental. Therefore, the second overtone is the third harmonic.

Therefore, in general, the frequency of the p^th overtone (p = 1, 2, 3, ...) is

n_p=(p+1)n  ….(4)

i.e., the p^th overtone is the (p + 1)th harmonic.

Equations (1), (2) and (3) show that allowed frequencies in an air column in a pipe open at both ends are n, 2n, 3n. That is, all the harmonics are present as overtones.

Given: n = 500 Hz, v = 350 m/s

V = n x λ

∴ λ  = 350/500 = 0.7 m

(a) in t = 1.0 ms = 0.001 s, the path difference is the distance covered

v x t = 350 x 0.001 = 0.35 m

∴ Phase difference = \(\frac{2π}{λ}\)x Path difference

                          = \(\frac{2π}{0.7}\) x 0.35 = π rad

(b) Phase difference = 45° = π/4 rad

∴ Path difference = \(\frac{λ}{2π}\)x Phase difference =\(\frac{0.7}{2π}×\frac{π}{4}\)  = 0.0875 m

Given : Distance between two successive nodes = λ/2 = 3.75 x 10^-2 m, v =1500 m/s

∴ λ = 7.5 x 10^-2 m

v = n x λ

∴ n = v/λ = \(\frac{1500}{3.75×10^{-2}}\)  = 20 kHz

∴ λ  = v/n = 330/330 = 1 m

Directly at the cenre of two sources of sound, path difference is zero. But since the waves are 180° out of phase, two maxima on either sides should be at a distance of λ /4  from the point at the centre. Other maxima will be located each λ/2   further along.

Thus, the sound intensity will be maximum at ± 0.25, ± 0.75, ± 1.25 and ±1.75 m from the point at the centre.

Given: n₁ = n₂ = 540 Hz, v = 330 m/s

V = n x λ

∴ λ  = 330/540 = 0.61 m

Here the path difference = 3.5 – 3 = 0.5 m

∴ Phase difference = \(\frac{2π}{λ}\) x Path difference

                          = \(\frac{2π}{0.61}\) x 0.5 = 1.64 p rad

Given : m₁ = m₂ = m, L₁ = 60 cm = 0.6 m, T₁ = 1.5 kg=14.7 N, T₂ =6 kg=58.8 N

n₁ = \(\frac{1}{2L_1}\sqrt{\frac{T_1}{m}}\) and  n₂ = \(\frac{1}{2L_2}\sqrt{\frac{T_2}{m}}\)

But n₁ = n₂

∴ \(\frac{1}{2L_1}\sqrt{\frac{T_1}{m}}\)  = \(\frac{1}{2L_2}\sqrt{\frac{T_2}{m}}\)

L₂ = \(\sqrt{\frac{T_2}{T_1}}×L_1\)  = \(\sqrt{\frac{58.8}{14.7}}×0.6\)= 1.2 m

The vibrating length of the second wire is 1.2 m.

The difference between the given frequencies of the overtones is 256 Hz. This implies that they are consecutive overtones. Let n_c be the fundamental frequency of the closed pipe and n_q, n_q+1, n_q+2 = the frequencies of the qth, (q + 1)th and

(q + 2)th consecutive overtones, where q is an integer.

Given : n_q = 640 Hz, n_q+1 = 896 Hz, n_q+2 = 1152 Hz

Since only odd harmonics are present as overtones,

n_q = (2q + 1)n_c  and n_q+1 = [2(q+ 1) + 1] n_c = (2q +3)n_c

∴ \(\frac{n_{q+1}}{n_q}=\frac{2q+3}{2q+1}=\frac{896}{640}=\frac{7}{5}\)

∴ 14q +7 = 10q + 15

∴ 4q = 8, ∴   q=2

Therefore, the three given frequencies correspond to the second, third and fourth overtones, i.e., the fifth, seventh and ninth harmonics, respectively.

∴ 5n_c = 640  ∴ n_c = 128 Hz

Given : For the tube open at both the ends,

n = 300 Hz and v = 340 m / s ignoring end correction, the fundamental frequency of the tube is

n = v/2L

∴ L =  v/2n = \(\frac{340}{2×300}\) = 0.5667 m

The length of the tube open at both the ends is 0.5667 m. -

Given : Open pipe, L= 25 cm = 0.25 m, v = 330 m/s

The fundamental frequency of an open pipe ignoring end correction,

n_O = v/λ = v/2L

∴ n_O = \(\frac{330}{2×0.25}\)  = 660 Hz

Since all harmonics are present as overtones, the first overtone is,

n₁ = 2n_O = 2 x 660 = 1320 Hz

The second overtone is

n₂ = 3n_O = 3 x 660 = 1980 Hz

Given : Open pipe, n_o = 600 Hz, n_c 1 = n_o 1 (first overtones)

For an open pipe, the fundamental frequency,

n_O = v/2Lo

 The length of the open pipe is

Lo = v /2n_O = \(\frac{330}{2×600}\) = 0.275 m

For the open pipe, the frequency of the first overtone is

2n_O = 2 x 600 = 1200 Hz

For the pipe closed at one end, the frequency of the first overtone is 3v/Lo

By the data 3v/4L = 1200

∴ Lc = \(\frac{3×330}{4×1200}\)  = 0.206 m

The length of the pipe open at both ends is 27.5 cm and the length of the pipe closed at one end is 20.6 cm.

Given : L = 1 m, n = 15 Hz.

The string is fixed only at one end. Hence, an antinode will be formed at the free end. Thus, with four and half loops on the string, the length of the string is

L = \(\frac{λ}{4}+4\frac{λ}{2}=\frac{9}{4}λ\)

∴ λ = \(\frac{4L}{9}\) =\(\frac{4}{9}×1\)

v = n λ

∴ Speed of the progressive wave

v =15 x\(\frac{4}{9}\)= 6.667 m/s

Given : n = 440 Hz

The first overtone, n₁ = 2n = 2 x 400 = 880 Hz

The second overtone, n₂ = 3n = 3 x 400 = 1320 Hz

Given : n₈ = 2 n₁, beat frequency = 4 Hz

The set of tuning fork is arranged in the increasing order of their frequencies.

∴ n₂ = n₁ + 4

n₃ = n₂ + 4 = n₁+ 2 x 4

n₄ = n₃ + 4 = n₁ +3 x 4

∴ n₈ = n₇ + 4 = n₁ + 7 x 4 = n₁ + 28

Since n₈ =2n₁,

2n₁ = n₁ + 28

 The frequency of the first fork, n₁ = 28 Hz

 The frequency of the last fork,

n₈ = n₁ + 28 = 28 + 28 = 56Hz

Given : T₁ = 40 N, n₁ = 384 Hz, T₂ = 40 − 1.24 = 38.76 N

n₁ = \(\frac{1}{2l}\sqrt{\frac{T_1}{m}}\) and n₂ = \(\frac{1}{2l}\sqrt{\frac{T_2}{m}}\)

∴ \(\frac{n_2}{n_1}\sqrt{\frac{T_2}{T_1}}\)

∴ \(\frac{n_2}{384}\sqrt{\frac{38.76}{40}}\)

∴ n₂ = 384 x 0.9844 = 378.0 Hz

 n₁ − n₂ = 384 − 378 = 6 Hz

 ∴ The number of beats produced in two seconds = 2 x 6 =12

Given : L= 0.5 m, T = 5kg = 5 x 9.8 = 49N, n = 100 Hz

n = \(\frac{1}{2L}\sqrt{\frac{T}{m}}\)

∴ Linear density, m = \(\frac{T}{4L^2n^2}\) = \(\frac{49}{4(0.5)^2(100)^2}\)= 4.9 x 10⁻³ kg/m

Given: L₁ = 80 cm n₁ = 112 Hz, n₂ =160 Hz

According to the law of length, n₁L₁ = n₂L₂.

 The vibrating length to produce the fundamental frequency of 160 Hz,

L₂ = \(\frac{n_1L_1}{n_2}\) = \(\frac{(112)(80)}{160}\)= 56 cm