A standing wave is formed when two identical waves travelling in the opposite directions along the same line, interfere. On the path of a stationary wave, the amplitude of vibration varies simple harmonically with respect to the distance from one end of the path.
On the path of the stationary wave, there are points where the amplitude is zero, they are known as NODES. On the other hand there are points where the amplitude is maximum, they are known as ANTINODES.
The distance between two consecutive nodes or two consecutive anitnodes is \(\frac{\lambda }{2}\)
The distance between a node and the next antinode is \(\frac{\lambda }{4}\)
Consider two waves of the same frequency, speed and amplitude, which are travelling in opposite directions along a string. Two such waves may be represented by the equations
y1 = a sin (kx - wt ) and
y2 = a sin (kx + wt)
Hence the resultant may be written as
y = y1 + y2 = a sin (kx - wt) + a sin (kx + wt)
y = 2 a sin kx cos wt
This is the equation of a standing wave.
Note:
(i) In this equation, it is seen that a particle at any particular point ‘x’ executes simple harmonic motion and all particles vibrate with the same frequency.
(ii) The amplitude is not the same for different particles but varies with the location ‘x’ of the particle.
(iii) The points having maximum amplitudes are those for which 2asinkx, has a maximum value of 2a, these are at the positions,
kx = p/2, 3p/2, 5p/2, ..............
or x = l/4, 3l/4, 5l/4, ...............
These points are called antinodes.
(iv) The amplitude has minimumn value of zero at positions where
kx = p, 2p, 3p,...........
or x = l/2, l, 3l/2 ,2l............
These points are called nodes.
(v) Energy is not transported along the string to the right or to the left, because energy can not flow past the nodal points in the string which are permanently at rest.
Lowest frequency standing wave: fundamental
The lowest standing wave frequency is called the fundamental or first harmonic. For this mode, all parts of the string vibrate together, up and down. Of course, the ends of the string are fixed in place and are not free to move. We call these positions nodes: a node is a point on the string that does not move. As we move along the string, the amplitude of oscillations at each position we look at changes, but the frequency of oscillation is the same. Near a node, the oscillation amplitude is very small.
In the middle of the string, the oscillation amplitude is largest; such a position is defined as an antinode. We assign a wavelength to the fundamental (and each higher harmonic discussed below) standing wave. At a fixed moment in time, all we observe is either a crest or a trough, but we never observe both at the same time for the lowest frequency standing wave. From this, we determine that half a standing wave length fits along the length of the string for the fundamental. Alternatively, we say that the wavelength of the fundamental is twice the length of the string, or
1. The amplitude of vibration of any particle in a standing wave, produced along a stretched string depends on
Solution:
location of particle
2. Assertion (A): A standing wave is produced in a string clamped at one end and free at the other end. The length of string is odd multiples of \(\frac{\lambda }{4}\)
Reason (R): Distance between successive node and antinode is \({}^{\lambda }/{}_{4}\)
Solution:
Both assertion and reason are true and reason is the correct explanation of assertion.
1. Standing waves are produced in 10m long stretched wire. If the wire vibrates in 5 segments and wave velocity is 20m/s , then the frequency is
Solution:
\(\nu =\frac{PV}{2l}\)
= 5 Hz
2. Transverse waves are generated in two uniform wires A and B of the same material by attaching their free ends to a vibrating source of frequency 200Hz. The area of cross section of A is half that of B while tension on A is twice than on B. The ratio of wavelengths of transverse waves in A and B is.
Solution:
\(\frac{{{V}_{1}}}{{{V}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}X\frac{{{A}_{2}}\,{{\rho }_{2}}}{{{A}_{1}}\,{{\rho }_{1}}}}\)
= 2:1
1. The equations of two sound waves are given by y1 = 3 sin 100 p t, and y2 = 4 sin 150 p t. The ratio of intensities of sound produced in the medium.
Solution:
Here \({{a}_{1}}=3,\ {{n}_{1}}=\frac{100\pi }{2\pi }=50\)
\ \({{I}_{2}}\propto {{(3)}^{2}}\times {{(50)}^{2}}\)
Again, a2 = 4 and \({{n}_{2}}=\frac{150\pi }{2\pi }=75\)
\({{I}_{1}}\propto {{(4)}^{2}}\times {{(75)}^{2}}\)
\ \(\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{(3)}^{2}}\times {{(50)}^{2}}}{{{(4)}^{2}}\times {{(75)}^{2}}}\)
= 1 : 4.
2. A weight of 10 kg suspended form lower end of a uniform steel wire of area of cross-section 0.0045 sq. cm. Find out the ratio of the fundamental frequencies of the wire in the cases, when it is rubbed with a resigned cloth and when it is plucked in the middle. Young’s modulus of the material is 19.6 x 1011dyne/cm2.
Solution:
The frequency of the longitudinal vibration when it is rubbed with resined cloth is given by
n1 = \(\frac{1}{2l}\,\,\sqrt{\frac{Y}{\rho }}\)
Where l and r are the length of the wire and density of the material. When the wire is plucked, frequency is given by
n2 = \(\frac{1}{2l}\,\,\sqrt{\frac{T}{\mu }}\)
Where m is the mass per unit length.
From these two equations
\(\frac{{{n}_{1}}}{{{n}_{2}}}\,\,\,=\,\,\,\,\sqrt{\frac{Y}{T}\frac{\mu }{\rho }}\,\,=\,\,\sqrt{\frac{Y}{Mg}\frac{a\rho }{\rho }}\)
Where a is the area of cross-section.
\(\frac{{{n}_{1}}}{{{n}_{2}}}\,\,\,=\,\,\,\,\sqrt{\frac{Ya}{Mg}}\,\,=\,\,\sqrt{\frac{19.6\,x\,{{10}^{11}}\,x\,0.0045}{{{10}^{4}}\,x\,980}}\) = 30
1. The following equation represents transverse wave
Z1 = A Cos (kx – w t) ; Z2 = A Cos (kx + w t) ; Z3 = A Cos (ky – w t)
Identify the combination (s) of the waves which will produce "Standing wave(s)"
Solution:
In case of standing wave, two component wave must travel along the same straight line in opposite directions.
Such two waves are
Z1 = A Cos (kx-wt) and Z2 = A Cos (kx+wt)
The resultant wave Z = Z1 + Z2
Z = A Cos (kx – wt) + A Cos (kx + wt)
= 2 A Cos kx. Cos wt.
The resultant amplitude of standing wave
Aresultant = 2 A Cos kx
I = 4A2 I Cos2 kx
Intensity is zero when
Cos2 kx = 0, kx = \(\frac{\pi }{2},\ \ \frac{3\pi }{2},---\frac{(2n+1)\ \pi }{2}\)
i.e. x = \(\frac{\pi }{2k},\ \ \frac{3\pi }{2k},\ \ \frac{5\pi }{2k},----\frac{(2n+1)\ \pi }{2k}\)
2. The following equation represents transverse wave
Z1 = A Cos (kx – w t) ; Z2 = A Cos (kx + w t) ; Z3 = A Cos (ky – w t)
Identify the combination (s) of the waves which will produce "A wave travelling in the direction making an angle of 45° with positive x and positive y axes."
Solution:
For a wave propagating in a direction making an angle 45° with positive X and Y axis, the component waves must be of equal amplitude, one traveling along x-axis and the other along Y axis. So component waves are
Z1 = A Cos (kx – wt)
Z2 = A Cos (ky – wt)
The resultant wave Z = Z1 + Z2
Z = 2 A Cos \(\frac{k(x-y)}{2}\ \ \operatorname{Cos}\ \left[ k\left( \frac{x+y}{2} \right)\ -\omega t \right]\)
The amplitude of resultant disturbance is
Aresultant = 2 A Cos \(\frac{k(x-y)}{2}\)
The intensity is zero at position where
Cos \(\frac{k(x-y)}{2}\ =\ 0\)
Therefore (x-y)/2 = \(\frac{(2n+1)\ \pi }{2k}\) Þ x - y = \(\frac{(2n+1)\pi }{2}\)
Where n = 0, 1, 2….
A standing wave is formed when two identical waves travelling in the opposite directions along the same line, interfere. On the path of a stationary wave, the amplitude of vibration varies simple harmonically with respect to the distance from one end of the path.
On the path of the stationary wave, there are points where the amplitude is zero, they are known as NODES. On the other hand there are points where the amplitude is maximum, they are known as ANTINODES.
The distance between two consecutive nodes or two consecutive anitnodes is \(\frac{\lambda }{2}\)
The distance between a node and the next antinode is \(\frac{\lambda }{4}\)
Consider two waves of the same frequency, speed and amplitude, which are travelling in opposite directions along a string. Two such waves may be represented by the equations
y1 = a sin (kx - wt ) and
y2 = a sin (kx + wt)
Hence the resultant may be written as
y = y1 + y2 = a sin (kx - wt) + a sin (kx + wt)
y = 2 a sin kx cos wt
This is the equation of a standing wave.
Note:
(i) In this equation, it is seen that a particle at any particular point ‘x’ executes simple harmonic motion and all particles vibrate with the same frequency.
(ii) The amplitude is not the same for different particles but varies with the location ‘x’ of the particle.
(iii) The points having maximum amplitudes are those for which 2asinkx, has a maximum value of 2a, these are at the positions,
kx = p/2, 3p/2, 5p/2, ..............
or x = l/4, 3l/4, 5l/4, ...............
These points are called antinodes.
(iv) The amplitude has minimumn value of zero at positions where
kx = p, 2p, 3p,...........
or x = l/2, l, 3l/2 ,2l............
These points are called nodes.
(v) Energy is not transported along the string to the right or to the left, because energy can not flow past the nodal points in the string which are permanently at rest.
Lowest frequency standing wave: fundamental
The lowest standing wave frequency is called the fundamental or first harmonic. For this mode, all parts of the string vibrate together, up and down. Of course, the ends of the string are fixed in place and are not free to move. We call these positions nodes: a node is a point on the string that does not move. As we move along the string, the amplitude of oscillations at each position we look at changes, but the frequency of oscillation is the same. Near a node, the oscillation amplitude is very small.
In the middle of the string, the oscillation amplitude is largest; such a position is defined as an antinode. We assign a wavelength to the fundamental (and each higher harmonic discussed below) standing wave. At a fixed moment in time, all we observe is either a crest or a trough, but we never observe both at the same time for the lowest frequency standing wave. From this, we determine that half a standing wave length fits along the length of the string for the fundamental. Alternatively, we say that the wavelength of the fundamental is twice the length of the string, or
1. The amplitude of vibration of any particle in a standing wave, produced along a stretched string depends on
Solution:
location of particle
2. Assertion (A): A standing wave is produced in a string clamped at one end and free at the other end. The length of string is odd multiples of \(\frac{\lambda }{4}\)
Reason (R): Distance between successive node and antinode is \({}^{\lambda }/{}_{4}\)
Solution:
Both assertion and reason are true and reason is the correct explanation of assertion.
1. Standing waves are produced in 10m long stretched wire. If the wire vibrates in 5 segments and wave velocity is 20m/s , then the frequency is
Solution:
\(\nu =\frac{PV}{2l}\)
= 5 Hz
2. Transverse waves are generated in two uniform wires A and B of the same material by attaching their free ends to a vibrating source of frequency 200Hz. The area of cross section of A is half that of B while tension on A is twice than on B. The ratio of wavelengths of transverse waves in A and B is.
Solution:
\(\frac{{{V}_{1}}}{{{V}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}X\frac{{{A}_{2}}\,{{\rho }_{2}}}{{{A}_{1}}\,{{\rho }_{1}}}}\)
= 2:1
1. The equations of two sound waves are given by y1 = 3 sin 100 p t, and y2 = 4 sin 150 p t. The ratio of intensities of sound produced in the medium.
Solution:
Here \({{a}_{1}}=3,\ {{n}_{1}}=\frac{100\pi }{2\pi }=50\)
\ \({{I}_{2}}\propto {{(3)}^{2}}\times {{(50)}^{2}}\)
Again, a2 = 4 and \({{n}_{2}}=\frac{150\pi }{2\pi }=75\)
\({{I}_{1}}\propto {{(4)}^{2}}\times {{(75)}^{2}}\)
\ \(\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{(3)}^{2}}\times {{(50)}^{2}}}{{{(4)}^{2}}\times {{(75)}^{2}}}\)
= 1 : 4.
2. A weight of 10 kg suspended form lower end of a uniform steel wire of area of cross-section 0.0045 sq. cm. Find out the ratio of the fundamental frequencies of the wire in the cases, when it is rubbed with a resigned cloth and when it is plucked in the middle. Young’s modulus of the material is 19.6 x 1011dyne/cm2.
Solution:
The frequency of the longitudinal vibration when it is rubbed with resined cloth is given by
n1 = \(\frac{1}{2l}\,\,\sqrt{\frac{Y}{\rho }}\)
Where l and r are the length of the wire and density of the material. When the wire is plucked, frequency is given by
n2 = \(\frac{1}{2l}\,\,\sqrt{\frac{T}{\mu }}\)
Where m is the mass per unit length.
From these two equations
\(\frac{{{n}_{1}}}{{{n}_{2}}}\,\,\,=\,\,\,\,\sqrt{\frac{Y}{T}\frac{\mu }{\rho }}\,\,=\,\,\sqrt{\frac{Y}{Mg}\frac{a\rho }{\rho }}\)
Where a is the area of cross-section.
\(\frac{{{n}_{1}}}{{{n}_{2}}}\,\,\,=\,\,\,\,\sqrt{\frac{Ya}{Mg}}\,\,=\,\,\sqrt{\frac{19.6\,x\,{{10}^{11}}\,x\,0.0045}{{{10}^{4}}\,x\,980}}\) = 30
1. The following equation represents transverse wave
Z1 = A Cos (kx – w t) ; Z2 = A Cos (kx + w t) ; Z3 = A Cos (ky – w t)
Identify the combination (s) of the waves which will produce "Standing wave(s)"
Solution:
In case of standing wave, two component wave must travel along the same straight line in opposite directions.
Such two waves are
Z1 = A Cos (kx-wt) and Z2 = A Cos (kx+wt)
The resultant wave Z = Z1 + Z2
Z = A Cos (kx – wt) + A Cos (kx + wt)
= 2 A Cos kx. Cos wt.
The resultant amplitude of standing wave
Aresultant = 2 A Cos kx
I = 4A2 I Cos2 kx
Intensity is zero when
Cos2 kx = 0, kx = \(\frac{\pi }{2},\ \ \frac{3\pi }{2},---\frac{(2n+1)\ \pi }{2}\)
i.e. x = \(\frac{\pi }{2k},\ \ \frac{3\pi }{2k},\ \ \frac{5\pi }{2k},----\frac{(2n+1)\ \pi }{2k}\)
2. The following equation represents transverse wave
Z1 = A Cos (kx – w t) ; Z2 = A Cos (kx + w t) ; Z3 = A Cos (ky – w t)
Identify the combination (s) of the waves which will produce "A wave travelling in the direction making an angle of 45° with positive x and positive y axes."
Solution:
For a wave propagating in a direction making an angle 45° with positive X and Y axis, the component waves must be of equal amplitude, one traveling along x-axis and the other along Y axis. So component waves are
Z1 = A Cos (kx – wt)
Z2 = A Cos (ky – wt)
The resultant wave Z = Z1 + Z2
Z = 2 A Cos \(\frac{k(x-y)}{2}\ \ \operatorname{Cos}\ \left[ k\left( \frac{x+y}{2} \right)\ -\omega t \right]\)
The amplitude of resultant disturbance is
Aresultant = 2 A Cos \(\frac{k(x-y)}{2}\)
The intensity is zero at position where
Cos \(\frac{k(x-y)}{2}\ =\ 0\)
Therefore (x-y)/2 = \(\frac{(2n+1)\ \pi }{2k}\) Þ x - y = \(\frac{(2n+1)\pi }{2}\)
Where n = 0, 1, 2….