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WAVES
A wave which travels continuously in a medium in the same direction without any change in its amplitude is called a progressive wave or a traveling wave.
A progressive wave may be transverse or longitudinal in nature. Suppose that a plane simple harmonic wave travels from origin O along the positive direction of X-axis.
As we know wave motion is accompanied by the transmission of energy which is communicated by the source to the medium. Energy distribution of the wave along the direction of propagation is not uniform. However energy density of the medium is independent of the position as well as time.
Consider an elastic medium and a wave
y = A sin 2p/l(vt – x)
Consider an elementary oscillating layer of the medium of thickness dx. If r be the average density of the medium. Then mass of the particles contained in the layer is m = rdx, for unit area of cross-section.
Now kinetic energy of the layer = \(\frac{1}{2}\,m{{u}^{2}}\,=\,\frac{1}{2}\,\rho \,dx\,{{\left( \frac{dy}{dt} \right)}^{2}}\,\)
= \(\frac{1}{2}{{\left( \frac{2\pi \,vA}{\lambda } \right)}^{2}}\,\rho \,{{\cos }^{2}}\,\frac{2\pi }{\lambda }\,\left( vt-x \right)dx\) (I)
Also the force acting on the layer
F = ma = \(\left( \rho dx \right)\frac{{{d}^{2}}y}{d{{t}^{2}}}\)
The work done for a displacement dy of layer against this force will be dW = Fdy
Then potential energy = - dW = - Fdy
Hence
P.E. = \(-\int_{0}^{y}{F.dy}\,\,=\,\,\int_{0}^{y}{{{\left( {}^{2\,\pi \,v}/{}_{\lambda } \right)}^{2}}\,\rho dx\,\,ydy}\)
= \(\frac{\rho }{2}\,{{\left( \frac{2\pi v}{\lambda } \right)}^{2}}\,\,{{y}^{2}}dx\)
= \(\frac{2{{\pi }^{2}}\,{{v}^{2}}\,{{A}^{2}}\,\rho }{\,{{\lambda }^{2}}}\,{{\sin }^{2}}\frac{2\pi }{\lambda }\left( vt-x \right)\,dx\) (II)
Hence total energy of the layer
E = K.E. + P.E.
= \(\frac{1}{2}{{\left( \frac{2\pi \,vA}{\lambda } \right)}^{2}}\,\rho \,{{\cos }^{2}}\,\frac{2\pi }{\lambda }\,\left( vt-x \right)dx\) + \(\frac{2{{\pi }^{2}}\,{{v}^{2}}\,{{A}^{2}}\,\rho }{\,{{\lambda }^{2}}}\,{{\sin }^{2}}\frac{2\pi }{\lambda }\left( vt-x \right)\,dx\)
= \(\frac{2{{\pi }^{2}}\,{{v}^{2}}\,{{A}^{2}}\,\rho }{\,{{\lambda }^{2}}}\,\,dx\)
The total energy per unit length = \(\frac{2{{\pi }^{2}}\,{{v}^{2}}\,{{A}^{2}}\,\rho }{\,{{\lambda }^{2}}}\,\) = 2p2rA2f2
Amplitude :
Amplitude is the magnitude of maximum displacement of a particle in a wave from the equilibrium position.
The image above shows the positive and negative amplitude in the case of a sinusoidal wave. But as we consider the magnitude of the displacement, the amplitude is always a positive quantity.
Phase :
The argument (kx – ωt + φ) of the oscillatory term sin (kx – ωt + φ) is defined as the phase of the function. It describes the state of motion of the wave. Points on a wave which travel in the same direction, rising a falling together, are said to be in phase with each other. Points on a wave which travel in opposite directions to each other, such that, one is rising while the other is falling, are said to be in anti-phase with each other.
Wavelength :
Wavelength (λ) is the distance between two identical points such as a crest or a trough on a wave parallel to the direction of propagation of the wave. It can also be defined as the distance over which the wave shape repeats itself. It is measured in metres (m).
1. A plane progressive wave cannot be represented by
Solution:
y - Alogx + Blogx
2. The phase difference between the particle at one compression and another particle in third compression is
Solution:
4
radians 3. The phase change between incident and reflected sound wave from a free end is
Solution:
0
1. A wire of uniform cross-section is stretched between two points 100 cm apart. The wire fixed at one end and a weight is hung over a pulley at the other end. A weight of 9 kg produces a fundamental frequency of 750. What is the velocity of wave in wire ?
Solution:
L = 100 cm, f1 = 750 Hz.
v1 = 2Lf1 = 2 ´ 100 ´ 750 = 150000 cms-1 = 1500 ms-1
2. A wire of uniform cross-section is stretched between two points 100 cm apart. The wire fixed at one end and a weight is hung over a pulley at the other end. A weight of 9 kg produces a fundamental frequency of 750. If the weight is reduced to 4 kg, what is the velocity of wave ? What is the wavelength and frequency of wave ?
Solution:
\({{v}_{1}}=\sqrt{\frac{{{T}_{1}}}{m}}and{{v}_{2}}=\sqrt{\frac{{{T}_{2}}}{m}}\)
\(\frac{{{v}_{2}}}{{{v}_{1}}}=\sqrt{\frac{{{T}_{2}}}{{{T}_{1}}}}\Rightarrow \frac{{{v}_{2}}}{1500}=\sqrt{\frac{4}{9}}\)
\ v2 = 1000 m s-1
l2 = wave length = 2L = 200 cm = 2 m
f2 = \(\frac{v}{{{\lambda }_{2}}}=\frac{1000}{2}=500\,\,Hz\)
1. Two simple harmonic motions are represented by the equations \({{y}_{1}}=0.1\sin \left( 100\pi t+\frac{\pi }{3} \right)\) and \({{y}_{2}}=0.1\cos \pi t.\) The phase difference of the velocity of particle 1 with respect to velocity of particle 2 at t=0 is
Solution:
\({{V}_{1}}=\frac{d{{y}_{1}}}{dt}=10\pi \sin \left[ 100\pi t+\frac{\pi }{3}+\frac{\pi }{2} \right]\)
\({{V}_{2}}=\frac{d{{y}_{2}}}{dt}=0.1\pi \sin \left[ \pi t+\pi \right]\)
\(\Delta \theta ={{\theta }_{1}}-{{\theta }_{2}}\) \(=\left( 100\pi t+\frac{\pi }{3}+\frac{\pi }{2} \right)-\left( \pi t+\pi \right)\)
Put t = 0 to get the answer
2. The amplitude of a wave disturbance propagating in the positive x direction is given by at in metres and at where x, y are in metres. If the shape of the wave does not changing with time, the velocity of the wave is
Solution:
At \(t=0,\,\,\,x=\sqrt{\frac{1-y}{y}}\)
At \(t=2,\,\,x=1+\sqrt{\frac{1-y}{y}}\)
\(\therefore \) Velocity \(V=\frac{{{x}_{2}}-{{x}_{1}}}{{{t}_{2}}-{{t}_{1}}}\)
1. A wave of angular frequency \(\omega \) propagates so that a certain phase of oscillation moves along x-axis, y-axis and z-axis with speeds \({{c}_{1}},\,\,{{c}_{2}}\) and \({{c}_{3}}\) respectively. The propagation constant k is
Solution:
\(\overline{k}={{k}_{x}}\widehat{i}+{{k}_{y}}\widehat{j}+{{k}_{z}}\widehat{k}=\frac{\omega }{{{c}_{1}}}\widehat{i}+\frac{\omega }{{{c}_{2}}}\widehat{j}+\frac{\omega }{{{c}_{3}}}\widehat{k}\)
2. Two sound waves are represented by \({{y}_{1}}=\sin \omega t+\cos \omega t\) and \({{y}_{2}}=\frac{\sqrt{3}}{2}\sin \omega t+\frac{1}{2}\cos \omega t\). The ratio of their amplitudes is
Solution:
\({{y}_{1}}=\sin \omega t+\cos \omega t=\sqrt{2}\sin \left( \omega t+\frac{\pi }{4} \right)\)
\({{y}_{2}}=\frac{\sqrt{3}}{2}\sin \omega t+\frac{1}{2}\cos \omega t=\sin \left( \omega t+\frac{\pi }{6} \right)\)
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WAVES
A wave which travels continuously in a medium in the same direction without any change in its amplitude is called a progressive wave or a traveling wave.
A progressive wave may be transverse or longitudinal in nature. Suppose that a plane simple harmonic wave travels from origin O along the positive direction of X-axis.
As we know wave motion is accompanied by the transmission of energy which is communicated by the source to the medium. Energy distribution of the wave along the direction of propagation is not uniform. However energy density of the medium is independent of the position as well as time.
Consider an elastic medium and a wave
y = A sin 2p/l(vt – x)
Consider an elementary oscillating layer of the medium of thickness dx. If r be the average density of the medium. Then mass of the particles contained in the layer is m = rdx, for unit area of cross-section.
Now kinetic energy of the layer = \(\frac{1}{2}\,m{{u}^{2}}\,=\,\frac{1}{2}\,\rho \,dx\,{{\left( \frac{dy}{dt} \right)}^{2}}\,\)
= \(\frac{1}{2}{{\left( \frac{2\pi \,vA}{\lambda } \right)}^{2}}\,\rho \,{{\cos }^{2}}\,\frac{2\pi }{\lambda }\,\left( vt-x \right)dx\) (I)
Also the force acting on the layer
F = ma = \(\left( \rho dx \right)\frac{{{d}^{2}}y}{d{{t}^{2}}}\)
The work done for a displacement dy of layer against this force will be dW = Fdy
Then potential energy = - dW = - Fdy
Hence
P.E. = \(-\int_{0}^{y}{F.dy}\,\,=\,\,\int_{0}^{y}{{{\left( {}^{2\,\pi \,v}/{}_{\lambda } \right)}^{2}}\,\rho dx\,\,ydy}\)
= \(\frac{\rho }{2}\,{{\left( \frac{2\pi v}{\lambda } \right)}^{2}}\,\,{{y}^{2}}dx\)
= \(\frac{2{{\pi }^{2}}\,{{v}^{2}}\,{{A}^{2}}\,\rho }{\,{{\lambda }^{2}}}\,{{\sin }^{2}}\frac{2\pi }{\lambda }\left( vt-x \right)\,dx\) (II)
Hence total energy of the layer
E = K.E. + P.E.
= \(\frac{1}{2}{{\left( \frac{2\pi \,vA}{\lambda } \right)}^{2}}\,\rho \,{{\cos }^{2}}\,\frac{2\pi }{\lambda }\,\left( vt-x \right)dx\) + \(\frac{2{{\pi }^{2}}\,{{v}^{2}}\,{{A}^{2}}\,\rho }{\,{{\lambda }^{2}}}\,{{\sin }^{2}}\frac{2\pi }{\lambda }\left( vt-x \right)\,dx\)
= \(\frac{2{{\pi }^{2}}\,{{v}^{2}}\,{{A}^{2}}\,\rho }{\,{{\lambda }^{2}}}\,\,dx\)
The total energy per unit length = \(\frac{2{{\pi }^{2}}\,{{v}^{2}}\,{{A}^{2}}\,\rho }{\,{{\lambda }^{2}}}\,\) = 2p2rA2f2
Amplitude :
Amplitude is the magnitude of maximum displacement of a particle in a wave from the equilibrium position.
The image above shows the positive and negative amplitude in the case of a sinusoidal wave. But as we consider the magnitude of the displacement, the amplitude is always a positive quantity.
Phase :
The argument (kx – ωt + φ) of the oscillatory term sin (kx – ωt + φ) is defined as the phase of the function. It describes the state of motion of the wave. Points on a wave which travel in the same direction, rising a falling together, are said to be in phase with each other. Points on a wave which travel in opposite directions to each other, such that, one is rising while the other is falling, are said to be in anti-phase with each other.
Wavelength :
Wavelength (λ) is the distance between two identical points such as a crest or a trough on a wave parallel to the direction of propagation of the wave. It can also be defined as the distance over which the wave shape repeats itself. It is measured in metres (m).
1. A plane progressive wave cannot be represented by
Solution:
y - Alogx + Blogx
2. The phase difference between the particle at one compression and another particle in third compression is
Solution:
4
radians 3. The phase change between incident and reflected sound wave from a free end is
Solution:
0
1. A wire of uniform cross-section is stretched between two points 100 cm apart. The wire fixed at one end and a weight is hung over a pulley at the other end. A weight of 9 kg produces a fundamental frequency of 750. What is the velocity of wave in wire ?
Solution:
L = 100 cm, f1 = 750 Hz.
v1 = 2Lf1 = 2 ´ 100 ´ 750 = 150000 cms-1 = 1500 ms-1
2. A wire of uniform cross-section is stretched between two points 100 cm apart. The wire fixed at one end and a weight is hung over a pulley at the other end. A weight of 9 kg produces a fundamental frequency of 750. If the weight is reduced to 4 kg, what is the velocity of wave ? What is the wavelength and frequency of wave ?
Solution:
\({{v}_{1}}=\sqrt{\frac{{{T}_{1}}}{m}}and{{v}_{2}}=\sqrt{\frac{{{T}_{2}}}{m}}\)
\(\frac{{{v}_{2}}}{{{v}_{1}}}=\sqrt{\frac{{{T}_{2}}}{{{T}_{1}}}}\Rightarrow \frac{{{v}_{2}}}{1500}=\sqrt{\frac{4}{9}}\)
\ v2 = 1000 m s-1
l2 = wave length = 2L = 200 cm = 2 m
f2 = \(\frac{v}{{{\lambda }_{2}}}=\frac{1000}{2}=500\,\,Hz\)
1. Two simple harmonic motions are represented by the equations \({{y}_{1}}=0.1\sin \left( 100\pi t+\frac{\pi }{3} \right)\) and \({{y}_{2}}=0.1\cos \pi t.\) The phase difference of the velocity of particle 1 with respect to velocity of particle 2 at t=0 is
Solution:
\({{V}_{1}}=\frac{d{{y}_{1}}}{dt}=10\pi \sin \left[ 100\pi t+\frac{\pi }{3}+\frac{\pi }{2} \right]\)
\({{V}_{2}}=\frac{d{{y}_{2}}}{dt}=0.1\pi \sin \left[ \pi t+\pi \right]\)
\(\Delta \theta ={{\theta }_{1}}-{{\theta }_{2}}\) \(=\left( 100\pi t+\frac{\pi }{3}+\frac{\pi }{2} \right)-\left( \pi t+\pi \right)\)
Put t = 0 to get the answer
2. The amplitude of a wave disturbance propagating in the positive x direction is given by at in metres and at where x, y are in metres. If the shape of the wave does not changing with time, the velocity of the wave is
Solution:
At \(t=0,\,\,\,x=\sqrt{\frac{1-y}{y}}\)
At \(t=2,\,\,x=1+\sqrt{\frac{1-y}{y}}\)
\(\therefore \) Velocity \(V=\frac{{{x}_{2}}-{{x}_{1}}}{{{t}_{2}}-{{t}_{1}}}\)
1. A wave of angular frequency \(\omega \) propagates so that a certain phase of oscillation moves along x-axis, y-axis and z-axis with speeds \({{c}_{1}},\,\,{{c}_{2}}\) and \({{c}_{3}}\) respectively. The propagation constant k is
Solution:
\(\overline{k}={{k}_{x}}\widehat{i}+{{k}_{y}}\widehat{j}+{{k}_{z}}\widehat{k}=\frac{\omega }{{{c}_{1}}}\widehat{i}+\frac{\omega }{{{c}_{2}}}\widehat{j}+\frac{\omega }{{{c}_{3}}}\widehat{k}\)
2. Two sound waves are represented by \({{y}_{1}}=\sin \omega t+\cos \omega t\) and \({{y}_{2}}=\frac{\sqrt{3}}{2}\sin \omega t+\frac{1}{2}\cos \omega t\). The ratio of their amplitudes is
Solution:
\({{y}_{1}}=\sin \omega t+\cos \omega t=\sqrt{2}\sin \left( \omega t+\frac{\pi }{4} \right)\)
\({{y}_{2}}=\frac{\sqrt{3}}{2}\sin \omega t+\frac{1}{2}\cos \omega t=\sin \left( \omega t+\frac{\pi }{6} \right)\)