Explanation
Lets first find resultant of four waves using phase diagram
first wave is representated along$$+ve$$ X-axis with vector of length equal to amplitude of the wave.
resultant of those four can be written as (by adjusting X and Y component)
$$R=\sqrt { { 4 }^{ 2 }+{ 4 }^{ 2 } } =4\sqrt { 2 } $$
and $$\sin { \theta =\cfrac { 4 }{ 4\sqrt { 2 } } } =\cfrac { 1 }{ \sqrt { 2 } } \\ \therefore \theta =\cfrac { \pi }{ 4 } \\ \therefore { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 }=4\sqrt { 2 } \sin { \left( wt+kx+\cfrac { \pi }{ 4 } \right) } $$
now add fifth wave
$$\therefore y=\left( { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 } \right) +{ y }_{ 5 }\\ =4\sqrt { 2 } \sin { \left( wt+kx+\cfrac { \pi }{ 4 } \right) } +4\sqrt { 2 } \sin { \left( wt-kx+\cfrac { \pi }{ 4 } \right) } \\ =4\sqrt { 2 } \left[ 2\times \sin { \left( wt+\cfrac { \pi }{ 4 } \right) cos\left( kx \right) } \right] $$
By using $$\sin { C } +\sin { D } =2\sin { \cfrac { C+D }{ 2 } \cos { \cfrac { C+D }{ 2 } } } $$
$$\therefore y=8\sqrt { 2 } \sin { \left( wt+\cfrac { \pi }{ 4 } \right) } \cos { \left( kx \right) } $$
Travelling wave is the temporary wave that creates a disturbance and moves along the transmission line at a constant speed. These waves occur for a short duration, but cause a much disturbance in the line. The travelling waves can be represented mathematically in many ways.
The basic requirements for a wave function to represent a travelling wave is that for all values of x and t, wave function must have a finite value. Out of the given functions for y, no one satisfies this condition ( i.e. tan, log and inverse). Therefore, none can represent a travelling wave.
Thus the correct option is D.
At a particular position the velocity of a particle in SHM with amplitude a is $$\dfrac{{\sqrt 3 }}{2}$$ that at its mean position. In this position, its displacement is:
Given that,
The displacement is
$$ x=4\left( \cos \pi t+\sin \pi t \right) $$
$$ x=4\cos \pi t+4\sin \pi t $$
Let,
$$ R\cos \phi =4....(I) $$
$$ R\sin \phi =4.....(II) $$
Now,
$$ x=R\sin \phi \cos Pt+R\cos \phi \sin \pi t $$
$$ x=Rsin\left( \pi t+\phi \right) $$
So, R is the amplitude
Now, squaring and adding equation (I) and (II)
$$ R=\sqrt{{{4}^{2}}+{{4}^{2}}} $$
$$ R=4\sqrt{2} $$
Hence, the amplitude is $$4\sqrt{2}$$
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