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Question

Of the following, the equation of plane progressive wave is:


A
y=rsinωt
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B
y=rsin(ωtkx)
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C
y=arsin(ωtkx)
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D
y=arsin(ωtkt)
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Solution

The correct option is B $$ y=r\sin(\omega t-kx) $$
For simple harmonic motion, the displacement at any instant of time is given by
$$y=A \sin \omega t$$ .......(1)

Where A is the amplitude and $$\omega$$ is the angular frequency of the wave. Consider a particle P at a distance x from the particle O on its right. Let the wave travel with a velocity v from left to right. Since it takes some time for the disturbance to reach P, its displacement can be written as 

$$y=A \sin (\omega t-\phi)$$ .......(2)

Where $$\phi$$ is the phase difference between the particles O and P.

We know that a path difference of $$\lambda$$ corresponds to a phase difference of $$2\pi$$ radians. Hence a path difference of x corresponds to a phase difference of 

$$\phi = \frac{2\pi}{\lambda}x$$. 

Now the propagation vector k is defined as $$k = \frac{2\pi}{\lambda}$$....(3)

Substituting equation (3) in equation (2)

We get, $$y=A\sin(\omega t-kx)$$


Physics

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