Question

5. To make a stationary wave having node at x = 0,
the equation y = Acos(wt + kx) is superimposed to
another wave of equation
(1)-Acos(kx – wt) (2) -Asin(kx – wt)
(3) Acos(wt - kx) (4) Acos(kx + @t)
dor identical​

Answer 1

To make a stationary wave having node at x = 0,  the equation  y = Acos(ωt + kx) is superimposed to another wave of equation   -Acos(kx – ωt)

Option (1) is correct.

Explanation:

The given wave

$y=A\cos(\omega t+kx)$

It can be written as

$y=A[\cos \omega t.\cos kx-\sin \omega t.\sin kx]$

$\implies y=A\cos \omega t.\cos kx-A\sin \omega t.\sin kx$

If the superimposed wave has a node at x = 0 then the term containing cos kx should be zero

i.e. $A\cos\omega t\cos kx$ should be zero

Thus, the other wave can have the equation of the form

$y=-A\cos \omega t.\cos kx-A\sin \omega t.\sin kx$

or, $y=-A\cos(\omega t-kx)$

Therefore, option (1) is correct.

Hope this answer is helpful.

Know More:

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