Question

A wave $y=a\sin (\omega t-kx)$ on a string meets with another wave producing a node at $x=0$. Then the equation of the unknown wave is:

• (A) $y=a\sin (\omega t+kx)$
• (B) $y=-a\sin (\omega t+kx)$
• (C) $y=a\sin (\omega t-kx)$
• (D) $y=-a\sin (\omega t-kx)$

Answer 1

The Correct Option is B

Explanation:
Given:Equation of wave, $y=a\sin (\omega t-kx)$......(i)Since the coefficient of $x$ is negative in the above wave equation, it suggests that the wave is traveling in a positive $x$ direction. The other wave must travel in the opposite direction to form a node.Let the two possible equations of waves travelling in negative direction be$y_1=a\sin (\omega t+kx)$......(ii)and$y_2=-a\sin (\omega t+kx)$.......(iii)Using the property$\sin (A+B)+\sin (A-B)=2\sin A\cos B$If waves (i) and (ii) superimpose, the net displacement is$y_3=(y+y_1)=2a\sin (\omega t)\cos (kx)$Similarly,$\sin (A-B)-\sin (A+B)=-2\sin B\cos A$So, if waves (i) and (iii) superimpose the net displacement is$y_3=(y+y_2)=-2a\sin (kx)\cos (\omega t)$Since at node, displacement $y_3=0$ so$y_3=0=-2a\sin (kx)\cos (\omega t)$$\Rightarrow \sin (kx)=0$ $\Rightarrow x=0$Therefore, eq. (iii) satisfies the property of a node at $x=0$.So, the equation of the unknown wave is$y=-a\sin (\omega t+kx)$Hence, the correct option is (B).