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Chapter 10: Problem 1467

A wave \(\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})\) on a string meets with another wave producing a node at \(\mathrm{x}=0 .\) Then the equation of the unknown wave is \(\ldots \ldots \ldots\) (A) \(y=a \sin (\omega t+k x)\) (B) \(\mathrm{y}=-\mathrm{a} \sin (\omega \mathrm{t}+\mathrm{kx})\) (C) \(\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})\) (D) \(\mathrm{y}=-\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})\)

Short Answer

Expert verified
The equation of the unknown wave is \(y = -a \sin(\omega t - kx)\) (option D).

Step by step solution

01

Superposition of waves

When two waves interfere, their amplitudes add together by the principle of superposition. Let the unknown wave be Y(x, t), then the resulting wave after the interference will be y(x, t) + Y(x, t). #Step 2: Apply condition of node at x = 0#
02

Node condition

If there is a node at x = 0, then the total displacement at this point must be zero. We can write this as y(0, t) + Y(0, t) = 0. #Step 3: Substitute x = 0 into the condition and solve for Y(0, t)#
03

Substitute and solve

Using the equation of the first wave, we can write y(0, t) = a sin(ωt - k * 0) = a sin(ωt). Substitute this into the node condition: a sin(ωt) + Y(0, t) = 0 ⇒ Y(0, t) = -a sin(ωt) #Step 4: Identify the matching wave equation#
04

Compare with the given options

Now we compare our result, Y(0, t) = -a sin(ωt), with the given options to find the most suitable option. (A) y=a sin(ωt + kx): The sine term is not inverted, so this is not the correct option. (B) y=-a sin(ωt + kx): The sine term is inverted, but the wave is traveling in the same direction as the first wave, so this is not the correct option. (C) y=a sin(ωt - kx): The equation is the same as the first wave, so this is not the correct option. (D) y=-a sin(ωt - kx): The sine term is inverted, and the wave travels in the same direction as the first wave, so this is the correct option. Thus, the equation of the unknown wave is \(y = -a \sin(\omega t - kx)\) (option D).

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Most popular questions from this chapter

When two sound waves having amplitude A, angular frequency \(\omega\) and a phase difference of \(\pi / 2\) superposes, the maximum amplitude and angular frequency of the resultant wave is \(\ldots \ldots \ldots \ldots\) (A) \(\sqrt{2} \mathrm{~A}, \omega\) (B) \((\mathrm{A} / \sqrt{2}),(\omega / 2)\) (C) \((\mathrm{A} / \sqrt{2}), \omega\) (D) \(\sqrt{2} \mathrm{~A},(\omega / 2)\)

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