A wave represented by a given equation $y (x, t) = a sin (ω t - k x)$ superimposes on another wave giving a stationary wave having antinode at $x = 0$ then the equation of the another wave is

y = - a sin (ω t - k x)

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y = a sin (ω t + k x)

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y = - a sin (ω t + k x)

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y = - a cos (ω t + k x)

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Solution

The correct option is B $y = a sin (ω t + k x)$
$\begin{matrix} a sin (w t - k x) + y_{1} = 2 a sin cot cos k x \Rightarrow y = a sin (w t + k x) A n s . (B) \end{matrix}$

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Similar questions

y = a cos (ω t - k x)

x = 0

y = A sin (ω t - k x)

y = A sin (ω t + k x)

Which of the following two wave equations can form stationary wave?

(i) $Y = A s i n (ω t - k x)$

(ii) $Y = A c o s (ω t - k x)$

(iii) $Y = A s i n (ω t + k x)$

(iv) $Y = A c o s (ω t + k x)$

y = a s i n (ω t - k x)

y = a c o s (ω t - k x)

y = a \sin (ω t - k x)

y = a \cos (ω t - k x)

\sqrt{2} a

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