A wave equation represented by y = 0.5 $sin (ω t - 5 x)$ is super-imposed with another wave such that a stationary wave is formed and antinode is formed at $x = 0 .$ The equation for the second wave should be:

y = - 0.5 sin (ω t - 5 x)

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y = 0.5 cos (5 x + ω t)

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y = - 0.5 cos (ω t + 5 x)

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y = 0.5 sin (5 x + ω t)

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Solution

The correct option is B $y = 0.5 cos (5 x + ω t)$
$y = 0.5 sin (ω t - 5 x)$
let the equation of another wave be
$y_{2} = a_{2} sin (ω t - k x)$
$y = 0.5 (sin ω t cos 5 x - cos ω t sin 5 x) + a_{2} (sin ω t cos k x - cos ω t sin k x)$
$y = (0.5 + a_{2}) (sin ω t cos 5 x - cos ω t sin 5 x) + (sin ω t cos k x - cos ω t sin k x)$
$y = 1$ for antinode
$(0.5 + a_{2}) = 1$
$a_{2} = 0.5$
for argument of $y$ to be $1$ . at $x = 0$
$+ sin 5 x = + sin k x$
$R = 5$
also the phase difference should be $π$
so $y = 0.5 sin (ω t + 5 x)$

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