Adjacent antinodes of a standing wave on a string are $15.0 c m$ apart. A particle at an antinode oscillates in simple harmonic motion with amplitude $0.650 c m$ and period $0.0750$ S . The string lies along the $+ x$ -axis and is fixed at $x = 0$ . Find the displacement of a point on the string as a function of position and time.

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$y (x, t) = (0.85 c m) \sin (\frac{πx}{30 c m}) \sin (\frac{πt}{0.075 s})$

$y (x, t) = (0.65 c m) \sin (\frac{2 πx}{30 c m}) \sin (\frac{2 πt}{0.075 s})$

$y (x, t) = (0.85 c m) \sin (\frac{2 πx}{20 c m}) \sin (\frac{πt}{0.075 s})$

$y (x, t) = (0.55 c m) \sin (\frac{2 πx}{10 c m}) \sin (\frac{2 πt}{0.075 s})$

answer is C.

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Detailed Solution

$\frac{λ}{2} = 15 c m$

$∴ λ = 30 c m$

$k = \frac{2 π}{λ} = (\frac{x}{15}) s^{- 1}$

$ω = \frac{2 π}{T} = \frac{2 π}{0.075} s^{- 1}$

Since, $x = 0$ is a node we will write sin equation.

$A_{s} = A_{m a x} \sin k x$ …(i)

$y = \sin ω t$

$∴$ $y (x, t) = (0.85 c m) \sin (\frac{2 πx}{30 c m}) \sin (\frac{2 πt}{0.075 s})$

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