For a certain transverse standing wave on a long string, an antinode is formed at $x = 0$ and next to it, a nearest node is formed at $x = 0.10 m .$ The displacement $y (t)$ of the string particle at $x = 0$ is shown in figure.

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Transverse displacement of the particle at $x = 0.05 m a n d t = 0.05 s$ is $- 2 \sqrt{2} c . m .$

Transverse displacement of the particle at $x = 0.04 m a n d t = 0.025 s$ is $- 2 \sqrt{2} c . m .$

The transverse velocity of the string particle at $x = \frac{1}{15} m$ and $t = 0.1 s$ is $20 π c m / s$

Speed of the travelling waves that interfere to produce this standing wave is $2 m / s .$

answer is A, C, D.

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Detailed Solution

$\frac{λ}{4} = 0.1 \Rightarrow λ = 0.4 m$
from graph $\Rightarrow T = 0.2 \sec .$ and amplitude of standing wave is $2 A = 4 c m .$
Equation of the standing wave
$y (x, t) = - 2 A \cos (\frac{2 π}{0.4} x) . \sin (\frac{2 π}{0.2} t) c m y (x = 0.05, t = 0.05) = - 2 \sqrt{2} c m y (x = 0.04, t = 0.025) = - 2 \sqrt{2} \cos 36 ° speed = \frac{λ}{T} = 2 m/sec. V_{y} = \frac{d y}{d t} = - 2 A \times \frac{2 π}{0.2} \cos (\frac{2 π x}{0.4}) . \cos (\frac{2 π t}{0.2}) Vy = (x = \frac{1}{15} m, t = 0.1) = 20 π cm/sec.$