A wave disturbance in a medium is described b | vedclass

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(D) The given equation is $y(x, t) = 0.02 \cos(10 \pi x) \cos(50 \pi t + \frac{\pi}{2})$.This represents a stationary wave of the form $y = A \cos(kx)

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All of the above.

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(D) The given equation is $y(x, t) = 0.02 \cos(10 \pi x) \cos(50 \pi t + \frac{\pi}{2})$.This represents a stationary wave of the form $y = A \cos(kx) \cos(\omega t + \phi)$,where $k = 10 \pi \ m^{-1}$.$1$. For nodes,the amplitude must be zero: $\cos(10 \pi x) = 0$.$10 \pi x = (2n + 1) \frac{\pi}{2} \Rightarrow x = \frac{2n + 1}{20} \ m$.For $n = 1$,$x = \frac{3}{20} = 0.15 \ m$. Thus,option $A$ is correct.$2$. For antinodes,the amplitude is maximum: $|\cos(10 \pi x)| = 1$.$10 \pi x = n \pi \Rightarrow x = \frac{n}{10} \ m$.For $n = 3$,$x = 0.3 \ m$. Thus,option $B$ is correct.$3$. The wavelength $\lambda$ is given by $k = \frac{2 \pi}{\lambda}$.$10 \pi = \frac{2 \pi}{\lambda} \Rightarrow \lambda = \frac{2 \pi}{10 \pi} = 0.2 \ m$. Thus,option $C$ is correct.Since all statements are correct,the answer is $D$.

$A$ wave disturbance in a medium is described by $y(x, t) = 0.02 \cos(50 \pi t + \frac{\pi}{2}) \cos(10 \pi x)$,where $x$ and $y$ are in metres and $t$ is in seconds.

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