Find the wrong statement form the following about the equation of stationary wave given by $Y=0.04\cos \left(\pi x\right)\sin \left(50\pi t\right)$ m where t is in second. Then for the stationary wave.

Key Idea – The displacement of a wave in term of time period is given by
$y=A\mathrm{s}\mathrm{i}\mathrm{n}\left[2\pi \left(\frac{t}{T}-\frac{x}{\lambda }\right)\right]$
This equation in terms of speed of wave (v),
Becomes $y=A\mathrm{s}\mathrm{i}\mathrm{n}\left[\frac{2\pi }{\lambda }\left(vt-x\right)\right]$
The given equation of wave is
$y=0.04\cos \left(\pi x\right)\sin \left(50\pi t\right)$
$=0.02\sin \left(50\pi t+\pi x\right)+0.02\sin \left(50\pi t-\pi x\right)$
$\left[\because 2\mathrm{s}\mathrm{i}\mathrm{n}A.\mathrm{c}\mathrm{o}\mathrm{s}B=\sin \left(A+B\right).\cos \left(A-B\right)\right]$
Thus, the given wave is the combination of two waves,
$y_1=0.02\sin \left(50\pi t+\pi x\right)\left(in-vex-direction\right)$
and $y_2=0.02\sin \left(50\pi t-\pi x\right)$ (in + $ve$ x-direction)
Comparing them with the general equation of wave $\mathrm{asin}\left(\omega t+kx\right)$ , we get
Amplitude, a = 0.02m
Time period, T = $\frac{2\pi }{50\pi }=\frac{1}{25}=0.04s$
Wavelength, $\lambda =\frac{2\pi }{\pi }=2m$
Velocity, v = $\frac{50\pi \times \lambda }{2\pi }$
$=\frac{100}{2}=50m{s}^{-1}$
So, option (a) is wrong.