Answer:
Option A
Explanation:
Key idea . The displacement of a wave in term of time period is given by
$y= A\sin\left[ 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)\right]$
This equation in terms of speed of wave (v) becomes $y= A\sin\left[ \frac{2\pi}{\lambda}(vt-x)\right]$
The given equation of wave is
$y= 0.04 \cos (\pi x) \sin (50 \pi t)$
$y= 0.02 \sin(50 \pi t+ \pi x)+0.02 \sin (50 \pi t- \pi x)$
[$\because$ $ 2 \sin A\cos B= \sin (A+B) .\cos (A-B)]$
Thus, the given wave is the combination of two waves.
$y_{1}=0.02 \sin (50 \pi t+\pi x)$ ( in -ve x- direction)
and $y_{2}=0.02 \sin (50 \pi t-\pi x)$ (in +ve x direction )
Comparing them with the general equation of wave $a\sin (\omega t+kx)$, we get
Amplitude , a= 0.02 m
Time period , T= $\frac{2 \pi}{50 \pi}= \frac{1}{25}=0.04 s$
Wave length , $\lambda= \frac{2 \pi}{\pi}=2 m$
velocity , v= $ \frac{50 \pi\times \lambda}{2 \pi}=\frac{100}{2}=50 ms^{-1}$
So, option (a) is wrong
