Question

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

• A. Time period $= 0.02 \,s$
• B. Wavelength $= 2 $
• C. Velocity $= 50 \,m/s $
• D. Amplitude $= 0.02 \,m$

Answer 1

Answer: (A)

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 $y=A \sin \left[\frac{2 \pi}{\lambda}(v t-x)\right]$
The given equation of wave is
$y= 0.04 \cos (\pi x) \sin (50 \pi t) $
$= 0.02 \sin (50 \pi t+\pi x)+0.02 \sin (50 \pi t-\pi x) $
$[\because 2 \sin A \cdot \cos B=\sin (A+B) \cdot \cos (A-B)]$
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 $a \sin (\omega t+k x)$, we get
Amplitude, $a=0.02\, m$
Time period, $T=\frac{2 \pi}{50 \pi}=\frac{1}{25}=0.04 \,s$
Wavelength, $\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.