Question

For the wave shown in figure, write the equation of this wave if its position is shown at t = 0. Speed of wave is v = 30 m/s.

• A. $y=\left(0.06m\right)cos\left[78.5m^{-1}\right)x+\left(2356.2s^{-1}\right)t]m$
• B. $y=\left(0.06m\right)sin\left[\right(78.5m^{-1})x-(2356.2s^{-1}\left)t\right]m$
• C. $y=\left(0.06m\right)sin\left[\right(78.5m^{-1})x+(2356.2s^{-1}\left)t\right]m$
• D. $y=\left(0.06m\right)cos\left[\right(78.5m^{-1})x-(2856.2s^{-1}\left)t\right]m$

Answer 1

The correct option is B $y=\left(0.06m\right)sin\left[\right(78.5m^{-1})x-(2356.2s^{-1}\left)t\right]m$

The amplitude, A = 0.06 m
$\frac{5}{2}\lambda =0.2m\therefore \lambda =0.08mf=\frac{v}{\lambda }=\frac{30}{0.080}=375Hzk=\frac{2\pi }{\lambda }=78.5m^{-1}and\omega =2\pi f=2356.2rad/sAtt=0,x=0,\frac{dy}{dx}=positive$
And the given curves is a since curve.
Hence, equation of wave travelling in positive x – direction should have the form,
$y(x,t)=Asin(kx-\omega t)$
Substituting the values, we have
$y=\left(0.06m\right)sin\left[\right(78.5m^{-1})x-(2356.2s^{-1}\left)t\right]m$