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 = 300m//s`.

Options

A. `y=(0.06 m)cos[(78.5 m^(-1))x+(23562 s^(-1))t]m`
B. `y=(0.06 m)sin[(78.5 m^(-1))x-(23562 s^(-1))t]m`
C. `y=(0.06 m)sin[(78.5 m^(-1))x+(23562 s^(-1))t]m`
D. `y=(0.06 m)cos[(78.5 m^(-1))x-(23562 s^(-1))t]m`

Answer

b

Explanation

The amplitude, `A=0.06 m`
`(5)/(2)lambda=0.2 m`
`:. Lambda=0.08 m`
`f=(v)/(lambda)=(300)/(0.08)=3750 Hz`
`k=(2pi)/(lambda)=78.5 m^(-1)` and `omega=2pi f=23562 rad//s`
At `T=0,x=0, (dy)/(dx)=`positive
and the given curve is a sine curve.
Hence, equation of wave travelling in positive `x-`direction should have the form,
`y(x,t)=A sin (kx-omegat)`
substituting the values, we have
`y=(0.06 m) sin [(78.5 m^(-1))x-(23562 s^(-1))t] m`