Question

Find the equation of a plane progressive wave travelling along positive: $x-axis$ having amplitude $0.04m$, frequency $120Hz$ and speed $360m/s$.

• A. $y=0.04\sin 2\pi (120t-x/3)$
• B. $y=-0.04\sin 2\pi (110t-x/3)$
• C. $y=0.04\sin 2\pi (120t+x/3)$
• D. $y=0.04\sin \pi (120t-x/3)$

Answer 1

Answer: (A)

The equation of a plane progressive wave moving in the positive direction is given by $y=a\sin (\omega t-kx)$
where, a is amplitude, angular frequency $\omega =2\pi n,n$, in, $n$ is frequency and propagation constant $k=\frac{2\pi }{\lambda },\lambda$ is wavelength.
Here, $a=0.04m,n=120Hz\left(s^{-1}\right)$ and wave speed $v=360m/s$.
Hence, $\omega =2\pi n=2\pi \times 120=240\pi$ and $v=\frac{\omega }{k}$
or $k=\frac{\omega }{v}=\frac{240\pi }{360}=\frac{2\pi }{3}$
Substituting these values of $a,\omega$ and $k$ in the above equation
$y=0.04\sin \left(240\pi t-\frac{2\pi }{3}x\right)$