A) \[y=0.04\sin 2\pi (120\,t-x/3)\]
B) \[y=-0.04\sin 2\pi (110\,t-x/3)\]
C) \[y=0.04\sin 2\pi (120\,t+x/3)\]
D) \[y=0.04\sin \,\pi (120\,t-x/3)\]
Correct Answer: A
Solution :
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, \[a=0.04\,m,\,n=120\,Hz\,({{s}^{-1}})\] and wave speed v = 360 m/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 \[y=0.04\sin (240\pi t-\frac{2\pi }{3}\,x)\] or \[y=0.04\sin 2\pi \left[ 120\,t-\frac{x}{3} \right]\]
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