A) \[4\,\pi \,m/s\]
B) \[0.5\,\pi \,m/s\]
C) \[\frac{\pi }{4}\,m/s\]
D) 8 m/s
Correct Answer: D
Solution :
Key Idea: The standard transverse wave propagating along x-axis can be written as \[y=a\sin (kx-\omega t+\phi )\] The given equation is \[y(x,\,t)=8.0\,\sin \,\left( 0.5\,\pi \,x-4\pi t-\frac{\pi }{4} \right)...(i)\] The standard wave equation can be written as, \[y=a\sin (kx-\omega t+\phi )\] ...(iii) where a is amplitude, k the propagation constant and \[\omega \] the angular frequency, comparing the Eqs. (i) and (ii), we have \[k=0.5\,\pi ,\,\omega =4\pi \] \[\therefore \] Speed of transverse wave \[v=\frac{\omega }{k}=\frac{4\,\pi }{0.5\,\pi }\] = 8 m/s
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