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 :
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, \[y=a\,\sin (k\,x-\omega t+\phi )\] ...(ii) where a is amplitude, k the propagation constant and co the angular frequency, comparing the Eqs. (i) and (ii), we have \[k=0.5\pi ,\omega =4\pi \] \[\therefore \] Speed of transverse wave \[v=\frac{\omega }{k}=\frac{4\pi }{0.5\pi }\] \[=8m/s\]
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