A) 3.6
B) \[3.6\times {{10}^{-4}}\]
C) \[3.6\times {{10}^{-6}}\]
D) \[3.6\times {{10}^{11}}\]
Correct Answer: B
Solution :
Key Idea: The first idea is that the standard equation of travelling wave is \[y=A\,\cos \,\,(\omega t-kx)\] Compare given equation \[y=60\times {{10}^{-6}}\,\cos \,(1800t-6x)\mu m\] with standard equation, we have \[\omega =1800\,rad/s\] \[A=60\times {{10}^{-6}}m,\,k=6{{m}^{-1}}\] Key Idea: The second idea is that the velocity of wave propagation is just the ratio of angular velocity and propagation constant and maximum particle velocity is \[A\omega \]. \[\text{Velocity}\,\,\text{of}\,\,\text{propagation=}\frac{\text{Angular}\,\text{vevlocity}}{\text{Propagtion}\,\text{constant}}\]\[v=\frac{\omega }{k}\] \[=\frac{1800}{6}=300m/s\] Maximum particle velocity \[{{v}_{\max }}=A\omega \] \[=60\times {{10}^{-6}}\times 1800m/s\] \[\therefore \] \[\frac{{{v}_{\max }}}{v}=\frac{60\times {{10}^{-6}}\times 1800}{300}\] \[=3.6\times {{10}^{-4}}\]
You need to login to perform this action.
You will be redirected in
3 sec
