A) \[zero\]
B) \[\pi /2\]
C) \[\pi /6\]
D) \[\pi /3\]
Correct Answer: B
Solution :
\[{{y}_{1}}=a\sin 2\pi \,\,vt\] and \[{{y}_{2}}=a\sin (2\pi vt+\phi )\] \[y={{y}_{2}}-{{y}_{1}}=a[\sin (2\pi \,\,vt+\phi )-\sin 2\pi \,\,vt]\] \[=2a\sin \frac{\phi }{2}\cos \left[ 2\pi \,\,vt+\frac{\phi }{2} \right]\] \[\therefore \]Maximum value of\[y=2a\sin \frac{\phi }{2}\] Now,\[2a\sin \frac{\phi }{2}=a\sqrt{2}\]You need to login to perform this action.
You will be redirected in
3 sec