A) 1 : 1
B) 2 : 1
C) 1 : 3
D) \[\sqrt{3}\]: 1
Correct Answer: B
Solution :
\[{{y}_{1}}=5[sin2\pi t+\sqrt{3}cos2\pi t]\] \[=10\left[ \frac{1}{2}sin2\pi t+\frac{\sqrt{3}}{2}cos2\pi t \right]\] \[=10\left[ \cos \frac{\pi }{3}sin2\pi t+\sin \frac{\pi }{3}cos2\pi t \right]\] \[=10\left[ \sin \left( 2\pi t+\frac{\pi }{3} \right) \right]\] \[\Rightarrow \]\[{{A}_{1}}=10\] Similarly,\[{{y}_{2}}=5\sin \left( 2\pi t+\frac{\pi }{4} \right)\] \[\Rightarrow \] \[{{A}_{2}}=5\] Hence, \[\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{10}{5}=\frac{2}{1}\]
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