A) \[{{T}_{1}}<{{T}_{2}}\]
B) \[{{T}_{1}}>{{T}_{2}}\]
C) \[{{T}_{1}}={{T}_{2}}\]
D) \[{{T}_{1}}=2{{T}_{2}}\]
Correct Answer: A
Solution :
Using \[x=A\sin \omega t\] For \[x=A/2,\ \ \sin \omega {{T}_{1}}=1/2\Rightarrow {{T}_{1}}=\frac{\pi }{6\omega }\] For \[x=A,\ \sin \omega ({{T}_{1}}+{{T}_{2}})=1\Rightarrow {{T}_{1}}+{{T}_{2}}=\frac{\pi }{2\omega }\] \[\Rightarrow {{T}_{2}}=\frac{\pi }{2\omega }-{{T}_{1}}=\frac{\pi }{2\omega }-\frac{\pi }{6\omega }=\frac{\pi }{3\omega }i.e.\ {{T}_{1}}<{{T}_{2}}\] Alternate method : In S.H.M., velocity of particle also oscillates simple harmonically. Speed is more near the mean position and less near the extreme position. Therefore the time taken for the particle to go from 0 to \[\frac{A}{2}\] will be less than the time taken to go from \[\frac{A}{2}\] to A. Hence \[{{T}_{1}}<{{T}_{2}}.\]
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