A) \[a/T\]
B) \[2a/T\]
C) \[3a/T\]
D) \[a/2T\]
Correct Answer: C
Solution :
[c] \[x=a\cos \omega t\] \[a/2=a\cos \omega {{t}_{0}}\] Or \[\cos \frac{\pi }{3}=\cos \omega {{t}_{0}}\] or \[\frac{\pi }{3}=\omega {{t}_{0}}\]\[\therefore {{t}_{0}}=\frac{\pi }{3\omega }\] \[v=\frac{dx}{dt}=-a\omega \sin \omega t\] \[\therefore \bar{v}=\frac{\int_{0}^{\pi /3\omega }{vdt}}{\int_{0}^{\pi /3\omega }{dt}}\] But \[\omega =\frac{2\pi }{T}\]\[\therefore \bar{v}=\frac{3a}{T}\]You need to login to perform this action.
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