A) 0.6 sec
B) 0.4 sec
C) 0.3 sec
D) 0.2 sec
Correct Answer: B
Solution :
Time taken by particle to move from x=0 (mean position) to x = 4 (extreme position)\[=\frac{T}{4}=\frac{1.2}{4}=0.3\ s\] Let t be the time taken by the particle to move from x=0 to x=2 cm \[y=a\sin \omega t\Rightarrow 2=4\sin \frac{2\pi }{T}t\]\[\Rightarrow \frac{1}{2}=\sin \frac{2\pi }{1.2}t\] \[\Rightarrow \frac{\pi }{6}=\frac{2\pi }{1.2}t\Rightarrow t=0.1\ s\]. Hence time to move from x = 2 to x = 4 will be equal to 0.3 ? 0.1 = 0.2 s Hence total time to move from x = 2 to x = 4 and back again \[=2\times 0.2=0.4\sec \]
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