A) 0.5s
B) 1.0s
C) 1.5s
D) 2.0s
Correct Answer: A
Solution :
In order to find the time taken by the particle from \[-12.5\,cm\]to \[+\text{ }12.5\,cm\]on either side of mean position, we will find the time taken by particle to go from \[x=-12.5\,cm\]to \[x=\text{0}\]and to go from \[x=0\]to\[~x=+\text{ }12.5\,cm.\] Let the equation of motion be \[x=A\sin \omega t.\] First, the particle moves from \[x=-12.5\,cm\]to\[x=0\] \[\therefore \] \[12.5=25\sin \omega t\] \[(\because \,A=25\,cm)\] \[\Rightarrow \] \[\frac{1}{2}=\sin \omega t\] \[\Rightarrow \] \[\omega t=\frac{\pi }{6}\] \[\therefore \] \[t=\frac{\pi }{6\omega }\] Similarly to go from \[x=0\]to \[x=12.5\,cm\] \[\omega t=\frac{\pi }{6}\] \[\Rightarrow \] \[t=\frac{\pi }{6\omega }\] \[\therefore \] Total time taken from \[x=-12.5\text{ }cm\]to \[x=12.5\,cm\] \[t=\frac{\pi }{6\omega }+\frac{\pi }{6\omega }=\frac{\pi }{3\omega }\] \[=\frac{\pi }{3\left( \frac{2\pi }{T} \right)}=\frac{T}{6}=\frac{3}{6}=0.5\,s\]
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