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