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 +12.5 cm on either side of mean position, we will find the time taken by particle to go from\[x=-12.5\text{ }cm\]to\[x=0\]and to go from\[x=0\]to\[x=+12.5\text{ }cm\]. Let the equation of motion be,\[x=A\text{ }sin\text{ }\omega \text{t}\] First, the particle moves from\[x=-12.5\text{ }cm\]to \[x=0\] \[\therefore \] \[12.5=25\sin \omega t\] \[(\because A=25\,cm)\] \[\Rightarrow \] \[\frac{1}{2}=\sin \omega t\] \[\Rightarrow \] \[\omega t={{\sin }^{-1}}\left( \frac{1}{2} \right)=\frac{\pi }{6}\] \[\Rightarrow \] \[t=\frac{\pi }{6\omega }\] Similarly, to go from\[x=0\text{ }to\text{ }x=12.5\text{ }cm,\] \[\omega t=\frac{\pi }{6}\] \[\Rightarrow \] \[t=\frac{\pi }{6\omega }\] Hence, total time taken from \[x=-12.5\text{ }cm\text{ }to\text{ }x=12.5cm\] \[t'=\frac{\pi }{6\omega }+\frac{\pi }{6\omega }=\frac{\pi }{3\omega }\] \[=\frac{\pi }{3\left( \frac{2\pi }{T} \right)}=\frac{T}{6}\] but \[T=3s\] \[\therefore \] \[t'=\frac{3}{6}=0.5\,s\]You need to login to perform this action.
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