A) \[\frac{1}{2}\]
B) \[\frac{1}{\sqrt{2}}\]
C) \[\frac{1}{\sqrt{2}-1}\]
D) \[\frac{1}{\sqrt{3}}\]
Correct Answer: C
Solution :
Time period of particle is given by \[T=\frac{2\pi }{\omega }\] \[(\because \,T=8s)\] So, \[\frac{2\pi }{\omega }=8\] \[\Rightarrow \] \[\omega =\frac{2\pi }{8}=\frac{\pi }{4}\] Equation of SHM is \[y=a\sin \omega t\] ?(i) (\[\because \] At mean position t = 0) Hence, \[y=0\] For t = 1 s Eq. (i) becomes \[y=a\sin \frac{\pi }{4}\times 1=\frac{a}{\sqrt{2}}\] ?(ii) For t = 2 s Eq. (i) becomes \[y=a\sin \frac{\pi }{4}\times 2=a\sin \frac{\pi }{2}=a\] ?(iii) Now, the distance covered in 2 s is given by \[a-\frac{a}{\sqrt{2}}=a\left( \frac{\sqrt{2}-1}{\sqrt{2}} \right)\] Again the ratio of the distance covered in first and second seconds is \[=\frac{\frac{a}{\sqrt{2}}}{a\left( \frac{\sqrt{2-1}}{\sqrt{2}} \right)}=\frac{1}{(\sqrt{2}-1)}\]
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