A) 14, 2\[\sqrt{2}\]
B) 16, 2\[\sqrt{2}\]
C) 14, 3\[\sqrt{2}\]
D) 16, 3\[\sqrt{2}\]
Correct Answer: A
Solution :
\[x=4\cos \left[ 88\,t+\frac{\pi }{4} \right]\] ?(i) Comparing Eq. (i) with standard equation of SHM of a particle, given by \[x=A\cos (\omega t+\text{o }\!\!|\!\!\text{ })\] ...(ii) We have, \[\omega =88\] \[\Rightarrow \] \[f=\frac{88}{2\pi }=14\,Hz.\] From Eq. (i), we have \[x=4\left[ \cos 88t\,\cos \frac{\pi }{4}-\sin 88t\sin \frac{\pi }{4} \right]\] \[=4\left[ \cos 88t\times \frac{1}{\sqrt{2}}-\sin 88t\times \frac{1}{\sqrt{2}} \right]\] So, initial displacement of the particle is \[2\sqrt{2}\,m.\]
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