A) \[\sqrt{2}\pi \]
B) \[16\sqrt{2}\pi \]
C) \[24\sqrt{2}\pi \]
D) \[\frac{32\sqrt{2}}{\pi }\]
Correct Answer: D
Solution :
At \[t=2\text{ }sec,\]the particle crosses mean position. At \[t=4\text{ }sec,\]its velocity is \[4\,m{{s}^{-1}}\] For simple harmonic motion, \[y=a\,\sin \,\omega t\] \[\therefore \,\,y=a\,\sin \left( \frac{2\pi }{T} \right)t\] \[{{y}_{1}}=a\,\sin \left[ \left( \frac{2\pi }{16} \right)\times 2 \right]=a\sin \left( \frac{\pi }{4} \right)=\frac{a}{\sqrt{2}}\] ?.(i) After 4 sec or after 2 sec from mean position, \[{{y}_{1}}=\frac{a}{\sqrt{2}},\] velocity \[=4\text{ }m{{s}^{-1}}\] \[\therefore \] Velocity \[=\omega \sqrt{{{a}^{2}}-y_{1}^{2}}\] \[\Rightarrow \,\,4=\left( \frac{2\pi }{16} \right)\sqrt{{{a}^{2}}-\frac{{{a}^{2}}}{2}}\] from (i)] \[\Rightarrow \,\,4=\frac{\pi }{8}\times \frac{a}{\sqrt{2}}\] or \[a=\frac{32\sqrt{2}}{\pi }\]metre.
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