A) \[\frac{\pi }{6}\sqrt{\frac{m}{k}}\]
B) \[\frac{2\pi }{3}\sqrt{\frac{m}{k}}\]
C) \[\frac{\pi }{4}\sqrt{\frac{m}{k}}\]
D) \[\frac{11\pi }{9}\sqrt{\frac{m}{k}}\]
Correct Answer: D
Solution :
[d] When spring is compressed by \[3{{x}_{0}}\]Amplitude, \[A=3{{x}_{0}}\]the time taken form extreme compressed position to mean position, \[{{t}_{1}}=T/4.\] If time taken \[({{t}_{2}})\]from mean position to \[x={{x}_{0}}\]is given by \[x=A\sin \frac{2\pi {{t}_{2}}}{T}\times {{x}_{0}}=3{{x}_{0}}\sin \frac{2\pi {{t}_{2}}}{T}\] \[\sin \frac{2\pi {{t}_{2}}}{T}=\frac{1}{3}\times \frac{2\pi {{t}_{2}}}{T}=\frac{\pi }{9}\times {{t}_{2}}=\frac{T}{18}\] \[{{t}_{1}}+{{t}_{2}}=\frac{T}{4}+\frac{T}{18}=\frac{11}{18}T\]\[=\frac{11}{18}2\pi \sqrt{\frac{m}{k}}=\frac{11}{9}\pi \sqrt{\frac{m}{k}}\]You need to login to perform this action.
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