A) 0.05
B) 0.06
C) 0.01
D) None of these
Correct Answer: B
Solution :
[b] \[E=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\Rightarrow e=\frac{1}{2}m{{(2\pi f)}^{2}}{{A}^{2}}k\Rightarrow A=\frac{1}{2\pi f}\sqrt{\frac{2E}{m}}\] Putting \[E=K+U\]we obtain, \[A=\frac{1}{2\pi \left( \frac{25}{\pi } \right)}\sqrt{\frac{2\times (0.5+0.4)}{0.2}}\Rightarrow A=0.06m\]
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