A) \[\frac{2\pi }{{{b}^{2}}}\sqrt{\frac{m}{{{U}_{0}}}}\]
B) \[\frac{\pi }{b}\sqrt{\frac{m}{{{U}_{0}}}}\]
C) \[\frac{{{\pi }^{2}}}{2b}\sqrt{\frac{m}{{{U}_{0}}}}\]
D) \[\frac{2\pi }{b}\sqrt{\frac{m}{{{U}_{0}}}}\]
Correct Answer: D
Solution :
According to the question, given that \[U(x)={{U}_{0}}(1-\cos bx)=2{{U}_{0}}{{\sin }^{2}}\frac{bx}{2}\] For small oscillations, \[\sin \frac{bx}{2}\approx \frac{bx}{2}\] Thus,\[U(x)=2{{U}_{0}}{{\left( \frac{bx}{2} \right)}^{2}}=\frac{{{U}_{0}}{{b}^{2}}{{x}^{2}}}{2}\] Comparing with\[U=\frac{1}{2}k{{x}^{2}}\], we get \[k={{U}_{0}}{{b}^{2}}\] \[\therefore \]Time period,\[T=2\pi \sqrt{\frac{m}{k}}\] \[=2\pi \sqrt{\frac{m}{{{U}_{0}}{{b}^{2}}}}=\frac{2\pi }{b}\sqrt{\frac{m}{{{U}_{0}}}}\]You need to login to perform this action.
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