A) 33
B) 87
C) 67
D) 50
Correct Answer: B
Solution :
The relation for kinetic energy of S.H.M. is given by \[=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})\] ...(1) Potential energy is given by \[=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] ...(2) Now for the condition of question and from eqs. (1) and (2) \[\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})=\frac{1}{3}\times \frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] Or \[\frac{4}{6}m{{\omega }^{2}}{{y}^{2}}=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\] Or \[{{y}^{2}}=\frac{3}{4}{{a}^{2}}\] So, \[y=\frac{a}{2}\sqrt{3}=0.866\,a\] \[\approx 87%\]of amplitude
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