A)
B)
C)
D)
Correct Answer: D
Solution :
The KE of a particle at a distance x from its mean position is given by \[K=\frac{1}{2}m({{A}^{2}}-{{x}^{2}}){{\Omega }^{2}}\] It means curve between K and x will be parabola. Hence options (a) and (b) wrong. The PE of a particle is given by \[U=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] But acceleration\[a=-{{\omega }^{2}}x\]. Hence, PE is proportional to\[{{a}^{2}}\]. At mean position, if will be equal to zero and maximum at extreme position. Hence (c option is correct). The sum of PE and KE is always constant. It is equal to \[\frac{1}{2}m{{A}^{2}}{{\omega }^{2}}\] Hence, \[U+K=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] \[U=-K+\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] It means curve between U and K will be straight line having positive intercept on y-axis and negative slop. Hence, (d) option is correct.
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