A) \[\frac{E}{4}\]
B) \[\frac{E}{16}\]
C) \[\frac{E}{2}\]
D) \[\frac{E}{8}\]
Correct Answer: A
Solution :
If a particle executes SHM, its kinetic energy is given by \[KE=\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{x}^{2}})\] Or \[KE=\frac{1}{2}k({{A}^{2}}-{{x}^{2}})\] where,\[k=m{{\omega }^{2}}=\]constant Its potential energy is given by \[PE=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] \[=\frac{1}{2}k{{x}^{2}}\] Thus, total energy of particle \[E=KE+PE\] \[=\frac{1}{2}k({{A}^{2}}-{{x}^{2}})+\frac{1}{2}k{{x}^{2}}\] \[=\frac{1}{2}k{{A}^{2}}\] Hence,\[PE=\frac{1}{2}k{{x}^{2}}=\frac{1}{2}k{{\left( \frac{A}{2} \right)}^{2}}\] \[(\because x=A/2)\] \[=\frac{1}{4}\left( \frac{1}{2}k{{A}^{2}} \right)\] \[=\frac{1}{4}E\] Hence, potential energy is one-fourth of total energy.
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