Question

A body executes simple harmonic motion. At a displacement $x$, its potential energy is $U_1$. At a displacement $y$, its potential energy is $U_2$. What is the potential energy of the body at a displacement $(x+y)$?

• A. $U_1+U_2$
• B. ${(\sqrt{U_1}+\sqrt{U_2})}^2$
• C. $\sqrt{{U_1}^2+{U_2}^2}$
• D. $\sqrt{U_1{U}_2}$

Answer 1

The correct option is B ${(\sqrt{U_1}+\sqrt{U_2})}^2$

$P.E.=\frac{1}{2}m{\omega }^2{(x+y)}^{2}P.E.=\frac{1}{2}m{\omega }^2{x}^2{U}_1=\frac{1}{2}m{\omega }^2{x}^2⟹x=\sqrt{\frac{2U_1}{m{\omega }_1}}{U}_2=\frac{1}{2}m{\omega }^2{y}^2⟹y=\sqrt{\frac{2U_2}{m{\omega }_2}}x+y=\frac{\sqrt{2}}{\omega \sqrt{m}}(\sqrt{U_1}+\sqrt{U_2})P.E.=\frac{1}{2}m{\omega }^2{\left(\frac{\sqrt{2}}{\omega \sqrt{m}}(\sqrt{U_1}+\sqrt{U_2})\right)}^2{P.E.}_{(x+y)}=\frac{1}{2}m{\omega }^2\times \frac{2}{{\omega }^{2}m}{(\sqrt{U_1}+\sqrt{U_2})}^2={(\sqrt{U_1}+\sqrt{U_2})}^2$