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Question: Answered & Verified by Expert
A body is executing simple harmonic motion. At a displacement $x$ its potential energy is $E_1$ and at a displacement $y$ it potential energy is $E_2$. The potential energy $(E)$ at displacement $(x+y)$ is
PhysicsOscillationsTS EAMCETTS EAMCET 2001
Options:
  • A $\sqrt{E}=\sqrt{E_1}-\sqrt{E_2}$
  • B $\sqrt{E}=\sqrt{E_1}+\sqrt{E_2}$
  • C $E=E_1+E_2$
  • D $E=E_1-E_2$
Solution:
2558 Upvotes Verified Answer
The correct answer is: $\sqrt{E}=\sqrt{E_1}+\sqrt{E_2}$
Potential energy of a body executing SHM.
$\begin{aligned} & U=\frac{1}{2} m \omega^2 y^2 \\ & U_1=\frac{1}{2} m \omega^2 x^2 \Rightarrow x=\frac{\sqrt{2 U_1}}{m \omega^2} \\ & \text { and } \quad U_2=\frac{1}{2} m \omega^2 y^2 \\ & \Rightarrow \quad y=\frac{\sqrt{2 U_2}}{m \omega^2} \\ & \end{aligned}$
At displacement $(x+y)$
$\begin{aligned} & U=\frac{1}{2} m \omega^2(x+y)^2 \\ & (x+y)=\sqrt{\frac{2 U}{m \omega^2}} \\ & \frac{\sqrt{2 U_1}}{m \omega^2}+\sqrt{\frac{2 U_2}{m \omega^2}}=\sqrt{\frac{2 U}{m \omega^2}} \\ & \therefore \quad \sqrt{U_1}+\sqrt{U_2}=\sqrt{U} \\ & \text { i.e. } \quad \sqrt{E}=\sqrt{E_1}+\sqrt{E_2} \\ & \end{aligned}$

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