Question

A body is executing simple harmonic motion. At a displacement x, its potential energy is $E_1$ and at displacement y, its potential energy is $E_2$ The potential energy E at a displacement ( x + y) is

• A. $E_1+E_2$
• B. $\sqrt{E_1^2+E_2^2}$
• C. $E_1+E_2+2\sqrt{E_1{E}_2}$
• D. $\sqrt{E_1{E}_2}$

Answer 1

The correct option is C $E_1+E_2+2\sqrt{E_1{E}_2}$

We have expression for potential energy

$R.E=\frac{1}{2}m{\omega }^2{x}^2$ X=displacement $\omega$= angular frequency

Given $E_1=\frac{1}{2}m{\omega }^2{x}^2{E}_2=\frac{1}{2}m{\omega }^2{y}^2$

Now E is potential Energy at (x+y)

$E=\frac{1}{2}m{\omega }^2(x+y{)}^2\Rightarrow E=\frac{1}{2}m{\omega }^2{x}^2+\frac{1}{2}m{\omega }^2{y}^2+\frac{1}{2}m{\omega }^{2}xy.2\Rightarrow E=E_1+E_2+2\left(\sqrt{\frac{m{\omega }^2{x}^2}{2}}\sqrt{\frac{m{\omega }^2{y}^2}{2}}\right)\Rightarrow E=E_1+E_2+2\sqrt{E_1{E}_2}$