Question

A body is executing simple harmonic motion. at a displacement x its potential energy is e1 and at a displacement-y its potential energy is e2. the potential energy e at displacement (x + y) is

Answer 1

hope it's clear

I assume you know how the formula for potential energy is derived


all I did was
1) first write what was given in the question

2)then write down what's X+Y in terms of whatever is left and hope that the mass,angular frequency get cancelled out!!!!

now put it back in the answer to get your answer

the final answer is

$( \sqrt{e1 } + \sqrt{e2} ) {}^{2}$

Answer 2

Answer:

potential energy at a displacement  (x + y) is  $e=(e_{1} +e_{2} })^{2}$

Explanation:

Given a body is executing simple harmonic motion

In simple harmonic system, the potential energy at displacement $r$ is given by

                                     $\frac{1}{2} m\omega^{2} r^{2}=\frac{1}{2}kr^{2}$          

Therefore, potential energy at a displacement x is

                     $e_{1}=\frac{1}{2} kx^{2}\implies x=\sqrt{\frac{2e_{1} }{k} }$        

         and  potential energy at a displacement y is

                     $e_{2}=\frac{1}{2} ky^{2}\implies y=\sqrt{\frac{2e_{2} }{k} }$

potential energy at a displacement (x + y) is

                      $e=\frac{1}{2} k (x+y)^{2}$

                        $=\frac{1}{2} k\Big (\sqrt{\frac{2e_{1} }{k} }+\sqrt{\frac{2e_{2} }{k} }\Big)^{2}$

                        $= (e_{1} +e_{2} })^{2}$

Therefore, potential energy at a displacement  (x + y) is $(e_{1} +e_{2} })^{2}$