Question

What happens to the energy of a simple harmonic oscillator if its amplitude is doubled?

Answer 1

total energy of simple harmonic oscillator is given by, $\bf{T.E=\frac{1}{2}m\omega^2A^2}$

where m is mass of particle in simple harmonic motion, $\omega$ is angular frequency and A is amplitude of motion of particle.

here you can see that, total energy of particle is directly proportional to square of amplitude of it. e.g., $T.E\propto A^2$
or, $\frac{T.E_1}{T.E_2}=\frac{A_1^2}{A_2^2}$

if we assume Initial amplitude is A. e.g., $A_1=A$
and initial total energy of simple harmonic oscillator is E. e.g., $T.E_1=E$

a/c to question, final amplitude, $A_2=2A$
then,
$\frac{E}{T.E_2}=\frac{A^2}{(2A)^2}=\frac{1}{4}$

or, $T.E_2=4E$

hence, if amplitude is doubled , total energy of simple harmonic oscillator becomes four times.


[ note :- total energy = kinetic energy + potential energy , in SHM , kinetic energy is given by, K.E = $\frac{1}{2}m\omega^2(A^2-y^2)$ and potential energy is given by, $P.E=\frac{1}{2}m\omega^2y^2$, so total energy = $\frac{1}{2}m\omega^2A^2$]

Answer 2

Total energy of an object executing SHM is given by the relation

E=1/2(mA^2)

Where

m-> mass of the object

A->amplitude of the oscillation

so if amplitude =A

E1=1/2(MA^2)

NOW if amplitude= 2A

E2=1/2(m)(2A)^2=2mA^2

E2/E1=4;

E2=4E1;

S0 , IF AMPLITUDE IS DOUBLED ENERGY WILL INCREASE BY FOUR TIME;

Because ENERGY IS PROPORTIONAL TO SQUARE OF AMPLITUDE.