Question

When the displacement in SHM is equal to 1/3 of the amplitude xm, what fraction of the total energy is (a) kinetic energy and (b) potential energy? (c) At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy? (Give the ratio of the answer to the amplitude)

Answer 1

Answer:

Explanation:

Given

Displacement is $\frac{1}{3}$ of Amplitude

i.e. $x=\frac{A}{3}$ , where A is maximum amplitude

Potential Energy is given by

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}k(\frac{A}{3})^2$

$U=\frac{1}{18}kA^2$

Total Energy of SHM is given by

$T.E.=\frac{1}{2}kA^2$

Total Energy=kinetic Energy+Potential Energy

$K.E.=\frac{1}{2}kA^2 -\frac{1}{18}kA^2$

$K.E.=\frac{8}{18}kA^2$

Potential Energy is $\frac{1}{8}$ th of Total Energy

Kinetic Energy is $\frac{8}{9}$ of Total Energy

(c)Kinetic Energy is $0.5\times \frac{1}{2}kA^2$

$P.E.=\frac{1}{4}kA^2$

$\frac{1}{2}kx^2=\frac{1}{4}kA^2$

$x=\frac{A}{\sqrt{2}}$                  

Answer 2

Answer:

Explanation:

Let the amplitude is A.

Displacement, x = one third of the amplitude = A/3

The total energy of the body executing Simple Harmonic Motion is given by

$T=\frac{1}{2}KA^{2}$

(a) Kinetic energy of the particle executing SHM is given by

$K=\frac{1}{2}K\left (A^{2} -x^{2} \right )$

$K=\frac{1}{2}K\left (A^{2} -\frac{A}{9}^{2} \right )$

$K=\frac{1}{2}\times \frac{8A^{2}}{9}$

So, the ratio of kinetic energy to the total energy is given by

K / T = 8 / 9

(b) Potential energy of a particle executing SHM is given by

$U=\frac{1}{2}Kx^{2}$

$T=\frac{1}{2}\times \frac{A^{2}}{9}$

So, the ratio of potential energy to the total energy is given by

U / T = 1 / 9

(c) Let at a displacement y the kinetic energy is equal to the potential energy

$\frac{1}{2}\times K\times \left ( A^{2}-y^{2} \right )=\frac{1}{2}\times K\times y^{2}$

$y=\frac{A}{\sqrt{2}}$