Question

When the displacement in SHM is equal to 1/4 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?

Answer 1

Answer:

$\frac{KE}{TE} = \frac{15}{16}$

b)

$\frac{U}{TE} = \frac{1}{16}$

Explanation:

Let the amplitude of SHM is given as A

so the total energy of SHM is given as

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

now we know that

a)

kinetic energy is given as

$KE = \frac{1}{2}k(A^2 - x^2)$

here

$x = \frac{A}{4}$

so now we have

$KE = \frac{1}{2}k(A^2 - \frac{A^2}{16})$

$KE = \frac{15}{32}kA^2$

now its fraction with respect to total energy is given as

$\frac{KE}{TE} = \frac{15}{16}$

b)

Potential energy is given as

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

so we have

$U = \frac{1}{32}kA^2$

so fraction of energy is given as

$\frac{U}{TE} = \frac{1}{16}$