Question

When the displacement of a mass on a spring is 1/2a the half of the amplitude, what fraction of the mechanical energy is kinetic energy?

Answer 1

Total energy is a spring:
$E = \frac{1}{2} kx^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2$

At x = 0.5a:
$\frac{1}{2} k \frac{a}{2} ^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2 \\ \frac{1}{2} mv^2 = \frac{1}{2} ka^2 - \frac{1}{8} ka^2 = \frac{3}{8} ka^2$

The ration:
$\frac{ \frac{3}{8}ka^2 }{ \frac{1}{2} ka^2} = \frac{3}{4}$

Answer 2

Answer:

$KE : TE = 3 : 4$

Explanation:

As we know that the total mechanical energy of the object which is executing SHM is given by

$E_{total} = \frac{1}{2}KA^2$

here we know that

A = amplitude of SHM

K = spring constant

now we know that total mechanical energy of the spring is always constant so here we can say

kinetic energy + Potential energy = total mechanical energy

we know that potential energy of the spring at any given distance is

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

at given position x = A/2

$U = \frac{1}{2}K(\frac{A}{2})^2 = \frac{1}{8}KA^2$

now we have

$KE + \frac{1}{8}KA^2 = \frac{1}{2}KA^2$

$KE = \frac{3}{8}KA^2$

now ratio of kinetic energy and total mechanical energy will be given as

$KE : TE = \frac{3}{8}KA^2 : \frac{1}{2}KA^2$

$KE : TE = 3 : 4$