Question

Problem14.38
Part A When the displacement of a mass on a spring is the half of the amplitude, what fraction of the energy is kineticenergy? Part B At what displacement, as a fraction of , is the energy half kinetic and half potential?

Answer 1

Concept and reason

The concept used in this question is Simple Harmonic Oscillations. First, write the expression for the kinetic energy and total energy of the particle and then find the ratio of these energies by dividing them. Finally, calculate the displacement of the particle using the energy equation.

Fundamentals

Simple harmonic motion is a periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. When a motion repeats itself, it is called as a periodic motion.

The kinetic energy (K) of a particle in a simple harmonic oscillation, when it is at a displacement x from the equilibrium position is as follows:

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

Here, m is the mass of the particle, $\omega$ is the angular velocity, x is the displacement of the particle, and A is the amplitude of the oscillation.

The total energy (E) of a particle in a simple harmonic oscillation is as follows:

$E = \frac{{m{\omega ^2}{A^2}}}{2}$

(a)

The fraction of the total energy which is kinetic energy is given as follows:

$f = \frac{K}{E}$

Substitute $\frac{{m{\omega ^2}}}{2}\left( {\frac{{3{A^2}}}{4}} \right)$ for K and $\frac{{m{\omega ^2}{A^2}}}{2}$ for E in the above expression.

$\begin{array}{c}\\f = \frac{{\frac{{m{\omega ^2}}}{2}\left( {\frac{{3{A^2}}}{4}} \right)}}{{\frac{{m{\omega ^2}{A^2}}}{2}}}\\\\ = \frac{{\left( {\frac{{3{A^2}}}{4}} \right)}}{{{A^2}}}\\\\ = \frac{3}{4}\\\\ = 0.75\\\end{array}$

Thus, the fraction of the kinetic energy is 0.75.

(b)

The expression for the kinetic energy is as follows:

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

The expression of the total energy can be written as follows:

$E = 2K$

Substitute $\frac{{m{\omega ^2}}}{2}\left( {{A^2} - {x^2}} \right)$ for K in the above expression.

$\begin{array}{c}\\E = 2\left( {\frac{{m{\omega ^2}}}{2}\left( {{A^2} - {x^2}} \right)} \right)\\\\ = m{\omega ^2}\left( {{A^2} - {x^2}} \right)\\\end{array}$

The expression for the total energy is as follows:

$E = \frac{{m{\omega ^2}{A^2}}}{2}$

Substitute $m{\omega ^2}\left( {{A^2} - {x^2}} \right)$ for E in the above expression.

$\begin{array}{c}\\m{\omega ^2}\left( {{A^2} - {x^2}} \right) = \frac{{m{\omega ^2}{A^2}}}{2}\\\\\left( {{A^2} - {x^2}} \right) = \frac{{{A^2}}}{2}\\\\{x^2} = {A^2} - \frac{{{A^2}}}{2}\\\\ = \frac{{{A^2}}}{2}\\\end{array}$

From the above equation, the value of the displacement is,

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

The fraction of the amplitude which is displacement is given as follows:

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

Substitute $\frac{A}{{\sqrt 2 }}$ for x in the above expression.

$\begin{array}{c}\\f = \frac{A}{{\sqrt 2 A}}\\\\ = \frac{1}{{\sqrt 2 }}\\\\ = 0.707\\\end{array}$

Therefore, the displacement, as a fraction of amplitude is the energy half kinetic and half potential is 0.71.

Ans: Part a

The fraction of the kinetic energy is 0.75.

Part b

The displacement of the particle is 0.71.