Question

As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horizontally moving block attached to a spring. Note that, since the gravitational potential energy is not changing in this case, it can be excluded from the calculations.
For such a system, the potential energy is stored in the spring and is given by

U=1/2kx^2,

where k is the force constant of the spring and x is the distance from the equilibrium position.

The kinetic energy of the system is, as always,

K=1/2mv^2,

where m is the mass of the block and v is the speed of the block.

We will also assume that there are no resistive forces; that is, E=constant.

a. Find the total energy of the object at any point in its motion.
b. Find the amplitude of the motion.
c. Find the maximum speed attained by the object during its motion.

Answer 1

a)E= U + K = $\frac{1}{2}$kx² +  $\frac{1}{2}$mv²

The total energy of the system at any point in the motion is equal to the sum of the elastic potential energy of the spring, U, and of the kinetic energy of the mass, K:

E= U + K = $\frac{1}{2}$kx² +  $\frac{1}{2}$mv²

where

'k' represents the spring constant

'x' is the compression/stretching of the spring with respect to its equilibrium position

'm' is the mass of the block attached to the spring

and 'v' is the speed of the block

b) A=$\sqrt{\frac{2E}{k}}$

The amplitude of the motion compares to the most extreme displacement of the mass-spring system. The displacement of the system, x(t), at time t, for a simple harmonic oscillator is given by,

x= Asin(ωt+∅)

where

amplitude  is 'A'

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency of the motion

t is the time

$\phi$ is the phase (we can take $\phi$=0 )

The amplitude of the motion occurs when the displacement of the motion is maximum: x=A. Regarding energy, the mass-spring system is at its maximum displacement (x=A) when all the mechanical energy of the framework is elastic potential energy, so when the kinetic energy is zero:

K=$\frac{1}{2}mv^2$=0

E=$\frac{1}{2}kA^2$ -->(1)

A=$\sqrt{\frac{2E}{k}}$

c)$v_{max}=\omega A$

When the elastic potential energy is zero, the maximum speed of the system occurs i.e U=0 and the kinetic energy is maximum, so:

U=0

E=$\frac{1}{2}mv_{max}^2$

According to the law of conservation of the mechanical energy, this energy must be equal to the energy of the system at its maximum displacement (1), so we can write

$\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2$

and solving for $v_{max}$we find an expression for the maximum speed:

$v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{k}{m}}A=\omega A$

$v_{max}=\omega A$