Lesson 8, Topic 4
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Energy of Simple Harmonic Motion

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Summary:

When force and displacement are involved in SHM, energy or work done is also involved.

The system, at any instant of the motion, may contain some energy as Kinetic energy (K.E) or potential energy (P.E) or both.

Consider the image above;

• The pendulum swings from the highest point through the centre of the swing to the other highest point.
• Let’s call the first highest point A, the centre O, and the other highest point B.
• The pendulum, swings from, A through, O to B
• At, point O, the Bob is at the lowest position, i.e height, h = 0, therefore the potential energy of the system is zero.
• Also at O, the speed of the pendulum, vmax is maximum and so the kinetic energy $$\left(\frac{1}{2} \scriptsize mv_{max}^2 \right)$$ is maximum at this point.
• As the bob moves from O to B, the kinetic energy at O is gradually transformed to potential energy with the potential energy becoming maximum at B, which is at a height “h” above O
• At point B, the energy is potential energy and is equal to “mgh”. Where m is the mass of the bob, h is the height above the lowest level and g is the acceleration due to gravity.
• This is also the potential energy at “A” which is the same height above O, as B is.

The K.E at A would be minimum, but maximum at O.

At O, h = 0, P.E = 0, K.E and Vmax are maximum

K.Emax = $$\frac{1}{2} \scriptsize mv_{max}^2$$

But from circular motion, V = ωr

∴ K.Emax = $$\frac{1}{2} \scriptsize m \left(\omega^2 r^2 \right)$$

At B, or A, the P.E is maximum as a result of the height to which the bob is raised when displaced.

Maximum P.E = mgh

From the principle of conservation of energy, the total energy must be constant.

∴ K.Emax = P.Emax

∴ mgh = $$\frac{1}{2} \scriptsize mv_{max}^2$$

or vmax = $$\scriptsize \sqrt{2gh}$$

If a mass is suspended from the end of a spring, stretched downwards vertically and released and it undergoes SHM. The force required to stretch the string through distance x is;

F = kx

where k is the force constant of the spring.

The workdone in extending the spring a distance x, is;

W = $$\scriptsize average \: force \: \times \: displacement$$

= $$\frac{1}{2} \scriptsize kx \: \times \: x \\ = \frac{1}{2} \scriptsize kx^2$$

If A = x, (A is the amplitude of oscillation) then,

W = $$\frac{1}{2} \scriptsize kA^2$$

At any stage of the oscillation, i.e any position other than A, O, B:

At D,

total energy, W = $$\underset{K.E} {\normalsize \frac{1}{2} \scriptsize mv^2} \: + \: \underset{P.E}{\normalsize \frac{1}{2} \scriptsize kx^2}$$

∴ $$\underset{total\: energy}{\normalsize \frac{1}{2} \scriptsize kA^2} = \underset{K.E} {\normalsize \frac{1}{2} \scriptsize mv^2} \: + \: \underset{P.E}{\normalsize \frac{1}{2} \scriptsize kx^2}$$

where v is the velocity of the suspended mass at a point x from the equilibrium position.

By making v the subject of the formula, we have;

v = $$\sqrt{\normalsize \frac{k}{m} \scriptsize (A^2 \: -\: x^2)}$$

At any position other than A, O or B, position = A – x

where x is the distance of mass from equilibrium position and A (amplitude) is the maximum displacement.

At the extreme position , i.e at x = A the block has only potential energy:

∴ $$\scriptsize W = \frac{1}{2} \scriptsize kx^2 \\ \scriptsize or \: W = \: \frac{1}{2} \scriptsize kA^2 \\ \scriptsize or \: \underset{total\: energy}{\normalsize \frac{1}{2} \scriptsize kA^2} \scriptsize = P.E_{max}$$

At the equilibrium position, i.e at x = O, the block has only kinetic energy:

∴ W = $$\frac{1}{2} \scriptsize mv_{max}^2$$

Now we know, vmax = ωA ,so at x = O

W = $$\frac{1}{2} \scriptsize m \omega^2 A^2$$

Example 1:

A force of 300N is used to stretch a horizontal spring with a 0.5kg block attached to it. The block is then released from rest and undergoes simple harmonic motion. Calculate the;
(a) spring constant
(b) amplitude
(c) maximum accelration
(d) total work done
(e) maximum velocity
(f) velocity when x = 0.15m

Solution:

(a)

F = 300N

x = 0.25m

F = kx

k = $$\frac{F}{x}$$

k = $$\frac{300}{0.25}$$

k = 1200 N/m

(b)

from the diagram above

$$\scriptsize -0.25 \leq x \leq 0.25$$

∴ the maximum displacement, A = 0.25

(c) amax = $$\frac{kA}{m}$$

where A is the maximum displacement or amplitude

∴ amax = $$\frac{1200 \: \times \: 0.25}{0.5}$$

∴ amax = $$\scriptsize 600 \: m/s^2$$

(d) W = $$\frac{1}{2} \scriptsize kA^2$$

W = $$\frac{1}{2} \scriptsize \: \times \: 1200 \: \times \: (0.25)^2$$

W = 37.5 J

(e)

At the equilibrium position, i.e at x = 0, the block maximum kinetic energy, maximim velocity and zero potential energy.

∴ W = $$\frac{1}{2} \scriptsize mv _{max}^2$$

but W = $$\frac{1}{2} \scriptsize kA^2$$

∴ $$\normalsize\frac{1}{2} \scriptsize kA^2 = \normalsize \frac{1}{2} \scriptsize mv _{max}^2$$

make vmax the subject of the formula;

$$\scriptsize mv _{max}^2 = \normalsize \frac{\not{2}}{\not{2}} \scriptsize kA^2$$

$$\scriptsize v _{max}^2 = \normalsize \frac{kA^2}{m}$$

$$\scriptsize v _{max}^2 = \normalsize \frac{1200 \: \times \: 0.25^2}{0.5}$$

$$\scriptsize v _{max}^2 = 150$$

$$\scriptsize v _{max} = \sqrt{150}$$

$$\scriptsize v _{max} = 12.25 \: m/s$$

(f) At any stage of the oscillation, i.e any position other than equilibrium and the two extremes;

W = K.E + P.E

∴ at position D (x = 0.15m)

$$\underset{total\: energy}{\normalsize \frac{1}{2} \scriptsize kA^2} = \underset{K.E} {\normalsize \frac{1}{2} \scriptsize mv^2} \: + \: \underset{P.E}{\normalsize \frac{1}{2} \scriptsize kx^2}$$

make v the subject of the formula

v = $$\sqrt{\normalsize \frac{k}{m} \scriptsize (A^2 \: -\: x^2)}$$

v = $$\sqrt{\normalsize \frac{1200}{0.5} \scriptsize (0.25^2 \: -\: 0.15^2)}$$

v = $$\sqrt{ \scriptsize 2400 \: \times \: (0.0625 \: -\: 0.0225)}$$

v = $$\sqrt{ \scriptsize 2400 \: \times \: 0.04}$$

v = $$\sqrt{ \scriptsize 96}$$

v = $$\scriptsize 9.8 \: m/s$$

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