8. 1 Characteristics of Simple Harmonic Motion

> Simple harmonic motion is the periodic motion in which the acceleration of the body is

-directly proportional to its displacement from a fixed point and

– always directed towards that point.

> The equation relating acceleration and displacement can be written as

a **a** -x or a = -wx (Note that w = 2pf)

**8.2 Kinematics of Simple Harmonic Motion**

Simple harmonic motion equation in terms of time

x = x_{0} sin wt

Note that x is a sinusoidal function and the equation can be written as x = x_{0} cos wt or more generally as x = x_{o} sin (wt + q)

Simple Harmonic Motion equation in term of displacement

Derivation

x = x_{0} sin wt

v = wx_{0} cos wt = ± wx_{0} = ±w

a = -w^{2}x_{0} sin wt = -w^{2}x

a = -w^{2}x

Example 1

simple pendulum oscillates with period 2.0 s and amplitude 50 mm. Calculate

(a) Its frequency.

(b) Its acceleration when it is 20 mm from the mid point.

(c) Its maximum speed.

Solution

Example 2

particle moves with simple harmonic motion has an amplitude of 0.20 m and period 12 s.

Find

(a) The maximum speed.

(b) The maximum acceleration of the particle.

Solution

Example 3

A body of mass 0.30 kg moves with simple harmonic motion in a ,straight line. The relation between the force F acting on the body and its displacement x is shown in figure above. Find

(a) The amplitude of the motion.

(b) Its period.

(c) The maximum speed of the body.

Solution

**8.3 Energy in Simple Harmonic Motion**

Variation of energy with displacement

Total energy, E = Maximum kinetic energy

Potential energy,

U = Total energy – kinetic energy

Variation of energy with time

Kinetic Energy,

potential Energy,

P = mw^{2}x_{0}^{2} sin^{2} wt

**8.4 System in Simple Harmonic Motion**

> Procedure to show that an oscillating system is in simple harmonic motion:

– Find the restoring force for a given displacement x from the equilibrium position.

– Apply Newton’s second law to obtain the equation of motion.

– If this equation falls into the form a = -w^{2}x the motion must be simple harmonic.

– Furthermore, as the constant value w^{2} has been found, the period T can be determined with relation T = .

Simple Pendulum

• Restoring force = mg sin q

• Applying Newton’s Second Law

ma = -mg sin q (negative sign because the force is directed towards O and opposite the direction of displacement)

• When q is small, sin q » q

ma = -mgq

i.e. acceleration directly proportional to x, the motion is simple harmonic.

Note: The value of g can be determined by measuring T for different values of l and plotting a graph of T^{2 }against l

Mass attached to a horizontal spring

• When the spring is pulled with extension x and released,

Restoring force = kx

• Applying Newton’s Second Law

ma = -kx (negative sign because the acceleration a is to the left while the displacement x is to the right)

• ma =-kx

i.e. acceleration directly proportional to x, the motion is simple harmonic.

Mass on a spring

when the spring is in equilibrium,

mg = ke

When the spring is pulled with extra extension x and released,

Restoring force = k(e + x) – mg

= k(e + x) – ke = kx

Applying Newton’s Second Law

ma = -kx (negative sign because the acceleration a is upwards while the displacement x is downwards)

ma = -kx

i.e. acceleration directly proportional to x, the motion is simple harmonic.

Note: Since mg = ke

Mass attached to two horizontal spring with spring constant k_{1} and k_{2} When the springs are in equilibrium,

· When the spring is pulled to the right with displacement x and released,

T_{1} = T_{2} and k_{1}e_{1 }= k_{2}e_{2}

· When the spring is pulled to the right with displacement x and released.

Restoring force = T_{1}‘- T_{2}‘

= k_{1} (e_{1} + x) – k_{2}(e_{2} – x)

= k_{1}e_{1} + k_{1}x – k_{2}e_{2} + k_{2}x

= (k1 + k2)x

\k_{1}e_{1} = k_{2}e_{2}

· Applying Newton’s Second Law

ma = -(k_{1} + k_{2})x (negative sign because the acceleration a is to the left while the displacement x is to the right)

· ma = -(k_{1} + k_{2})x

i.e. acceleration directly proportional to x, the motion is simple harmonic.

Example 4

A light spiral spring having spring constant k = 10 N m^{-1} and loaded with a mass of 150 g is 1 oscillating in simple harmonic motion with amplitude 5.0 cm. Calculate

(a) The angular frequency.

(b) The period of oscillation.

(c) The total energy of the oscillating system.

(d) The kinetic energy of the particle when the displacement is 2.0 cm.

Solution

Example 5

A man who has a mass of 75 kg is doing a bungee jump. When he is at the bottom of the bungee jump, he bounces up and down on the end of the bungee rope through an amplitude of 2.0 m. The rope has a spring constant of 240 N m^{-1}. Assuming the bouncing is simple harmonic motion, calculate :

(a) The period of oscillation.

(b) The maximum acceleration during a bounce.

Solution

Filed under: Uncategorized

## Leave a Reply