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Chapter 8 Simple Harmonic motion

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

clip_image002

Simple harmonic motion equation in terms of time

x = x0 sin wt

clip_image004

clip_image006

Note that x is a sinusoidal function and the equation can be written as x = x0 cos wt or more generally as x = xo sin (wt + q)

Simple Harmonic Motion equation in term of displacement

Derivation

x = x0 sin wt

v = wx0 cos wt = ± wx0 clip_image008 = ±wclip_image010

a = -w2x0 sin wt = -w2x

v = ±w clip_image010[1]

a = -w2x

clip_image012

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

clip_image014

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

= clip_image016

= clip_image018

T = clip_image020

Kinetic energy, K = clip_image022

= clip_image024m(w clip_image010[2])2

= clip_image024[1]mw2(x02-x2)

Potential energy,

U = Total energy – kinetic energy

= clip_image020[1]clip_image024[2]mw2(x02-x2)

= clip_image027

clip_image029

Variation of energy with time

Total energy, E = clip_image020[2]

Kinetic Energy,

K = clip_image022[1]

= clip_image024[3]mw2x02 cos2 wt

potential Energy,

P = clip_image024[4]mw2x02 sin2 wt

clip_image031

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 = -w2x the motion must be simple harmonic.

– Furthermore, as the constant value w2 has been found, the period T can be determined with relation T = clip_image033 .

Simple Pendulum

clip_image035

• 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

= -mgclip_image037­

a = -g clip_image037[1]

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

•w2= clip_image040

T = clip_image033[1]= 2pclip_image043

Note: The value of g can be determined by measuring T for different values of l and plotting a graph of T2 against l

Mass attached to a horizontal spring

clip_image045

• 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

a = – clip_image047x

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

• w2 = clip_image047[1]

T = clip_image033[2]= 2pclip_image051

Mass on a spring

clip_image053

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

a = –clip_image047[2]x

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

w2 = clip_image047[3]

T = clip_image033[3]= 2pclip_image051[1]

Note: Since mg = ke

clip_image057= clip_image059

T = 2pclip_image051[2]

= 2pclip_image061

Mass attached to two horizontal spring with spring constant k1 and k2 When the springs are in equilibrium,

clip_image063

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

T1 = T2 and k1e1 = k2e2

clip_image065

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

Restoring force = T1‘- T2

= k1 (e1 + x) – k2(e2 – x)

= k1e1 + k1x – k2e2 + k2x

= (k1 + k2)x

\k1e1 = k2e2

· Applying Newton’s Second Law

ma = -(k1 + k2)x (negative sign because the acceleration a is to the left while the displacement x is to the right)

· ma = -(k1 + k2)x

· a = – clip_image067

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

· w2 = clip_image069

T = clip_image033[4]= 2pclip_image071

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

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