PHYSICS FORM FIVE-SIMPLE HARMONIC MOTION-DAMPING OF S.H.M

PHYSICS FORM FIVE-SIMPLE HARMONIC MOTION-DAMPING OF S.H.M

UNAWEZA JIPATIA NOTES ZETU KWA KUCHANGIA KIASI KIDOGO KABISA:PIGA SIMU: 07872327719

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PHYSICS FORM FIVE-SIMPLE HARMONIC MOTION-DAMPING OF S.H.M

SIMPLE HARMONIC MOTION

       Before discussing simple harmonic motion, it is desirable to discuss periodic motion and oscillatory motion.

1.  Periodic motion

Is the motion which repeats itself after a regular interval of time.

The regular interval of time is called time periodic of the periodic motion

Example of periodic motion

(i)   The revolution of earth around the sun is a periodic motion. Its period of revolution is 1 year

(ii)    The revolution of moon around the earth is a periodic motion. Its period of revolution is 1 year

(iii)   The motion of the hands of a clock is a periodic motion.

(iv)   Heart beats of person is a periodic motion. It is period of revolution is about 0.83s for a normal person

(v)    Motion of Halley’s Comet around the sun, period is 76 years.

2.   Oscillatory motion (vibratory motion)

is the motion which moves along the same path to and fro about an equilibrium           position.

For a body to oscillate or vibrate three conditions must be satisfied:

(i)  The body must have inertia to keep it moving across the midpoint of its path.

(ii)  There must be a restoring force (elastic) to accelerate the body towards the midpoint.

(iii)  The friction force acting on the body against its motion must be small.

 All oscillatory motions are periodic motions but all periodic motions are not oscillatory

Examples for oscillatory motion

(i)  Oscillation of a simple pendulum

(ii) Vibration of a mass attached to a spring

(iii)  Queering of the strings of musical instruments

PHYSICS FORM FIVE-SIMPLE HARMONIC MOTION-DAMPING OF S.H.M




Restoring force

Restoring force is one that tries to pull or push a displaced object back to its equilibrium position

Simple harmonic motion is a motion of a particle which moves to and fro about s fixed point under the action of restoring force which is direct proportional to the displacement from the fixed point and always directed towards the fixed point.

This fixed point is called mean position or equilibrium position.

It is called mean position because it lies in the middle of the line of oscillation.

It s called equilibrium position because at this point the resultant force acting on the particle is zero

Let the displacement of the particle from the mean position be y and F be the force acting on the particle.

Then

F ∝ -y

F = -ky

 

-ve sign shows that F and y are oppositely directed.

K is called spring constant because the restoring force F has the property of a spring force.

If the motion takes place under a restoring force it is called liner S.H.M. For restoring torque is called Angular S.H.M.


DAMPING OF S.H.M

We can now form a definition of simple harmonic motion . It is the motion of particle whose acceleration is always
(i)directed toward a fixed point,
(ii)directly proportional to its distance from that point

Under the condition when time t=0 the displacement along x-direction  is equal to amplitude A.

MECHANICAL OSCILLATIONS

  1.       Oscillating system – spring and mass

The figure below shows a mass connected to spring whose free end is connected to a rigid support. The mass and the spring are laid on a frictionless horizontal surface. The mass of the spring is assumed to be negligible.

spring-----F5_physics_simple_harmonic_motion1

When the mass is pulled so that it has a displacement X from (the) its equilibrium position then the spring is extended by X there is restoring force. If the spring obeys Hooke’s law then the force is directly proportion to the extension. F acts in opposite direction to  so F = -KX is a constant called the spring or force constant.

 

If M is the mass of the body A, F=Ma so that

p15

The motion of a simple harmonic motion and the period T is given by  p22

The acceleration is always directed forwards the equilibrium position.

  1.       The spiral spring.

spring-----F5_physics_simple_harmonic_motion3

Consider a spiral spring of natural length ‘L’ suspended at the upper end

Let a mass M attached to the lower end extends the spring by ‘e’

Assuming the spring obeys Hook’s law then

mg = Ke Where K is a spring constant.

p32

 (Force needed to produce unit extension of the spring)

Suppose ‘m’ is pulled further down distance ‘x’ from O and then released from A

The restoring force F = -Kx   (Force which produce the extra extension)

Ma = -kx

p43

Harmonic Motion (SHM) about the equilibrium

p51

The period of the oscillation is given by

p62

Note

  1. Oscillating spring is an exact S.H.M
  2. p71 is plotted it should be a straight lime passing through  the origin im practice the graph does not pas through the origin occurring that   the mass of the spring was not taken into account




PHYSICS FORM FIVE-SIMPLE HARMONIC MOTION-DAMPING OF S.H.M

Spring in series and parallel

Consider a situation shows in Fig (a) below where two

Identical springs are in series.

spring

 

The mass applied to the end of the springs until stretch each spring by the same amount as if it mere applied to each separately. There will be the twice extension that would have occurred with just the single spring. For one spring we have “mg = -kx” but total extension         where  force constant for the two spring in series

p81

p91

The simple pendulum

3358

The bob of length L when it is in equilibrium position at O. suppose the bob is iron displace d through angle θ. The position of A the work done in doing so is stored in the system as

P:E = mgh at A.

If the bob is now released from A it swing’s towards O changing all it PE F in mg to KE owing to the KE of the bob at O the bob over shoots from O and swing to A until all its KE change into P:E at A hence  the pendulum swings to and from indefinitely assuming no resistance to its motion

Suppose any movements of the right of O are positive suppose also the bob is released from A therefore that tend to restore the bob back to O is F = -mg sinÆŸ ( newton’s  second law of motion)

F is acting along the tangent to the Arc AO

ma = -mg sinÆŸ

a ≃ -gÆŸ (as ÆŸ → 0 sin ÆŸ = ÆŸ)

p101

NOTE

p111

 is a position constant…… acceleration  is directly proportional to the displacement from a fixed point 0

When X is positive ( between O and A) acceleration is negative ie. It is directed towards O

When X is negative ( between O and A’ ) the acceleration is positive ie. Directed towards ÆŸ

Thus the bob exciting simple (pendulum) ……………. C motion with angular velocity

p121

This is only on approximate simple harmonic motion which is true only when Ó¨ is very small

If T and L are measured g can be calculated




PHYSICS FORM FIVE-SIMPLE HARMONIC MOTION-DAMPING OF S.H.M

LIQUID  COLUMN

3357

Consider liquid column of length 2h containing in a U-tube of cross sectional area A

At equilibrium the liquid levels O and O’ will be along the same horizontal plane

If the surface is depressed through depth X to the A by blowing into the tube and then left it

The restoring force is =-2xAρg where ρ is the density of liquid.

From the 2nd Newton’s law of motion

p131

FLOATING CYLINDER

Consider a cylinder of cross section area a floating up right in a liquid of density ρ such that it is submerged to a depth L

The cylinder should be in equilibrium

Of the mass of the cylinder equals to the mass of liquid displaced.

CYLINDER


p151

Suppose the cylinder is pushed.  +VE direction further down through a extra (angle) distance X them released. The extra up thrust will tend to restore the cylinder to equilibrium assuming displacement vector from downwards are positive

p16

NOTE

  1.       This is an example of exact of
  2.       L is not the long of the cylinder column but depth submerged

Relation between linear S.H.M and uniform circular motion

A particle moving around a circle with constant speed is said to be in uniform circular motion.In uniform circular motion the speed remains constant but the velocity changes due to the change in direction. Hence the particles accelerate.

(i) They are both periodic motions
(ii)  Their accelerations are directed towards a fixed point i.e circular motion is directed towards center of a circle while simple harmonic motion is directed towards the mean positions.

Consider a particle P moving along the circumference of a circle of radius A with a uniform angular velocity w as shown in figure 1

3360


Fig 1. Description of S.H.M

O is called the mean position or equilibrium position

A is the maximum displacement of the particles executing S.H.M on either side of the equilibrium position.

that when the particle reaches point P, the displacement is maximum OP = A = radius of reference circle.

        The amplitude is equal to the radius of the reference circle.

         The displacement of a body which executes simple harmonic motion can be expressed in terms of x and y.

The displacement of a particle executing simple harmonic motion at any instant is the distance of the particle from the equilibrium position at that instant.

image003

image004

            image005

image006

 is the angular displacement, t is the time taken by the body to oscillate from point O to P and describe an angular displacement image007

image008

image009

    The displacement can be represented by the relation
    image010

              image011

Where A is the amplitude of oscillation


Fig. 2

FIG21


From fig.2

At equilibrium position y =0

The positions where y =  +A are called the extreme positions

Example oscillation of simple pendulum at t = 0, the body is at the maximum reach

2216


t = 0

image021

image022

image023

  Example of  oscillation of piston rings in engine cylinder

y image025


t=0

y = image026

y = image027

y = 0




PHYSICS FORM FIVE-SIMPLE HARMONIC MOTION-DAMPING OF S.H.M

VELOCITY OF A BODY EXECUTING S.H.M

The velocity of a particle executing S.H.M at any instant is the time rate of change of its displacement at that instant

Since the displacement of a S.H.M is a function of time, therefore its velocity will be a function of time (Instantaneous velocity)

V = image028


for image029

V = image030

V = image031

V = image032

Also

V = image033


for y = image034

V = image035

image036

image037

Consider the right angled triangle OPN in fig. 1

1221

image039


image040

2311

From equation (i)

V = image043

V = image044

From the triangle OPN

x2 + y2 = A2

image045

image046

————————– (iii)

Also

From equation (ii)

V = image047

V = image048

V = image049

From the triangle ONP

image050

image051

image052

—————–(iv)

 

Equation (i) and (ii) represents velocity as a function of time

Both equation (iii) and (iv) represents velocity as a function of displacement
Velocity of a particle executing S.H.M at any instant is

V = image053

At equilibrium position y = 0

image054

image055

 = image056




PHYSICS FORM FIVE-SIMPLE HARMONIC MOTION-DAMPING OF S.H.M

ACCELERATION OF A BODY EXECUTING S.H.M

The acceleration of a particle executing S.H.M at any instant is the time rate of change of velocity at that instant.

a = image057

 

247


a = image059

v =  image060

a = image061

a = image062


image029

a = image063

     2511

For y = image034

2612

At equilibrium position y = 0

a = image068

                           a = 0

At the extreme positions

y = + A

a = image068

a = image069

   (This is expression for maximum)
The maximum value of acceleration is called acceleration amplitude in S.H.M

The – ve sign means that the acceleration and displacement are directed in opposite direction ensuring that the motion is always directed to the center (fixed point)

GRAPHICAL REPRESENTATION OF SIMPLE HARMONIC MOTION
274

image070

Fig3.(a)

287

 

4418


Fig. 3(b)

298

4419

Fig 3.(c)

The displacement velocity and acceleration all vary sinusoidal with time but are not in phase




PHYSICS FORM FIVE-SIMPLE HARMONIC MOTION-DAMPING OF S.H.M

ENERGY OF A BODY EXECUTING S.H.M

A harmonic oscillation executes S.H.M under the action of a restoring force.
This force always opposes the displacement of the particle. So to displace the particle against this force work must be done.

The work done is stored in the particle in form of potential energy (P.E), as the particle is in motion it has kinetic energy (K.E)

The sum of P.E and K.E is always a constant provided that part of this energy is not used to overcome frictional resistance.

Expression for kinetic energy

K.E = ½ mv2

304

The kinetic energy as a function of displacement

3114

The K.E is maximum when its velocity is maximum

image055

 = image074
                 (At midpoint)

            K.E = ½ mω2A2

The K.E as a function of time

V = image075


image076

K.E = image077

Or

K.E = image078

Expression for potential energy

The P.E is the energy possessed by the body due to its position

By Hooke’s law

3211

The work done to extend the spring from  x = 0 to x = xo

336

image081

image082

image083

The P.E of a string extended by displacement x is given by

                P.E = image084

 

F = image085

395

345

image087

 = image088

image089

 = image090

355

Total energy in S.H.M

The total energy ET of the particle per displacement y is given by

image093

  image094
+P.E

385

386

Also

403

 

image096

 = image097

4111

ET = image099

ET = 2 image100




GRAPHS OF K.E VS TIME

For

image101

  image102

V = image060

Then

K.E = ½ m image096

2A2Sin2 image096
t

427

Fig. 4

GRAPH OF P.E Vs TIME

434

Fig 5.

Since

P.E = ½ m image096

2A2Cos2 image096
t

PHYSICS FORM FIVE-SIMPLE HARMONIC MOTION-DAMPING OF S.H.M

ENERGY EXCHANGE

The P.E and K.E for a body oscillating in  S.H.M causes the motion of the body.

image107


Fig.6 Energy of S.H.M

From the figure 6, the total energy of  vibrating system is constant. When the K.E of the mass m is maximum (energy= 1⁄2 mω2 A2 and mass passing through the centre O), the PE of the system is zero (x=0). Conversely, when the P.E of the system is a maximum (energy=1⁄2KA2=1⁄2 mω2 A2  and mass at end of the oscillation), the K.E of the mass is zero (V=0).

SOLVED PROBLEMS

1.The restoring force acting on a body executes simple harmonic motion is 16N when the body is 4cm away from the equilibrium position. Calculate the spring constant

Solution

Restoring force F = 16N

Displacement y =4 cm = 4 x 10 -2

Spring constant K =?

F = image108

K = image109

K = 400 Nm-1

K = 400 Nm -1

2.      A body is executing simple harmonic motion with an amplitude of 0.1m and frequency 4 Hz. Compute

(i)   Maximum velocity of the body

(ii)  Acceleration  when displacement is 0.09m and time required to move from mean position to a point 0.12 away from it

Solution

Amplitude A = 0.15m

Frequency f = 4Hz

Angular velocity = 2πf = 8 π image110

(i)    Maximum velocity of the body

V = A image096

= 0.15 x 8 π image110

V = 3.768 m/s

(ii)             Acceleration  a = – image111

= image112

                      = – 56.79 m/s2

The negative sign shows that the acceleration is directed towards the equilibrium position.

593

3.     A particle executes S.H.M with amplitude of 10 cm and a period of 5s. Find the velocity and acceleration of the particle of a distance 5cm from the equilibrium position

       Solution

A = 10 cm   = 10 image113

 10 -2

T = 5s           y = 5 cm = 5 image113

 10 -2m

Velocity V = image053

= image114

V = image115

V = 10.88 m/s

Acceleration a = image044

= image116

a = image117

image118

a= – 4 x 3.142 x 5 x 10 -2/25

                       a = 0.079m/s2

 

4.   A bob executes simple harmonic of period 20s. Its velocity is found to be 0.05m/s after 2s when it has passed through its mean position. Find the amplitude of the bob.

Solution

T = 20s   v = 0.05 m/s    t = 2s

image096

 = image119

V = image056

cos image120

0.05 = A image113

 0.314 image121
(0.314 image113
 2)

A = 0.16m





5.     A body describes S.H.M in a line 0.04m long. Its velocity at the center of the line is 0.12 m/s. Find the period also the velocity

Solution

Length of the line image122

A = 0.02m

image055

 = 0.12 m/s

image056

 = 0.12

A image123

 = 0.12

T = image124

 

T = 1.046S

Velocity at a displacement y = 10 -2 image125

V = image053

V= image114

V = image126

V = 0.06 m/s

6.      In what time after its motion began will a particle oscillating according to the equation
image127

(0.5 πt) move from the mean position to the maximum displacement?

Solution

image127

(0.5 image128
)

Maximum displacement image129

Time taken to move from the mean position to the extreme position is to be found out when image130

   t=?

image131

= image132
0.5πt

image133

 = 1

0.5πt = image134

0.5πt = image135

t = 1s

7.      A particle with a mass of 0.5 kg has a velocity of 0.3 image136

 after 1s starting from the mean position. Calculate the K.E and Total energy if its time period is 6s.

Solution

m   = 0.5 kg
T = 6s

V = 0.3 image136

t = 1s   image096

 = image137
 = image138

Velocity v = image139

V image140

image141

    A = 0.57m

image143

             K.E = 0.0225J

image145

image146


603

 



8.      A period of a particle executing S.H.M is 0.0786J. After a time π/4 s the displacement is 0.2m. Calculate the amplitude and mass of the particle.

Solution

T =2 π
t = image148


y = 0.2m

ET = 0.0786J

image096

 = image137
= image149
  = image150

Displacement after a time t = image151

 sec   is y = 0.2 m

y = image152

0.2m = image153

A = image154

A = 0.283m

ET = image155

0.04 m = 0.0786

m = 1.96 kg

9.      A simple harmonic oscillation is represented by

image156

 = 0.34cos (3000t + 0.74)

Where x and t are in mm and sec respectively. Determine

(i) Amplitude

(ii) The frequency and angular frequency

(iii)  The time period

Solution

image156

 = image157
 (3000t + 0.74) ———————– (i)

The standard displacement equation of S.H.M

image156

 = image158
 ( image096
t + image159
) ——————————- (ii)

Comparing equation (i) and (ii)

(i)  Amplitude A = 0.34 m

(ii)   Angular frequency w = 3000 image110

image096

 = 2π image160

image161

= image162

image163

 = image164
  Hz

(iii)           Time period T = image165

483




10. An object executes S.H.M with an amplitude of 0.17m and a period of 0.84s.Determine

(i)     The frequency

(ii)     The angular frequency of the motion
(iii) Write down the expression for the displacement equation

Solution

Amplitude A = 0.17 M

Period T = 0.84 s

(i)                image160

 = image167
 = image168
 = 1.19 Hz

(ii)             image169

= 2π image160

= 2π x 1.19

= 2.38 x 3.14

image096

= 7.48 Hz

The displacement equation of S.H.M is

image156

 = image170
( image096
t + image159
)

image156

 = 0.17 image171
(7.5t + 0)

494

11.The equation of S.H.M is given as image156

 = 6sin 10πt  + 8Cos 10πt where image156
 is in cm and t  in second Find

(i)       Period

(ii)       Amplitude

(iii)       Initial phase of motion

Solution

image156

 = 6 image174
  + 8 image175
  ———————- (i)

The standard displacement equation of S.H.M

image156

 = image170
 ( image096
t + image159
)

image156

 = image176
( image177
t image178
 + image179
  image180
)

image156

 = image181
 + image182
  image180
 ————— (ii)

Comparing equation (i) and equation (ii)

image120

 = 10πt

image096

 = 10π

image183

 = 6 image184

image185

 =6 ———————————— (iii)

image186

 =8 image175

image182

 = 8 ——————————— (iv)

(i)    Period

image096

 = image137

T = image187

T = 0.2 s

 

(ii)             Amplitude

Squaring equation (iii) and (iv) then add

image188

image189

image190

A = 10cm

(iii)Initial phase angle

image192

image193

image194

image195

12.The periodic time of a body executing S.H.M is 2sec. After how much time interval from t = 0 will its displacement be half of its amplitude.

Solution
  y = image196

image197

Here,
image198

           image199

image200

image201

image202

image203

503




13. A particle executes S.H.M of amplitude 25cm and time period 3sec. What is the minimum time required for the particle to move between two points 12.5 cm on either side of the mean position?

Solution

Let t be the time taken by the particle to move from mean position to a point 12.5cm from it

y = image205

t

5110

y = 12.5 cm,
A = 25 cm,
T = 3s

524

535

image208

 = sin-1 0.5

image209

 = image210

t = image211

Required time = 2t

= image212

Required time is 0.5 s

14. A particle in S.H.M is described by the displacement function

image156

 = image158
( image096
t + image159
)            image096
 = image137

If the initial (t = 0) position of the particle is 1cm and its initial velocity is π cm/s. What is its amplitude and phase angle? The angular frequency of the particle is π s-1

Solution

544

At   t = 0

image156

 = 1cm
v = π cm/s
w = π/s

1 = image214

(0 + image159
)

1 = image185

 ————————-(i)

V = image028

   = image215
 ( image096
t + image159
)

π   = image216

 (0 + image159
)

-1 = image170

  image159
————————(ii)

Squaring and adding equation (i) and (ii)

image217

 = image218

A2 = 2

A = √ 2

A = 1.41421 m

15.The time period of a particle executing S.H.M is 2 seconds and it can go to and fro from equilibrium position at a maximum distance of 5cm, if at the start of the motion the particle is in the position of maximum displacement towards the right of the equilibrium position, then write the displacement equation of the particle.

Solution

The general equation for the displacement in S.H.M is

y = image219

t + image159
)

image096

 = image137
 = image220

image096

 = image221

y = 5cm
A = 5 cm,
At,    t =0

556

1 = sin image159

image159

 = image135

y = image224

565

16 .In what time after its motion begins will a particle oscillating according to the equation  y = image226

 move from the mean position to maximum displacement?

Solution

y = image226

The standard displacement equation of S.H.M is

y = image227

Comparing equation (i) and (ii)

A =7 image096

 =0.5πt

Let t be the time taken by the particle in moving from mean position to maximum displacement position

y = A =7

y = image228

y = A = 7

7 = image226

image229

 = 1

0.5πt = sin -1 1

0.5πt = image135

image230

 = image231

t = 1 s

17.The vertical motion of a huge piston in a machine is approximately simple harmonic with a frequency of 0.5 s -1. A block of 10 kg is placed on the piston. What is the maximum amplitude of the piston’s executing S.H.M for the block and piston to remain together?

Solution

The block will remain in contact with the piston if maximum acceleration ( image232

) of S.H.M does not exceed g image233
 is at most equal to image234

image232

 = image087
A

573

A = image236

A = image237

A = 0.994 m

18. A particle executing S.H.M has a maximum displacement of 4cm and its acceleration at a distance of 1 cm from the mean position is 3m/s2. What will be its velocity when it is at a distance of 2 cm from its mean position

Solution

image238

 =  – image111

image238

 =3cm/s2
A = 4cm

y1=1cm
y2 = 2 cm

image096

 = image239

w = 1.78 image110

The velocity of a particle executing SHM is given by

V = image053

V = image240

V = 6 cm/s

19.A particle executing S.H.M along a straight line has a velocity of 4m/s when its displacement from mean position is 3m and 3m/s when the displacement is 4m. Find the time taken to travel 2.5m from the positive extremity of its oscillation.

Solution

V = image053

For the first case

V = 4m/s

y = 3m

V2 = image241

16 = image242

  —————– (i)

For the second case

V = 3m/s

y = 4m

V2 = image242

9 = image242

  ——————– (ii)

Take equation (i) divide by equation (ii)

image243

 = image244

image243

 = image245

9A2 -81 = image246

7A2 = 175

A2 = image247

 = 25

A = image248

A =5m

From equation (i)

16 = image249

16 = image250

16 = image251

image087

 =1

image096

 = 1 image110

When the particle is 2.5m from the positive extreme position, its displacement from the mean position is

y = 5 – 2.5

y = 2.5 m

Since the time is measured when the particle is at extreme position

y = image102

2.5 = image252

Cos t = image253

Cos t = 0.5

t = 0.5

t = cos -1 (0.5)

t = image254

t = 0.1 s



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