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

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

 

APPLICATION OF S.H.M

We shall consider the following cases of S.H.M

i)   Oscillation of a Loaded Spring

ii)  Oscillation of a Simple Pendulum

iii  Oscillation of a Liquid in a U – tube

iv)  Oscillation of a Floating Cylinder

v)    Body Dropped in a funnel along earth diameter

vi)    Oscillation of a ball placed in the Neck of Chamber Containing air

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

Oscillations of a Loaded Spring

If load attached to a spring is pulled a little from its mean position and then released the load will execute S.H.M

We shall consider the following two cases

1.      Vibrations of a Horizontal spring

2.      Vibrations of a Vertical spring

VIBRATIONS OF A HORIZONTAL SPRING

Consider a block of mass M attached to one end of a horizontal spring whi9le the other end of the spring is fixed to a rigid support

 

horizontal_spring1

Fig. 7

The Block is at rest but is free to move along a friction less horizontal surface

In figure 7 is displaced through a small distance x to the right, the spring gets stretched

BLOCK

Fig.8




According to Hooke’s law, the spring exerts a restoring force F to the left given by

F  = image085

………………………………. (i)

Here k is the force constant (spring constant) and image156

 is the displacement of mass m from the mean position.

Clearly equation (i) satisfies the condition to produce S.H.M

If the block is released from the displaced position and left , the block will execute S.H.M

The time period (T) and frequency (f) of the vibrations can be obtained from

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

F = image085

Ma = image085

image238

 = image257

image258

image259

image260

image261

From

image262

image263


2.      VIBRATION OF A VERTICAL SPRING

Consider unloaded vertical spring of spring constant k

 

image264

Fig. 9

 

Suppose the spring is loaded with a body of mass m and extended from its original length to an extension ‘e’

 

By Hookes law

mg = image265

Now suppose the load is displaced down to distance x and then released. The applied force is given by

F = image266

When realized the applied force is opposed by gravity force (weight)

Net result force = F – W

Ma = image267

Ma = image268

Ma = image085

a = image269

image096

2 = image088

image270

image271

image272

From

image263

image273

The period of oscillation depends on mass of the loaded body and the spring constant.

In many practical situation springs are connected in series as well as in parallel.




SERIES AND PARALLEL CONNECTION OF SPRINGS

1.  PARALLEL

Consider two  springs of spring constant K1 and K2 arranged in parallel and then both loaded with a body of mass m as shown in fig. 10

image274

Fig. 10

Suppose this body is displaced from its equilibrium position and the extension is x for the system to remain horizontal
image275

image276

image277

Let F1 and F2 be the restoring forces acting on the springs.

Total force = F1+ F2

ma = -K1 image156

 + (-K2 image156
)

ma = image278


Thus, the springs will execute S.H.M

From

633

4420

653

663

TP = Periodic time for parallel connection 

 

If     image283

 (for identical spring)

673

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

From

image263

682

OR

For

image283

 (for identical spring)

693

Alternative arrangment of parallel connection formula
715

Therefore if two identical sprinngs are arranged in parallel , their frequency increases by the factor of 724

; since
733
    is  the frequency for a single spring  for n-identical springs their frequency increases by a factor of 743
.

Consider two springs S1and S2 of force constants   k1 and k2 attached to a mass m and two fixed supports as shown in figure 11.

When the mass pulled downward, then length of the spring S1 will be extended by x while that of spring S2 will be compressed by x

Since the force constants of the two springs are different the restoring force exerted by each spring

 

Let F1 and F2 be the restoring forces exerted by springs S1 and S2 respectively

image315

                       image316

Both the restoring forces will be directed upward (opposite to displacement)




The resultant restoring force F

Fig .11

springx-----F5_physics_simple_harmonic_motiona

803

Effectively force constant of the system

834

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

2.    SERIES CONNECTION

Consider two springs S1 and S2 of force constant  K1, and K2 connected in series as shown below

702

Fig. 12

 

If x is the total displacement

image288

 

Where x1 is extension due to spring of spring constant k1 and x2 to that of k2

F1 = image289

F2 = image290

F1 = F2  =   F

image291


image292

image293

image288

image294

image295


image296

The effective spring constant image297

image298

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

image299

image300


image301

image302

image303

image304

4421

 = periodic time for series connection.
753

 

      For identical spring
763




Also since,
772


782

792

image311

image312

image313

image314



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