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CHAPTER 2SIMPLE HARMONIC MOTION

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Simple harmonic motion, occurs when the acceleration is proportional to displacement but they are in opposite directions.

Simple Harmonic Motion (SHM). The motion that occurs when an object is accelerated towards a mid-point. The size of the acceleration is dependent upon the distance of the object from the mid-point. Very common type of motion, eg. sea waves, pendulums, spring.

Simple harmonic motion occurs when the force F acting on an object is directly proportional to the displacement x of the object, but in the opposite direction.

Mathematical statement F = - kx

The force is called a restoring force because it always acts on the object to return it to its equilibrium position.

2.1 Definition of Simple Harmonic Motion

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a. The amplitude A is the maximum displacement from the equilibrium position.

b. The period T is the time for one complete oscillation. After time T the motion repeats itself. In general x(t) = x (t + T).

c. The frequency f is the number of oscillations per second. The frequency equals the reciprocal of the period. f = 1/T.

d. Although simple harmonic motion is not motion in a circle, it is convenient to use angular frequency by defining = 2pf = 2p/T

2.2 Descriptive terms

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A body at simple harmonic motion if :a)The acceleration always directed toward a fixed

points on their path .b) The acceleration is proportional to its

displacement from a fixed points and always directed toward that points .

Example for simple harmonic motiona) When a mass hanging from the spring is deflected

it will move with simple harmonic motion.b) Motion of the piston in a cylinder is close simple

harmonic motion.c) Weight of the pendulum moves with simple

harmonic motion if the angle is small.

2.3 Simple Harmonic Motion

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Example for simple harmonic motion

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2.4 Simple Harmonic Motion Diagram

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Figure 2.1 (i) shows a point P rotating with constant velocity , in a circular path of radius , a

Linear velocity , V = a

From figure 2.1 (ii) , the horizontal component of velocity V, the velocity of point Q

Vq = V sin = a sin   From space diagram,

PQ = (a2 – x2)

sin = (a2 - x2)a

 

Vq = (a) (a2 - x2)

a= () (a2 - x2)

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When x = 0 , Vq is maximum , Vq maks = (a2 – 0) = a (figure 2.2 i)

When x = a , Vq is minimum Vq min= (a2 – a2) = 0 (figure 2.2 ii)

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If P rotate with constant angular speed , it also has a central acceleration f, goes to center of rotation O.

f = a 2

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(from figure 2.3 ii : vector diagram ),central acceleration of the horizontal component f, the acceleration of point Qfq = a 2 kos

from figure 2.3 iii , kos = x a

  fq = (a 2) x

a 

= x 2

  When x = 0 , fq is minimum

fq min = 02= 0

When x = a , fq is maximum

fq maks = a 2

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Periodic time or period, T is the time taken by the point Q to make a swing back and forth to complete.And that time is equal to the time taken by the OP to turn a rotation 2 radians with angular speed rad/s .

  Periodic time ,T = 2

But , fq = x2 AND = fq

X  T = 2 x = 2 distance fq acceleration

2.5 Periodic Time , Frequency and Amplitud

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Frequency, n is the number of a complete cycle of oscillation in the penetration by the point Q in the second. The unit is the Hertz (Hz), ie one cycle per second.

Frequency , n = Hz 2

n = 1/T =1 Hz

2 (distance/acceleration )Amplitude, a is the maximum displacement of point Q from a fixed point O. The distance , 2a traveled by the point Q is known as a stroke or swing .

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A point moves with simple harmonic motion with the acceleration of 9 m/s2 and velocity of 0.92 m/s when it is 65 mm from the center of travel. Find:

a. amplitude, b. time of the periodic motion

EXAMPLE 1

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A particle moving with simple harmonic motion has a periodic time of 0.4 s and it was back and forth between two points is 1.22 m. Determine :

i. The frequency and amplitude of the oscillation.

ii. Velocity and acceleration of the particle when it is 400 mm from the center of oscillation.

iii. Velocity and maximum acceleration of the movement.

EXAMPLE 2

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A mass of body 1.5 kg moving with simple harmonic motion is towards to the end of the swing. At the time he was at A, 760 mm from the center of oscillation, velocity and acceleration is 9 m/s and 10 m/s2 , respectively. Find:

a)frequency and amplitude of the oscillation,

b) the maximum acceleration and the inertia of the body when it is at the end of the swing,

c) the time has elapsed for it to go and back to A.

EXAMPLE 3

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The figure shows a body of mass M kg supported by a spring of stiffness S N/m . Static spring deflection is d meters, then:

 Mg = Sd

  If the body is in the pull-down x meters from the

equilibrium position O (a fixed point) and then released, so 

Body weight + inertia force = total spring forceMg + Mf = Sd + SxWhere, Mg = Sd Mf = Sx

f = (S/M) x

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The reference S/M is constant for a system under consideration.

So, the acceleration f is proportional to the distance x from the equilibrium position O.

This indicates that the body is moving with simple harmonic motion.

From SHM,f = x2

= So, time periodic , T = 2

Mg = Sd dan =

   

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A body of mass 14 kg is suspended by springs up from the end attached to a rigid support. The body produces a static deflection of 25 mm. It is in the pull down as far as 23 mm and then released. Find:

a. Initial acceleration of the body, b. the periodic time oscillations, c. the maximum spring force,d. the velocity and acceleration of the body

when it is 12 mm from the equilibrium position.

EXAMPLE 4

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Two body, each type per 6.5 kg, suspended on a vertical spring with stiffness 2.45 kN/m.A body is removed and this causes the system to oscillate. Find:

a. The maximum extension spring,b. the periodic time and amplitude of the

oscillation,c. the velocity and acceleration of the mass

when it is at the center of the amplitude, d. the total energy of the oscillations.

EXAMPLE 5 

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2.7 Simple pendulum

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The diagram shows a simple pendulum of length cord, L and weight B with mass M. Amplitude of the oscillation is small, not exceeding 120, the angular displacement and the three forces, the heavy weight, cord tension and inertia forces, is in equilibrium( figure ii) From the force triangle.

  

= = 

    length of arc , x = L (s = r) 

=

for pendulum

f = x2 for SHM

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This shows that the oscillations of a simple pendulum is simple harmonic. 

2 = and = )

 

for SHM 

for pendulum 

Periodic time and frequency of a simple pendulum is independent of the mass of the pendulum and the angle oscillation (if this angle exceeds 120). If is the angular speed linegenerating SHM.

 

angular speed of the pendulum  = (2 - 2) and maximum =  

angular acceleration of the pendulum 

= 2 and maximum = 2 

Periodic time , T = 2 ( )= 2

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The amplitude of a simple pendulum is 70and the periodic time is 5s, find:

a. maximum linear velocity of the pendulum weight and the maximum angular speed of the pendulum cord,

b. maximum linear acceleration of the pendulum weight and maximumangular acceleration of the pendulum cord .

EXAMPLE 6

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A pendulum clock required rhythm second,

with periodic time 2 s, was found late 80 s a

day. The pendulum is shortened so that it

rhythm seconds exactly. Find the difference

in the length of the pendulum clock.

EXAMPLE 7

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2.8 SIMPLE CONE