PHYSICS FORM FIVE-SIMPLE HARMONIC MOTION-DAMPING OF S.H.M
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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
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
- 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.
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
The motion of a simple harmonic motion and the period T is given by 
The acceleration is always directed forwards the equilibrium position.
- The spiral spring.
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.
(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
Harmonic Motion (SHM) about the equilibrium
The period of the oscillation is given by
Note
- Oscillating spring is an exact S.H.M
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
Spring in series and parallel
Consider a situation shows in Fig (a) below where two
Identical springs are in series.
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
The simple pendulum
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)
NOTE
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
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
LIQUID COLUMN
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
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.
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
NOTE
- This is an example of exact of
- 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
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.
is the angular displacement, t is the time taken by the body to oscillate from point O to P and describe an angular displacement 
The displacement can be represented by the relation
Where A is the amplitude of oscillation
Fig. 2
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
t = 0
Example of oscillation of piston rings in engine cylinder
y 
t=0
y = 
y = 
y = 0
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 = 
for 
V = 
V = 
V = 
Also
V = 
for y = 
V = 
Consider the right angled triangle OPN in fig. 1
From equation (i)
V = 
V = 
From the triangle OPN
x2 + y2 = A2
————————– (iii)
Also
From equation (ii)
V = 
V = 
V = 
From the triangle ONP
—————–(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 = 
At equilibrium position y = 0
= 
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 = 
a = 
v = 
a = 
a = 
a = 
For y = 
At equilibrium position y = 0
a = 
a = 0
At the extreme positions
y = + A
a = 
a = 
(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
Fig3.(a)
Fig. 3(b)
Fig 3.(c)
The displacement velocity and acceleration all vary sinusoidal with time but are not in phase
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
The kinetic energy as a function of displacement
The K.E is maximum when its velocity is maximum
= 
(At midpoint)
K.E = ½ mω2A2
The K.E as a function of time
V = 
K.E = 
Or
K.E = 
Expression for potential energy
The P.E is the energy possessed by the body due to its position
By Hooke’s law
The work done to extend the spring from x = 0 to x = xo
The P.E of a string extended by displacement x is given by
P.E = 
F = 
= 
= 
Total energy in S.H.M
The total energy ET of the particle per displacement y is given by
+P.E
Also
= 
ET = 
ET = 2 
GRAPHS OF K.E VS TIME
For
V = 
Then
K.E = ½ m 
2A2Sin2 
t
Fig. 4
GRAPH OF P.E Vs TIME
Fig 5.
Since
P.E = ½ m 
2A2Cos2 
t
ENERGY EXCHANGE
The P.E and K.E for a body oscillating in S.H.M causes the motion of the body.
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 = 
K = 
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 
(i) Maximum velocity of the body
V = A 
= 0.15 x 8 π 
V = 3.768 m/s
(ii) Acceleration a = – 
= 
= – 56.79 m/s2
The negative sign shows that the acceleration is directed towards the equilibrium position.
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 
10 -2
T = 5s y = 5 cm = 5 
10 -2m
Velocity V = 
= 
V = 
V = 10.88 m/s
Acceleration a = 
= 
a = 
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
= 
V = 
cos 
0.05 = A 
0.314 
(0.314 
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 
A = 0.02m
= 0.12 m/s
= 0.12
A 
= 0.12
T = 
T = 1.046S
Velocity at a displacement y = 10 -2 
V = 
V= 
V = 
V = 0.06 m/s
6. In what time after its motion began will a particle oscillating according to the equation
(0.5 πt) move from the mean position to the maximum displacement?
Solution
(0.5 
)
Maximum displacement 
Time taken to move from the mean position to the extreme position is to be found out when 
t=?
= 
0.5πt
= 1
0.5πt = 
0.5πt = 
t = 1s
7. A particle with a mass of 0.5 kg has a velocity of 0.3 
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 
t = 1s 
= 
= 
Velocity v = 
V 
A = 0.57m
K.E = 0.0225J
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 = 
y = 0.2m
ET = 0.0786J
= 
= 
= 
Displacement after a time t = 
sec is y = 0.2 m
y = 
0.2m = 
A = 
A = 0.283m
ET = 
0.04 m = 0.0786
m = 1.96 kg
9. A simple harmonic oscillation is represented by
= 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
= 
(3000t + 0.74) ———————– (i)
The standard displacement equation of S.H.M
= 
( 
t + 
) ——————————- (ii)
Comparing equation (i) and (ii)
(i) Amplitude A = 0.34 m
(ii) Angular frequency w = 3000 
= 2π 
= 
= 
Hz
(iii) Time period T = 
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) 
= 
= 
= 1.19 Hz
(ii) 
= 2π 
= 2π x 1.19
= 2.38 x 3.14
= 7.48 Hz
The displacement equation of S.H.M is
= 
( 
t + 
)
= 0.17 
(7.5t + 0)
11.The equation of S.H.M is given as 
= 6sin 10πt + 8Cos 10πt where 
is in cm and t in second Find
(i) Period
(ii) Amplitude
(iii) Initial phase of motion
Solution
= 6 
+ 8 
———————- (i)
The standard displacement equation of S.H.M
= 
( 
t + 
)
= 
( 
t 
+ 
)
= 
+ 
————— (ii)
Comparing equation (i) and equation (ii)
= 10πt
= 10π
= 6 
=6 ———————————— (iii)
=8 
= 8 ——————————— (iv)
(i) Period
= 
T = 
T = 0.2 s
(ii) Amplitude
Squaring equation (iii) and (iv) then add
A = 10cm
(iii)Initial phase angle
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 = 
Here,
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 = 
t
y = 12.5 cm,
A = 25 cm,
T = 3s
= sin-1 0.5
= 
t = 
Required time = 2t
= 
Required time is 0.5 s
14. A particle in S.H.M is described by the displacement function
= 
( 
t + 
) 
= 
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
At t = 0
= 1cm
v = π cm/s
w = π/s
1 = 
(0 + 
)
1 = 
————————-(i)
V = 
= 
( 
t + 
)
π = 
(0 + 
)
-1 = 
————————(ii)
Squaring and adding equation (i) and (ii)
= 
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 = 
t + 
)
= 
= 
= 
y = 5cm
A = 5 cm,
At, t =0
1 = sin 
= 
y = 
16 .In what time after its motion begins will a particle oscillating according to the equation y = 
move from the mean position to maximum displacement?
Solution
y = 
The standard displacement equation of S.H.M is
y = 
Comparing equation (i) and (ii)
A =7 
=0.5πt
Let t be the time taken by the particle in moving from mean position to maximum displacement position
y = A =7
y = 
y = A = 7
7 = 
= 1
0.5πt = sin -1 1
0.5πt = 
= 
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 ( 
) of S.H.M does not exceed g 
is at most equal to 
= 
A
A = 
A = 
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
= – 
=3cm/s2
A = 4cm
y1=1cm
y2 = 2 cm
= 
w = 1.78 
The velocity of a particle executing SHM is given by
V = 
V = 
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 = 
For the first case
V = 4m/s
y = 3m
V2 = 
16 = 
—————– (i)
For the second case
V = 3m/s
y = 4m
V2 = 
9 = 
——————– (ii)
Take equation (i) divide by equation (ii)
= 
= 
9A2 -81 = 
7A2 = 175
A2 = 
= 25
A = 
A =5m
From equation (i)
16 = 
16 = 
16 = 
=1
= 1 
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 = 
2.5 = 
Cos t = 
Cos t = 0.5
t = 0.5
t = cos -1 (0.5)
t = 
t = 0.1 s
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