UNIT-III-A
ELASTICITY
DEFORMING FORCE: - On application of external force on a body, If there is a change in Length/Area/Volume of a body, then the force is called deforming force.
ELASTICITY:- The property of a body due to which it tends to remain its original shape, size or volume, on removing deforming force is called Elasticity. Example:- Rubber, Spunz ball.
(i) When the body does not regain its original shape size and volume in removing deforming force, is called plasticity.
(ii) Elastic limit is the upper limit of deforming force upto which, the body regains its original form completely on removing deforming force.
STRESS:- The restoring force per unit area, setup inside the body is called stress. S.I. unit ⇒ N/m2 . Stress is a vector quantity. Stress is of
two types.
1) Normal stress:- When the restoring force is perpendicular to the area, the Stress is called normal stress. The Stress is always normal in case of change in length of a wire or volume of body.
2) Tangential stress:-When the stress is applied tangential to the surface, it is called tangential or shearing stress. In this case, the volume is not changed.
STRAIN:- The fractional change in the configuration of a body is called strain. Strain is of three types
-1) Longitudnal strain:- if the deforming force produces a change in length alone, then the strain is called longitudinal strain.(L.S. =
/ )
2) Volumetric strain:- When the deforming force produces, a change in volume only, the strain is called Volumetric Strain. .(V.S. = / )
3) Shearing strain:- When the deforming force acting on
the body produces a change in shape, without changing its volume, the strain is called shearing strain.
Shearing Strain ⇒ θ = tanθ =
HOOKE’ LAW: - With is elastic limit, stress developed in a body is proportional to strain produced in it.
Stress ∝ Strain Stress = E ×Strain
Where ‘E’ is called Modulus of Elasticity.
There are three different types of modulus of elasticity- Young's Modulus of elasticity, Bulk
Modulus of elasticity and Modulus of Rigidity.
1) YOUNG’S MODULUS OF ELASTICITY (Y):- Within elastic limit, the ratio of normal stress and longitudinal strain is called Young’s Modulus of Elasticity.
Y = ; ⇒ Y = /
/ ⇒ S.I. Unit ⇒ N/m
2
Stress =
2) BULK MODULUS OF ELASTICITY (B) :- Within elastic limit, the ration of normal stress and volumetric strain is called bulk modulus of elasticity.
B = ; ⇒ B = /
/ ⇒ or
Here negative sign indicates that on increasing Pressure, Volume is decreasing.
3) MODULUS OF RIGIDITY ( ):- Within elastic limit, the ratio of shearing stress and shearing strain is called Modulus of Rigidity.
= ⇒ = =
POSITION’s RATIO:- The ration of lateral strain to longitudinal strain is called position’sratio. It is denoted by σ
σ =
Let ‘L’, and ‘d’ be the original length and diameter of the wire. Due to stress ‘ L’ be the increase in length and ‘ d’ be the small decrease in diameter correspondingly.
= =
σ = /
/ ⇒ σ = (-ve sign indicates diameter is difference)
STRESS-STRAIN CURVE: In fig, stress vs. strain curve is shown, from the curve,
(i). Portion OA is the straight line which clearly shows that stress produced is directly proportional to strain i.e., Hook's law is perfectly obeyed upto A and on removal of stress wire or bar will recover its original condition. Point A is called Proportionality limit.
(ii). As soon as proportionality limit is crossed beyond point A, the strain increases more rapidly than stress and curve AB in graph shows that extension of wire in
this limit is partly elastic and partly plastic and point B is the elastic limit of the material. Thus if we start decreasing load from point B the graph does not come to O via path BAO instead it traces straight line BG. So that there remains a residual strain. This is called permanent set.
(iii). If we continue to increases the stress beyond point B then for little or no increase in stress the strain increases rapidly up to point C.
(iv). Further increase of stress beyond point C produces a large increase in strain until a point E is reached at which fracture takes place and from B to D material is said to undergo plastic flow which is irreversible.
Notes: Coefficient of elasticity depends upon the material, its temperature and purity but not on stress or strain.
Q:-A wire of length 1 meter is stretched by a force of 10 N, the area of cross section, of the wire is 2×10-6 m2 and Young’s Modulus is 2×1011 N/m2 . Calculate (i) Stress (ii) The increase in length of wire.
L= 1 m; F= 10 N; A= 2×10-6 m2
(i) Stress = =
× = 0.5×10
7
= 5×106 N/m2
(ii) Y =
= × × =
×
× × × = 2.5×10
-5
m ⇒ Increase in length = 2.5×10-5 m
(iii) Strain = 2.5×10-5 m
Q:- Calculate the % increase in length of a wire of d=2.2mm. Stress by a load of 100 kg Y= 12.5×1010 N/m2
d= 2.2 mm = 2.2×10-3m ⇒ r = 1.1×10-3m m= 100 Kg
W = mg = 100×9.8 = 980; Stress =
. × . × × . × = 2.58×10
8
UNIT-III- B SIMPLE HARMONIC MOTION
PERIODIC MOTION:- Periodic motion of a body is defined as the motion which repeats itself on a definite path after the fixed interval of time.
SIMPLE HARMONIC MOTION:- Simple Harmonic Motion of a body is defined as the motion, which can be expressed in terms of single Harmonic function (Sine or co sine)
General equation of SHM is
y = sin
Amplitude:- It is the maximum displacement of a particle from mean position.
Phase:- Phase is a physical quantity which determines, the position and direction of the particle executing simple harmonic motion.
Velocity:- Rate of change of displacement is called velocity.
= = ( sin ) = ( cos ).
We know that
cos2 + sin2 =1
cos2 = 1- sin2 = 1 =
cos =
v= v=
Acceleration:- The rate of change of velocity is called Acceleration
A= = (aωcosωt)
A= aωsinωt. ω = (a sinωt ) ω 2 A= ω2y
Amax = ω2a
Characteristics of SHM and Restoring force:- The Characteristics of a particle executing SHM are- 1) The direction of acceleration is opposite to the direction of displacement.
2) for a constant angular velocity acceleration is directly proportional to displacement Aα –y 3) For a particle of mass m, executing SHM, the force is given by F=m A ∴ F=mw2y
This force is called restoring force. It always tries to bring back the particle to the mean position.
Q:- A Particle is doing SHM, acceleration to the particle at 3 cm displacement is 20 cm/s2. If maximum displacement of the particle is 10 cm, then find out the maximum acceleration of the particle.
Given that y= 3cm; A=20cm/sec2; a=10m/sec2; A= ω 2y
ω 2= =
Q:- Equation of displacement of a particle under SHM is given by y=0.1 sinπ (t+0.5)m where time t is in second. Determine and displace and start of the oscillation.
Given that y=0.1 sinπ (t+0.5)m = 0.1 sin (π t+ 0.5π)
Comparing with standard equation y= a sin (ω t+ϕ)
amplitude a= 0.1m
ω t=πt =
Time period T=2 s.
vmax=w √ 0 = = × 0.1 = 0.314 m/s
A= ω 2y = ω 2a = × = × . × 0.1 =0.985 m/s2 At start of oscillation t=0
y= a sin (ω t+ϕ) = 0.1 sin (π×0 +0.5π) = 0.1 sin =0.1×1 y= 0.1 m (oscillation)
SIMPLE PENDULUM: - When a heavy point mass is suspended from a light , and flexible string, this system is called simple pendulum.
Let ball of mass m is hanged with a string of length L, it is displaced by small angle θ and it execute SHM.
Decomposing Weight mg into horizontal and vertical components, restoring force is given by F= mg sinθ
for small angle sinθ ~ θ = = ∴ F= ---(1) If A is acceleration the force F = mA ---(2)
mA = mg = = ---(3) We know that, the time period is given by
T= 2π ( )
T= 2π
Q:- A simple pendulum completes 25 oscillation in 8 minutes velocity of the bob at lowest position is 50 cm/s. Calculate the amplitude and acceleration of oscillation.
n = 25/(8X60) = 0.052Hz v=50cm/s at y=0
v= ω
50 = ω√ 0 = ω a=
ω = = × . × . = . = 153.111m/s
2
