TEACHING NOTES
SHM (Complete in 5 Lecture) Introduction Our hearts beat, our lungs oscillate, we shiver when we are cold, we sometimes snore, we can hear and speak because our eardrums and larynges vibrate. We cannot even say “vibration” properly without the tip of the tongue oscillating. Periodic motion : Any motion in which all the parameters of motion are repeated after a definite time interval, is called as periodic motion. The time which elapses between successive passages in the same direction through any point is termed the periodic time, or the period, of the motion. The number of periods comprised in one second is called the frequency ; thus, if T is the period, and n is the frequency of a periodic motion, we have n = I/T. The motion of the hands of a clock is periodic, the period of the motion of the minute hand being one hour, or 3600 seconds. The bob of a pendulum moves periodically, the period being equal to the time of one complete (to and fro) oscillation. Periodic motion can be along any path. Oscillatory motion :– If in a case of periodic motion particle moves to and fro on the same path, the motion is said to be oscillatory motion. A body that undergoing oscillatory motion always does so about a stable equilibrium position. When it is moved away from this position and released, it experiences a net force or torque to pull it back toward equilibrium position. But by the time it gets there, the restoring force/torque would have done some positive work on it. Thus it must have gained some kinetic energy, so it overshoots, the equilibirium position. Now it stops somewhere on the other side, and is again pulled back toward equilibrium. Imagine a ball rolling back and forth in a round bowl or a pendulum that swings back and forth past its lowest position.
Asking Question : A particle hanging from a thread is launched horizontally such that it completes the circle. Is the motion periodic? Is the motion oscillatory? Sol.
Motion is periodic but not oscillatory as the particle is not moving to and fro.
Asking Question : A particle hanging from a thread is launched horizontally with small speed such that it does even reach to the horizontal level. Is the motion periodic? Is the motion oscillatory? Sol.
Motion is periodic and oscillatory as the particle is moving to and fro repeatedly. Page-1
Imp. characteristics of oscillatory motion :– 1.
When a particle in stable equilibrium is disturbed, then it has tendency to return to the position of equilibrium and this tendency is exhibited as oscillatory motion.
2.
The force on the body acts towards the mean position i.e. Force is always opposite to the displacement vector of the particle w.r.t. mean position. (This force is known as restoring force) Fˆ = – rˆ ,
ˆ = – ˆ
Where rˆ and ˆ are displacements and angular displacement from mean position. 3.
Energy is also conserved. If enrgy is not conserved than the particle will not be able to repeat the parameters of the motion.
SHM Optional Explanation of why we study SHM : Sinusoidal Vibrations : Our study will be mostly related to sinusoidal vibrations which we will show later on arise when net force experienced by an oscillating body has magnitude proptional to the distance from the mean position or torque is directly propotional to the angular displacement from mean position. There are two reasons for this – one physical, one mathematical, and both basic to the whole subject. The physical reason is that purely sinusoidal vibrations are common in many type of mechanical systems. Such motion is almost always possible if the displacements are small enough. If, for example, we have a body attached to a spring, the force exerted on it at a displacement x from equilibrium is actually F(x) = – (k1x + k2x2 + k3x3 + .........0 where k1, k2, k3, etc., are a set of constants, and we can always find a range of values of x within which the sum of the terms in x2, x3, etc., is negligible, compared to the term – k1x. If the body is of mass m and the mass of the spring is negligible, the equation of motion of the body then becomes m
d2x = – k1x dt 2
It is easy to verify, that above equation is satisfied by an equation of the form x = A sin (t + 0) where = (k1/m)1/2. Thus sinusoidal vibration or simple harmonic motion is likely possibility in small vibrations. But we should remember that in general it is only an approximation (although perhaps a very close one) to the true motion.
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The second reason is the mathematical one. The actual importance of purely sinusoidal vibrations is is proved a famous theorem given by the French mathematician J.B. Fourier in 1807. According to Fourier’s theorem, any periodic function with a period T can be considered as sum of of pure sinusoidal vibrations of periods T, T/2, T/3, etc., with appropriately chosen amplitudes. A thorough familiarity with sinusoidal vibrations will be stepping stone for our understanding of every conceivable problem involving periodic phenomena. TYPES OF SHM (a) Linear SHM : When a particle moves to and from about an equilibrium point, along a straight line. A and B are extreme positions. M is mean position AM = MB = Amplitude. M • •B A• (b) Angular SHM : When body / particle is free to rotate about a given axis executing angular oscillations. EQUATION OF SHM : The necessary and sufficient condition for SHM is F = –kx where k = positive constant for a SHM = Force constant x = displacement from mean position
d2x = –kx dt 2
or
m
k d2x x=0 2 + m dt
d2x 2 2 +x= 0 dt Its solution is x = A sin (t + )
[differential equation of SHM]
where =
k m
DERIVATION F = – kx a=–
k x m
k = 2 m a = – 2x
put
v
vdv 0
x
=–
2 xdx A
v2 2 2 =– [x – A2] 2 2
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v = ± A2 x2 dx =± dt
A2 x 2
dx
= ± dt A 2 2 sin–1 x/A = ± t + k at t = 0 x = x0 x = A sin {t + } CHARACTERISTICS OF SHM Note : In the figure shows, path of the particle is on a straight line. (a) Displacement : It is defined as the distance of the particle from the mean position at that instant. Displacement in SHM at time is given by x = A sin (t + ) (b) Amplitude : It is the maximum value of displacement of the particle from its equilibrium position
Amplitude = 1/2 (distance between extreme points/position) It depends on energy of the system. (c) (d)
2 = 2f and its units is rad/sec. T Frequency (f) : Number of oscillations completed in unit time interval is called frequency of oscillations,
Angular Frequency () : =
1 = , its units is sec–1 or Hz. T 2 Time period (T) : Smallest time interval after which the oscillatory motion gets repeated is called time
f= (e)
period, T =
2 m = 2 k
Ex.
d2x For a particle performing SHM, equation of motion is given as 2 + 4x = 0. Find the time period. dt
Sol.
d2x = –4x dt 2 Time period : T =
2 = 4
= 2
2 =
Objective : Determination of from basic SHM equation. Page-4
(f) (g) (h)
Phase : The physical quantity which represents the state of motion of particle (eg. its position and direction of motion at ay instant.) argument of sine function is called phase. Phase constant () : Constant in equation of SHM is called phase constant or initial phase. It depends on initial position and direction of velocity. Velocity (v) : It is the rate of change of particle’s displacement w.r.t. time at that instant. Let the displacement from mean position is given by x = A sin (t + ) dx d = [A sin (t + )] dt dt v = A cos (t + )
Velocity,
v=
or, v = A2 x2 At mean position (x = 0), velocity is maximum vmax = A At extreme position (x = A), velocity is minimum. vmin = zero GRAPH OF SPEED (v) VS DISPLACEMENT (x) : v2 = 2 (A2 – x2)
v = A2 x2
v2 x2 1 v +x =A A 2 A 2 Graph would be an ellipse 2
(i)
2 2
2
2
Acceleration : It is the rate of change of particle velocity w.r.t. time at that instant.
dv d = [A cos (t + )] dt dt a = –2A sin (t + ) a = –2x NOTE : Negative sign shows that acceleration is always directed towards the mean position. At mean position (x = 0), modulus of acceleration is minimum amin = zero At extreme position (x = A), acceleration is maximum amax = 2A
Acceleration,
a=
GRAPH OF ACCELERATION (A) VS DISPLACEMENT (x) : a = –2x
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GRAPHICAL REPRESENTATION OF DISPLACEMENT, VELOCITY AND ACCELERATION IN SHM Displacement, x = A sin t Velocity, v = A cos t = A sin (t + ) 2 or v = A2 x2 Acceleration, a = –2A sin t = 2 A sin (t + ) or a = –2x Note : v = A 2 x 2 a = –2x These relations are true for any equation of x.
Time, t 0 T/4 T / 2 3T / 4 T Displacement , x 0 A 0 A 0 Velocity, v A 0 A 0 A 2 2 Acceleration, a 0 A 0 A 0
1. 2.
All the three quantities displacement, velocity and acceleration vary harmonically with time, having same period. The velocity amplitude is times the displacement amplitude (vmax = A).
3.
The acceleration amplitude is 2 times the displacement amplitude (amax = 2A).
4.
In SHM, the velocity is ahead of displacement by a phase angle of
5.
In SHM, the acceleration is ahead of velocity by a phase angle of
. 2
. 2
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A particle executing simple harmonic motion has angular frequency s–1 and amplitude 1 cm. Find (a) the time period, (b) the maximum speed (c) the maximum acceleration, (d) the speed when the displacement is 0.6 cm from the mean position, (e) the speed at t = /6 s assuming that the motion starts from rest at t = 0. Objective : Application of basic formula of SHM. {Home Work : HCV – Q.1 to Q.7 } Phasor One of the best ways of describing simple harmonic motion is obtained by considering it as the projection of uniform circular motion. Imagine, for example, that a disk of radius A rotates about a horizontal axis at the rate of rad/sec. Suppose that a peg P is attached to the edge of the disk and that a horizontal beam of parallel light casts a shadow of the peg on a vertical screen, as shown in Fig. (a). Then this shadow perform simple harmonic motion with period 2/ and amplitude A along a vertical line on the screen.
Screen
Screen A x O –A
O x
A
P
P A
A
–A
Ex.
Parallel light
Parallel light (a)
(a)
P A O
X
x
(b)
More abstractly, we can imagine SHM as being the geometrical projection of uniform circular motion. (By geometrical projection we mean simply the process of drawing a perpendicular to a given line from the instantaneous position of the point P.) Let the tracing point, P, revolve uniformly at a constant distance OP from O, the direction of motion being anticlockwise. Through O we draw the rectangular axes X’OX and Y’OY. From P draw PQ perpendicular to OY. Thus OQ is the component of OP Page-7
resolved parallel to OY. As P revolves about O, the point Q moves up and down along Y’OY, its motion being of the simple harmonic type. When P pases across the axis X’OX, Q will pas through O. When P passes across the axis Y’OY, Q will be at its position of maximum displacement from its mean position, O. The maximum displacement of Q will thus be equal to the radius a of the circular path of P ; it is termed the amplitude of the s.h.m. The phase of the s.h.m. at any instant is equal to the angle which has been swept out by the line OP. If we measure the phase from the particular position of OP when Q is moving in a certain direction (say upwards) through O, the angle XOP = is equal to the phase of the s.h.m.
Y b P
Q X
O
R
p X
C
r
O
o
Y
The displacements of Q for various values of the phase angle , are shown by the curve to the right of Fig. ( ). The x coordinate, is equal to the circular measure of the angle XOP, and the distance OQ is plotted on y axis. Let the amplitude OP = a, while OQ = y. Then y/a = RP/OP = sin . Therefore y = a sin . plotted to the right of the figure is described by this equation. Further, if OP completes a rotation (i.e. rotate through an angle 2) in the period T ; then we have = 2t/T, and therefore y = a sin (2t/T). When the tracing point P crosses the axis OX, it is moving, for the instant, parallel to the axis OY. Consequently, at this instant the point Q is moving along OY with a velocity equal to that of the tracing point P in its circular path. The velocity of P is equal 2a/T. Hence, the velocity of Q as it passes through its mean position O is equal to 2a/T. When the tracing point P crosses the axis OY, it is moving, for the instant, in a direction perpendicular to OY. At this instant point Q will be stationary. Consequently, the point Q is stationary for an instant at the extreme, on either side of its mean position O.
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Y P
Q
S
Q X
P X
O
R R
Y
The velocity corresponding to any phase of a s.h.m. can be determined readily. Let the vector OP (Fig.), of length a, revolve uniformly about O in an anticlockwise direction, its perod being equal to T ; and let the projection Q of the point P on the axis YOYis the simple harmonic motion. Thus OQ = y = a sin (2t/T). Differentiating it with repect to time we get V=
2a 2 t cos ( ). T T
This gives the instantaneous velocity of the simple harmonic motion at the time t. In terms of the phase angle , the velocity is equal to
2a cos . When = O ; cos = 1, and the velocity is equal to 2a/ T
T, as already discussed ; when = /2, cos = O, and the velocity is equal to zero ; and so on. Thus, a point executing a s.h.m. moves with a maximum velocity equal to 2a/T on passing through its mean position. Its velocity diminishes as it moves away from its mean position, being equal to (2a/T) cos when the phase angle is , and attaining the value zero at the extreme position. Subsequently, as the point returns towards its mean position, its velocity increases, and once more attains the value 2a/ T on moving through the mean position. The use of a uniform circular motion as a purely geometrical basis for describing SHM embodies more than we have so far chosen to recognize. This circular motion, once we have set it up, defines SHM of amplitude A and angular frequency along any, straight line in the plane of the circle. Very Easy Ex.(a) A particle starts from mean position and moves towards positive extreme as shown. Find the equation of the SHM. Amplitude of SHM is A.
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Also write the equation of SHM for the situation shown below. (b) (c) (d) Ans.
(a) x = A cos t ; (b) x = –A cos t ; (c) x = A sin (t + 150°) Objective : To calculate initial phase angle & write SHM equation.
Ex.
Write equation of S.H.M of angular frequency and A amplitude if the particle is situated at
A at 2
t = 0 and is going towards mean position. Sol. At t=0 particle was at
A and was going towards mean position as shown in the figure(a) below.. 2
A/2 –A
M
A
The same situation is described by making reference point on acircle of radius A as shown in figure(b)
below.
A/2
The reference point can be located at any of the two positions G and H as shown in figure for having displacement
A , but to move towards mean position it should be H. Position H Corrospondes to angle 2
. Thus =
3 3 and x = A sin (t + ) 4 4
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Alternative Mathematical solution Putting t= 0 in equation x = A sin (t + ) We get
A = A sin 2
Thus, =
3 , 4 4
but at t = 0 ; V < 0 so x = A sin (t +
is rejected 4 3 ) 4
Objective : the ease of using phasor diagram is quite evident in this solution.
Ex.
If initial position of particle performing SHM is +A/2 & moving towards centre having angular frequency ‘’ find minimum time when it will be at (1) – A/2 (2) Mean postion (3) – A (4) + A
Ex.
x = – A cos (
– t) 4 Convert this equation into standard form
Objective : sign of A and should be positive Ex. Find distance traveled by a particle of time period T & amplitude A in time T/12 starting from rest. t=0
xt
Sol.
M
–A
l
A
Eqn of SHM x = A cos t 2 T xt = A cos T 12
xt =
A 3 2
l =A–
l=
A 3 2
A(2 3 ) 2
Objective : Explain SHM is not linear motion Page-11
Ex.
Two particles P and Q are executing SHM across same straight line whose equations are given as yP = A sin (t + 1) and yQ = A cos(t + 2). An observer, at t = 0, observers the particle P at a distance A
2 moving to the right from mean position O while Q at
3 A moving to the left from 2
mean position O as shown. Find, (2– 1)
Sol.
yP = A sin (t + 1) at t = 0
1 A = Asin1 sin1= 1 = /4, 3/4 2 2 & vP = Acos1 since vP = (+)ive so, 1 = /4 yQ = A cos(t + 2). at t = 0
3A = Acos2 2
cos2=
3 2
2 = – /6 = 5/6 + /6 = 7/6
& vQ = –Asin2, since vQ = (–)ive so, 2 = 5/6
= 5/6 – /4 =
10 3 = 7/12 Ans. 12
Objective : Application of phasor diagram NOTE :
Ex.
If mean position is not at the origin, then we can replace x by x – x0 and the equation becomes x – x0 = –A sin t, where x0 is the position co-ordinate of the mean position.
A particle is moving on Y-axis under a variable force F = – ky + c. (a) Is the motion of the particle can be called SHM ? Hence or otherwise write y position of particle as function of time.
Sol. We can write the force as F = – k (y –
C ) k
lets define s = y –
C k
,..........................................eqn() Page-12
Where s is displacement from mean position because at y =
Thus mean position is at y = (
C , F is zero. k
C ) k
Thus force can be written as F = -ks...................................() which is equation of SHM about y =
C k
Differentiating eqn() wrt time twice we get
d2y Thus F = m 2 dt
d2y d 2s = dt 2 dt 2
d 2s can also be written as F = m 2 dt
Substituting in eqn () m
d 2s = – ks dt 2
Comparing with standard eqn we know that the solution is s = A sin ( t + )
=
k m
Replacing with value of s = y -
We get y = A sin (t + ) +
C k
C k
(b)
If in the previuos question F = – 10y + 20, and mass of the particle is m = 2.5 kg and is released from rest from y = +3, at t = 0. Write the equation of SHM.
Sol.
From previuos discussion we can write y = A sin (t + ) +
C k
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where, k =10 and C = 20, thus =
k , gives us = 2 m
Mean position is 2 as force at y = 2 is zero
F = – 10 x 2 + 20 = 0
y = A sin (2t + ) + 2 at t = 0 the particle is at y = 3 and is at rest thus its extreme position is y = 3. the distance between the mean position and extreme position is 1m. Thus amplitude A = 1. at t = 0 `
1 sin + 2 = 3
sin = 1 =
2
Ans. cos 2t + 2 (Home Work : H.C.V. Q. 8 to 11) ENERGY OF SHM : Kinetic Energy (KE) 1 1 1 mv2 = m2 (A2 – x2) = k (A2 – x2) (as a function of x) 2 2 2
=
1 1 mA22 cos2 (t + ) = KA2 cos2 (t + ) (as a function of t) 2 2
1 2 1 KE 0 T kA 2 ; kA ; 2 4 Frequency of KE = 2 (frequency of SHM)
KEmax =
1 KE 0 A kA 2 3
Potential Energy (PE) 1 2 1 Kx (as a function of x) = kA2 sin2 (t + ) (as a function of time) 2 2
Total Mechanical Energy (TME) Total mechanical energy = Kinetic energy + Potential energy 1 1 1 k(A2 – x2) + Kx2 = KA2 2 2 2 Hence total mechanical energy is constant in SHM.
=
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Graphical Variation of energy of SHM
Ex.
A particle of mass 0.50 kg executes a simple harmonic motion under a force F = –(50 N/m)x. If it crosses the centre of oscillation with a speed of 10 m/s, find the amplitude of the motion.
SPRING-MASS SYSTEM
m k A particle of mass 200 g executes a simple harmonic motion. The restoring force is provided by a spring of spring constant 80 N/m. Find the time period. T = 2
Ex.
(1)
Sol.
(2)
The time period is T = 2
m 200 103 kg = 2 = 2 × 0.05 s = 0.31s. k 80 N / m
Explain that in this case along with variable force constant force of mg is acting but still the motion is SHM and for this SHM equation is F = Ax + B for above figure A = k and B = mg Objective : Application of basic formula. Ex.
Ans
A block of mass 100gm attached to a spring of spring constant 100N/m is lying on a frcitionless floor as shown. In natural length the block is moved to compress the spring by 10cm and then released. If the collisions with the wall in front are elastic then find the time period of the motion. 0.133 sec Objective : Example of incomplete SHM, application of phasor diagram.
Qus.
Block A of mass is performing SHM of amplitude a. Another block B of mass m is gently placed on A when it passes through mean position and B sticks to A. Find the time period and amplitude of new SHM.
Ans.
T = 2
a 2m amplitude = 2 K
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a from mean position. 2
Qus.
Repeat the above problem assuming B is placed on A at distance
Ans.
T = 2
Ex.
The left block in figure collides inelastically with the right block and sticks to it. Find the amplitude of the resulting simple harmonic motion.
Ex.
Two blocks of mass m1 and m2 are connected with a spring of natural length and spring constant k. The system is lying on a smooth horizontal surface. Initially spring is compressed by x0 as shown in figure. Show that the two blocks will perform SHM about their equilibrium position. Also (a) find the time period, (b) find amplitude of each block and (c) length of spring as a function of time.
2m 5 amplitude = a K 8
Objective : To explain SHM in non inertial frame. Optional Ex. A string of normal length l & force constant k is released from ceiling. Find its time period? M l/2
[Assume force constant k is not varying with extension is string]
{Home Work : HCV – Q.11 to Q.28 } METHOD’S TO DETERMINE TIME PERIOD, ANGULAR FREQUENCY IN SHM (a) Force / torque method (b) Energy method Ex.
The string, the spring and the pulley shown in figure are light. Find the time period of the mass m.
Sol.
(a) Force Method Let in equilibrium position of the block, extension in spring is x0. kx0 = mg ......... (1) Page-16
Now if we displace the block by x in the downward direction, net force on the block towards mean position is.
F = k(x + x0) – mg = kx using (1) Hence the net force is acting towards mean position and is also proportional to x. So, the particle will perform S.H.M. and its time period would be. T = 2 (b)
m k
Energy Method Let gravitational potential energy to be zero at the level of the block when spring is in its natural length. Now at a distance x below that level, let speed of the block be v. Since total mechanical energy is conserved in SHM 1 2 1 2 kx + mv = constant 2 2 Differentiating w.r.t. time, we get –mgv + kxv + mva = 0 where a is acceleration
–mgh +
F = ma = –kx + mg
or
mg F = –k x k
This shows that for the motion, force constant is k and equilibrium position is x = So, the particle will perform S.H.M. and its time period would be T = 2
mg . k
m . k
Objective : Write SHM equation & determine ‘’. Qus.
Solve the above problem if the pulley has a moment of inertia I about its axis and the string does not slip over it.
Ans.
2
(m I / r 2 ) k
SIMPLE PENDULUM If a heavy point-mass is suspended by a weightless, inextensible and perfectly flexible string from a rigid support, then this arrangement is called a simple pendulum. Time period of a simple pendulum T = 2
g
(some times we can take g = 2 for making calculation simple) Page-17
Note : (a)
If angular amplitude of simple pendulum is more, then time period 20 1 (For other exams) 16 where 0 is in radians. General formula for time period of simple pendulum.
T = 2 g
(b)
1 T = 2 g 1 1 R (c)
On increasing length of simple pendulum, time period increases, but time period of simple pendulum of infinite length is 84.6 min which is maximum and is equal to T = 2
(d) (e) (f)
(Where R is radius of earth) Time period of seconds pendulum is 2 sec and = 0.993 m. Simple pendulum performs angular S.H.M. but due to small angular displacement, it is considered as linear SHM. If time period of clock based on simple pendulum increases then clock will be slow but if time period decrease then clock will be fast. T 1 ×100 = × 100 T 2
(g)
If g remains constant and is change in length, then
(h)
If remains constant and g is change in acceleration then,
(i)
R . g
T 1 g × 100 = – ×100 T 2 g If is change in length and g is change in acceleration due to gravity then,
1 1 g T 100 100 T 2 2 g Time Period of simple Pendulum in accelerating Reference Frame :
T = 2 g where eff .
geff. = Effective acceleration due to gravity in reference system = | g a | a = acceleration of the point of suspension w.r.t. ground. Condition for applying this formula: | g a | = constant.
Ex.
A block of mass m is suspended from the ceiling of a stationary elevator through a spring of spring constant k it is in equilibrium. Suddenly, the cable breaks and the elevator starts falling freely. Show that block now executes a simple harmonic motion of amplitude mg/k in the elevator. Objective : Explain SHM in moving frame & discuss motion of particle from cabin frame.
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Ex.
Sol.
A simple pendulum is suspended from the ceiling of a car accelerating uniformly on a horizontal road. If the acceleration is a0 and the length of the pendulum is , find the time period of small oscillations about the mean position. We shall work in the car frame. As it is accelerated with respect to the road, we shall have to apply a psuedo force ma0 on the bob of mass m. For mean position, the acceleration of the bob with respect to the car should be zero. If be the angle made by the string with the vertical, the tension, weight and the psuedo force will add to zero in this position. Hence, resultant of mg and ma0 (say F = m g 2 a 02 ) has to be along the string.
ma 0 a0 = mg g Now, suppose the string is further deflected by an angle as shown in figure. Now, restoring torque can be given by (F sin ) = –m2 Substituting F and using sin = , for small .
tan 0 =
(m g 2 a 20 ) m 2 g 2 a 20 g 2 a 02 so, 2 = This is an equation of simple harmonic motion with time period
or,
=–
T=
2 = 2 2 (g a 02 )1 / 4
Compound pendulum / Physical pendulum When a rigid body is suspended from an axis and made to oscillate about that then it is called compound pendulum. C = Initial position of centre of mass C = Position of centre of mass after time t S = Point of suspension l = Distance between point of suspension and centre of mass (it remains constant during motion) for small angular displacement from mean position The restoring torque is given by = mgl or, I = – mgl where, I = Moment of inertia about point of suspension. or,
=–
mgl I
Time period, T = 2
or, I mgl
2 =
mgl I
I = ICM + ml2
where ICM = momentum of inertia relative to the axis which passes from the centre of mass & parallel to the axis of oscillation.
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ICM ml 2 mg l
T = 2
where ICM = mk2 k = gyration radius (about axis passing from centre of mass) Torsional pendulum In torsional pendulum, an extended object is suspended at the centre by a light torsion wire. A torsion wire is essentially inextensible, but is free to twist about its axis. When the lower end of the wire is rotated by a slight amount the wire applies a restoring torque causing the body to oscillate rotationally when released. The restoring torque produced is given by
or,
= – C I = – C
or,
=–
C I
where, C = Torsional constant where, I = Moment of inertia about the vertical axis.
Time period, T = 2
I C
Ex.
A uniform disc of radius 5.0 cm and mass 200g is fixed at its centre to a metal wire, the other end of which is fixed to a ceiling. The hanging disc is rotated about the wire through an angle and is released. If the disc makes torsional oscillations with time period 0.20s, find the torsional constant of the wire.
Sol.
The situation is shown in figure. The moment of inertia of the disc about the wire is mr 2 (0.200 kg) (5.0 10 2 m) 2 = = 2.5 × 10–4 kg-m2. 2 2 The time period is given by
I=
T = 2
I C
or,
4 2 (2.5 10 4 kg m 2 ) = (0.20 s) 2
C=
42 I T2
kg m 2 = 0.25 s2
Objective : Basic application of torsional pendulum. {Home Work : HCV – Q.28 to Q.54 }
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Superposition of two SHM’s In same direction and of same frequency x1 = A1 sin t x2 = A2 sin (t + ), then resultant displacement x = x1 + x2 = A1 sin t + A2 sin (t + ) = A sin (t + ) where A =
A12 A 22 2A1A 2 cos
&
A 2 sin = tan–1 A A cos 1 2
If = 0, both SHM’s are in phase and A = A1 + A2 If = , both SHM’s are out of phase and A = |A1 – A2| The resultant amplitude due to superposition of two or more than two SHM’s of this case can also be found by phasor diagram also. In same direction but are of different frequencies. x1 = A1 sin 1t x2 = A2 sin 2t then resultant displacement x = x1 + x2 = A1 sin 1t + A2 sin 2t this resultant motion is not SHM. In two perpendicular directions x = A sin t y = B sin (t + ) Case (i) : If = 0 or then y = ± (B/A) x. So path will be straight line & resultant displacement will be r = (x2 + y2)1/2 = (A2 + B2)1/2 sin t which is equation of SHM having amplitude Case (ii) : If
A 2 B2
then x = A sin t 2 y = B sin (t + /2) = B cos t
=
x2 y2 so, resultant will be 2 + 2 = 1 i.e. equation of an ellipse and if A = B, then superposition will be A B an equation of circle.
1. 2.
Superposition of SHM’s along the same direction (using phasor diagram) If two or more SHM’s are along the same line, their resultant can be obtained by vector addition by making phasor diagram. Amplitude of SHM is taken as length (magnitude) of vector Phase difference between the vectors is taken as the angle between these vectors. The magnitude of resultant of vector’s give resultant amplitude of SHM and angle of resultant vector gives phase constant of resultant SHM.
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For example : x1 = A1 sin t x2 = A2 sin (t + ) If equation of resultant SHM is taken as x = A sin (t + ) A=
A12 2A1A 2 cos
A1 sin tan = A A cos 1 2
Ex.
Sol.
x1 = 3 sin t x2 = 4 cos t Find (i) amplitude of resultant SHM. (ii) equation of the resultant SHM. First write all SHM’s in terms of sine functions with positive amplitude. x1 = 3 sin t x2 = 4 sin (t + /2) A=
32 4 2 2 3 4 cos
4 sin tan =
2
3 4 cos 2
= 2
9 16 =
25
=5
= 53°
equation x = 5 sin (t + 53°) Ex.
Sol.
x1 = 5 sin (t + 30°) x2 = 10 cos (t) Find amplitude of resultant SHM. x1 = 5 sin (t + 30°) x2 = 10 sin t 2 Phasor diagram
A= =
52 10 2 2 5 10 cos 60
25 100 50 = 175 = 5 7
{Home Work : HCV – Q.55 to Q.58 }
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