OSCILLATIONS PART-2LS-16/AS 26th,JAN,2010 7.30pm : OSCILLATIONS PART-2LS-16/AS 26th,JAN,2010 7.30pm ENERGY IN S.H.M
SIMPLE PENDULUM OSCILLATIONS
OF A SPRING
SIMPLE HARMONIC MOTION : SIMPLE HARMONIC MOTION S.H.M is the motion executed by a point mass particle subjected to a force that is proportional to the displacement of the particle but opposite in sign.
VELOCITY
In linear simple harmonic motion,the particle moves faster near the mean position and slower near the extreme position.
VELOCITY & ACCELERATION IN SHM : VELOCITY & ACCELERATION IN SHM In linear simple harmonic motion,at the extreme position,the velocity of the particle is minimum and its acceleration is maximum.
At the mean position,the velocity of the particle is maximum and its acceleration is minimum.
OSCILLATION : OSCILLATION 4. The acceleration in S.H.M. is always in opposite phase to that of displacement as the displacement of the particle in S.H.M. at an instant is directed away from the mean position and acceleration at that instant is directed towards the mean position.
OSCILLATION : OSCILLATION 5. The velocity in simple harmonic motion is in phase with acceleration,when particle is moving from extreme position to mean position and is in opposite phase with acceleration when particle is moving from mean position to extreme position.
TOTAL ENERGY IN S.H.M. : TOTAL ENERGY IN S.H.M. A particle executing S.H.M. possesses two types of energy.
(i) Potential energy.This is on account of the displacement of the particle from its mean position.
(ii)Kinetic energy.This is on account of the velocity of the particle.
TOTAL ENERGY IN S.H.M. : TOTAL ENERGY IN S.H.M.
TOTAL ENERGY IN S.H.M. : TOTAL ENERGY IN S.H.M. Neglecting the frictional forces, the total energy of the particle at any instant is equal to the sum of the two types of energies at that instant. Let us calculate each one of them.
POTENTIAL ENERGY : POTENTIAL ENERGY Consider a particle of mass m, executing linear S.H.M. with amplitude a and constant angular frequency ?. Suppose t second after starting from the mean position, the displacement of the particle is y, which is given by
y = a sin ?t
? Velocity of the particle at instant t,
POTENTIAL ENERGY : POTENTIAL ENERGY Acceleration of the particle at this instant,
Here –ve sign shows that the acceleration is always directed towards mean position.
Restoring force, F = mass x acceleration
= m ?2 y = - ky
POTENTIAL ENERGY : POTENTIAL ENERGY where, m?2 = k = force constant or spring factor of S.H.M.
The work done for the additional very small displacement dy against the restoring force will be
d W = F dy = - (-ky) dy = ky dy
? Total work done for displacing the particle from the mean position to a position of displacement y will be
POTENTIAL ENERGY : POTENTIAL ENERGY This work done appears as potential energy U of the particle at the given instant. Thus
KINETIC ENERGY : KINETIC ENERGY K.E.of the particle at the instant t,is given by
KINETIC ENERGY : KINETIC ENERGY
TOTAL ENERGY : TOTAL ENERGY Total energy of the particle at the instant t will be
TOTAL ENERGY : TOTAL ENERGY For a given particle in S.H.M.,m,?, a are constant. Therefore,the total energy of the particle executing S.H.M. remains constant at all times.
At the instant t = 0, the particle is at the mean position, then y = 0
? P.E. of the particle,
U = 1/2m?2(0)2=0
TOTAL ENERGY : TOTAL ENERGY K.E.of the particle,
TOTAL ENERGY : TOTAL ENERGY Thus at mean position, the total energy of the particle in S.H.M. is in the form of kinetic energy.
At the instant t = T/4, the particle is at the extreme position, so that y = a
? P.E. of the particle,
TOTAL ENERGY : TOTAL ENERGY K.E.of the particle, Thus at extreme position, the total energy of the particle in S.H.M. is in the form of its potential energy.
TOTAL ENERGY : TOTAL ENERGY The variation of P.E. and K.E. with time have been shown in Fig. By dotted parabolic curve and thin solid parabolic curve. The variation of total energy of the particle in S.H.M.with
time has been shown by thick solid straight line AB parallel to time axis.
TOTAL ENERGY : TOTAL ENERGY From the graph, we note that P.E. or K.E.completes two vibrations in a time during which S.H.M. completes one vibration. Thus the frequency of P.E.or K.E. is double than that of S.H.M.
IMPORTANT POINTS FORTOTAL ENERGY IN SHM : IMPORTANT POINTS FORTOTAL ENERGY IN SHM In simple harmonic motion,the total energy of the particle is constant at all instants which is totally kinetic when particle is passing from the mean position and is totally potential when particle is passing from the extreme position.
IMPORTANT POINTS FORTOTAL ENERGY IN SHM : IMPORTANT POINTS FORTOTAL ENERGY IN SHM The frequency of kinetic energy or potential energy of a particle in S.H.M.is double than that of a particle in linear S.H.M.
The frequency of total energy of particle in S.H.M.is zero because it remains constant.
IMPORTANT POINTS FORTOTAL ENERGY IN SHM : IMPORTANT POINTS FORTOTAL ENERGY IN SHM The average value of kinetic energy or potential energy of a particle in S.H.M.in a time
of one complete oscillation = Kav = Uav
But the average value of total energy of a particle in S.H.M. in one complete oscillation
FORCE LAW FOR SIMPLE HARMONIC MOTION AND EXPRESSION FOR TIME PERIOD : FORCE LAW FOR SIMPLE HARMONIC MOTION AND EXPRESSION FOR TIME PERIOD Suppose initially, a body is in equilibrium. The forces on the body balance. When a small displacement is given to the body, three possibilities arise:
FORCE LAW AND TIME PERIOD : FORCE LAW AND TIME PERIOD (i) If the forces still balance, then the body will come to rest in new position. In this case, the body is in NEUTRAL EQUILIBRIUM.
(ii) If the net force produced is acting in the direction of displacement,then the body will move onwards in the direction of force. In this case,the body is in UNSTABLE EQUILIBRIUM.
FORCE LAW AND TIME PERIOD : FORCE LAW AND TIME PERIOD (iii) If the net force produced is acting in a direction opposite to that of displacement,then the body will return to its initial position of equilibrium. In this case,the body is in STABLE EQUILIBRIUM.
The simple harmonic motion of a body takes place under the condition of stable equilibrium.
FORCE LAW AND TIME PERIOD : FORCE LAW AND TIME PERIOD Consider a body of mass m executing simple harmonic motion with a constant angular frequency ?,along a straight line with mean position at O. Let C and D be the extreme positions of the body during motion.
D----------.----y---.----C
O P
FORCE LAW AND TIME PERIOD : FORCE LAW AND TIME PERIOD Let at any instant, the body be at P where OP = y = displacement of S.H.M, The restoring force F acting on the body in S.H.M.at the given instant is
F = -ky
FORCE LAW AND TIME PERIOD : FORCE LAW AND TIME PERIOD where k is a force constant. It is the force required to give unit displacement to the body.
The relation F = -ky is called force law for S.H.M.
The acceleration of the body executing S.H.M. is
A= -?2y
Which is directed towards mean position
FORCE LAW AND TIME PERIOD : FORCE LAW AND TIME PERIOD ? Restoring force on the body,
F = m A = - m ?2y
From F = -ky and above eq.
ky = m ?2y
Or
FORCE LAW AND TIME PERIOD : In different types of S.H.Ms, the quantities m and k will go on taking different forms and names.
In general, m is called inertia factor and k is called the spring factor. Thus, in general S.H.M. FORCE LAW AND TIME PERIOD
FORCE LAW AND TIME PERIOD : FORCE LAW AND TIME PERIOD In linear S.H.M;The spring factor stands for force per unit displacement and inertia factor for mass of the body executing S.H.M.
FORCE LAW AND TIME PERIOD : FORCE LAW AND TIME PERIOD In linear S.H.M;The spring factor stands for force per unit displacement and inertia factor for mass of the body executing S.H.M.
In angular S.H.M.,the spring factor stands for Torque constant i.e. the moment of the couple to produce unit angular displacement or the restoring torque per unit angular
FORCE LAW AND TIME PERIOD : FORCE LAW AND TIME PERIOD displacement and inertia factor stands for moment of inertia of the body executing S.H.M.
Thus, in linear S.H.M,
FORCE LAW AND TIME PERIOD : FORCE LAW AND TIME PERIOD
SIMPLE PENDULUM : SIMPLE PENDULUM An ideal simple pendulum consists of a heavy point mass body suspended by a weightless inextensible and perfectly flexible string from a rigid support about which it is free to oscillate.
SIMPLE PENDULUM : SIMPLE PENDULUM The point of suspension S. Here O is called equilibrium position or mean position or
point of oscillation of simple pendulum. The distance between point of suspension and the point of oscillation is called the effective length of simple pendulum.
SIMPLE PENDULUM : SIMPLE PENDULUM When the bob is displaced to position P, through a small angle ? from the vertical, the various forces acting on the bob at P are
(i) the weight mg of the bob acting vertically downwards.
(ii) the tension T in the string acting along PS. Resolving mg into two rectangular components,we get
SIMPLE PENDULUM : SIMPLE PENDULUM (a) mg cos ? acts along PA,opposite to tension T.
(b) mg sin ? acts along PB,tangent to the arc OP and directed towards O.
If the string neither slackens nor breaks but remains taut, then T = mg cos ?
The force mg sin ? will provide the restoring torque, which tends to bring the
SIMPLE PENDULUM : SIMPLE PENDULUM Bob back to its mean position O. The restoring torque ? is given by
? = - (mg sin ?) l = - mg l sin ?
Here, negative sign shows that the torque acts to reduce ?. If ? is small (i.e., less than 10o),then sin ? can be replaced by ?)
Replacing sin ? by ? in above eq ,we have
? = -mgl ?
SIMPLE PENDULUM : SIMPLE PENDULUM From ? = -mgl ? we note that
will bring the bob back towards its equilibrium position, so if the bob is left free, it will execute angular simple harmonic motion.
Comparing ? = -mgl ? , with the equation
? = - k ?, we have
Spring factor, k = mgl
SIMPLE PENDULUM : SIMPLE PENDULUM Here, inertia factor = moment of inertia of bob about the point of suspension = ml2
In simple harmonic motion,
time period,
SIMPLE PENDULUM : SIMPLE PENDULUM From , we note that the time period of
simple pendulum
(i) is independent of the amplitude of vibration provided it is small.
(ii) is independent of the mass of the bob of pendulum.
(iii) depends upon the length of the pendulum and acceleration due to gravity at the place of doing the experiment.
SIMPLE PENDULUM : SIMPLE PENDULUM Note. 1. As , therefore T must increase with increase in the value of l (i.e.effective length of simple pendulum).It does not mean that as , T will become infinite. Infact, the relation (41) is not valid for pendulum whose effective length
(radius of the earth). This shows that T increases up to a certain limit only.
SIMPLE PENDULUM : SIMPLE PENDULUM 2. If the effective length l of simple pendulum is very large which is comparable with the radius of earth (Re), then it can be shown that, the time period T is given by the relation.
SIMPLE PENDULUM : SIMPLE PENDULUM
SIMPLE PENDULUM : SIMPLE PENDULUM 3. As ,therefore T will increase with decrease in the value of g. As the value of acceleration due to gravity is less at hills or in mines than that on surface of earth, hence the time period of simple pendulum increases at hills or inside the mines. Due to this, the pendulum clock will slowed down. It means,the pendulum clock will be losing the time at hills or inside the mines.
SIMPLE PENDULUM : SIMPLE PENDULUM 4. If the bob of the simple pendulum is suspended by a wire then effective length of pendulum will increase with the rise of temperature. Due to which the time period of simple pendulum will increase.
SIMPLE PENDULUM : SIMPLE PENDULUM 5. If bob of simple pendulum is made to oscillate in some fluid of density
(the density of the body) then time period of simple pendulum gets increased. Here the effective value of acceleration due to gravity acting on the bob will be
SIMPLE PENDULUM : SIMPLE PENDULUM
SECONDS PENDULUM : SECONDS PENDULUM Second’s Pendulum. It is that simple pendulum whose time period of vibrations is two seconds. The bob of such pendulum while oscillating passes through the mean position after every one second.i.e.it beats seconds.
Putting, T =2s and g = 9.8 ms-2 in (41),we get
SECONDS PENDULUM : SECONDS PENDULUM
SECONDS PENDULUM : SECONDS PENDULUM Hence length of a second’s pendulum (where g = 9.8 cm-2) is 99.3 cm. The length of second’s pendulum changes with the change in the value of acceleration due to gravity from place to place. For all your Physics Problems
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SIMPLE PENDULUM : SIMPLE PENDULUM 7. When a simple pendulum is mounted on a trolley which is moving down on an inclined plane of inclination ?, then the effective acceleration due to gravity perpendicular to the plane = g cos ? , then the time period of oscillations is given by
SIMPLE PENDULUM : SIMPLE PENDULUM If a simple pendulum is made to oscillate in a satellite or freely falling lift, then effective value of acceleration due to gravity,
This shows that the pendulum will not oscillate and will remain where left free.
SIMPLE PENDULUM : SIMPLE PENDULUM If a simple pendulum is in a carriage which is accelerating with acceleration a
(i) upwards, then g’=g +a; so
(ii) downwards, then g’= g –a ; so
SIMPLE PENDULUM : SIMPLE PENDULUM (iii) horizontally, then g’= For all your Physics Problems
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OSCILLATIONS OF A LOADED SPRING : OSCILLATIONS OF A LOADED SPRING Vibrations in the horizontal direction.
Consider a body of mass m attached to one end B of a light elastic massless spring of spring constant k. The other end A of the spring is fixed to a rigid support. The body is resting on a frictionless horizontal surface. The weight of the body is balanced
OSCILLATIONS OF A LOADED SPRING : OSCILLATIONS OF A LOADED SPRING by the reaction of the horizontal surface.
Let the body be displaced towards right through a small distance y. The spring gets stretched. A restoring force F comes into play due to elastic nature of the spring.
Then, F = -ky
OSCILLATION : OSCILLATION The negative sign indicates that the restoring force is directed towards the equilibrium position of the body.From (52),we note that F ? y and F is always directed towards the equilibrium position.
In the displaced position, if the body is left free, it will start executing linear S.H.M. on the smooth horizontal surface,
OSCILLATIONS OF A LOADED SPRING : OSCILLATIONS OF A LOADED SPRING with mean position as equilibrium position.
Here, inertia factor = mass of body = m.
Spring factor = force constant of spring = k
Vibrations of a vertical spring : Consider a light and highly elastic massless spring AB of spring constant k suspended from a rigid support at A, Fig. The spring is unstretched and is in relaxation state. Let a body of mass m be attached to the lower end B of the spring. The spring gets stretched and suffers an extension, BC=l, Vibrations of a vertical spring
OSCILLATION : OSCILLATION Let F1 be the restoring force set up in the spring, then F1 = - kl
Here, negative sign shows that the extension l is directed downwards and restoring force F is directed upwards. As the system is in equilibrium, therefore,
F1 + mg = 0 or F1 = - mg
Hence, mg = kl or k = mg/l
OSCILLATION : OSCILLATION Let the body be pulled downwards through a small distance, CD = y ( < l ) . Now the total extension in the spring is (l + y). If F2 is the restoring force in this position, then
F2 = - k (l + y)
The effective restoring force will be
F = F2 – F1
= - k (l + y) – ( -kl) = - ky
OSCILLATION : OSCILLATION From F=-ky, we note that F ? y and F is directed towards equilibrium position. Hence if the pull from the suspended body is released, it will start executing S.H.M.with C as mean position.
Here, spring factor = spring constant = k
Inertia factor = mass of body = m
OSCILLATION : OSCILLATION
OSCILLATIONS OF LOADED SPRING COMBINATIONS : OSCILLATIONS OF LOADED SPRING COMBINATIONS Consider two springs of spring constants k1 and k2. Let there be three spring combinations as shown in Fig.
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PARALLEL COMBINATION : PARALLEL COMBINATION Case I. Here, a body of weight mg is suspended by the two springs in parallel combination, Fig., Let the body be pulled downwards through a small distance y. Each spring gets stretched by a length y. If F1 and F2 are the restoring forces set up due to extension of springs,then
F1 = k1y
and F2 = - k2 y
Total restoring force,
F = F1 + F2 = - (k1 + k2)y
OSCILLATION : OSCILLATION If K is the spring constant of this combination of springs, then restoring force,
F = - ky
? K = k1 + k2
If the body is left free after pulling a little distance down, it will start executing S.H.M. of period T given by
OSCILLATION : OSCILLATION .
SERIES COMBINATION : SERIES COMBINATION Case.II. Here, the body of weight mg is suspended at the free end of the two springs in series combination. Fig. When the body is pulled downwards through a little distance y, the two springs suffer different extensions say y1 and y2. But the restoring force is same in each spring.
OSCILLATION : OSCILLATION .
OSCILLATION : OSCILLATION If K is the spring constant of series combination of springs, then F = -Ky
If the body is left free after pulling down, it will execute S.H.M. of period
OSCILLATION : OSCILLATION
PARALLEL -2 COMBINATION : PARALLEL -2 COMBINATION Case III. Here, the body of weight mg is connected in between the two springs, Fig, (c) . When the body is pulled to one side through a small distance y, one spring gets compressed by length y and the other spring gets stretched by length y. The restoring forces F1 and F2 set up in both the springs will act in the same direction.
OSCILLATION : OSCILLATION Then F1 = -k1y and F2 = - k2y
Total restoring force, F = F1 + F2
= - k1y – k2y = - (k1 + k2) y
If K is the spring constant of
this combination of springs, then
F = - ky ? K = (k1+k2)
If the body after pulling down a
little is left free,it will execute
S.H.M. of period
OSCILLATION : OSCILLATION .
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