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- 1. MODERN Physics PHY1005
- 2. Oscillations Module - 2
- 3. Introduction We come across various kinds of motions in our daily life. You have already studied some of them like linear and projectile motion. However, these motions are non-repetitive. Here, we are going to learn about periodic and oscillatory motion. Let’s find out what they are. Periodic Motion What is common in the motion of the hands of a clock, wheels of a car and planets around the sun? They all are repetitive in nature, that is, they repeat their motion after equal intervals of time. A motion which repeats itself in equal intervals of time is periodic motion.
- 4. A body starts from its equilibrium position (at rest) and completes a set of motion after which it will return to its equilibrium position. This set of motion repeats itself in equal intervals of time to perform the periodic motion. Circular motion is an example of periodic motion. Oscillatory Motion Oscillatory motion is the repeated to and fro movement of a system from its equilibrium position. Every system at rest is in its equilibrium position. At this point, no external force is acting on it. Therefore, the net force acting on the system is zero. Now, if this system is displaced a little from its fixed point, a force acts on the system which tries to bring back the system to its fixed point. This force is the restoring force and it gives rise to oscillations or vibrations.
- 5. For example, consider a ball that is placed in a bowl. It will be in its equilibrium position. If displaced a little from this point, it will oscillate in the bowl. Therefore, every oscillatory motion is periodic but all periodic motions are not oscillatory. Example, the circular motion is a periodic motion but not oscillatory. Moreover, there is no significant difference between oscillations and vibrations. In general, when the frequency is low, we call it oscillatory motion and when the frequency is high, we call it vibrations.
- 6. Simple Harmonic motion (SHM) A very common type of periodic motion is called simple harmonic motion (SHM). A system that oscillates with SHM is called a simple harmonic oscillator. All the Simple Harmonic Motions (SHMs) are oscillatory and also periodic but not all oscillatory motions are SHM. Now, consider mass 𝑚 is connected to one end of the spring of negligible mass whose other end is fixed to a rigid wall. There is no force applied to it, it is in equilibrium position. Now, pull it outwards (extension), there is a force exerted by the spring that is directed towards the equilibrium position. And, if we push the spring inwards (compression), there is a force exerted by the spring towards the equilibrium position.
- 7. In each case, the force exerted by the spring is towards the equilibrium position, this force is called the restoring force. Let the force be 𝐹 and the displacement of the spring from the equilibrium position be 𝑥. The restoring force 𝐹 = – 𝑘𝑥 (negative sign indicates that the force is in the opposite direction of displacement). Here, 𝑘 is the constant called the force constant. This force obeys Hooke’s law. If the net force can be described by Hooke’s law and there is no damping i.e., slowing down due to friction or other nonconservative forces. Then a simple harmonic oscillator oscillates with equal displacement on either side of the equilibrium position. Restoring Force : The simple harmonic motion of a body in which the restoring force is always directed towards the equilibrium position or mean position and its magnitude is directly proportional to the displacement from the equilibrium position.
- 8. Equation of motion of SHM The only force that acts parallel to the surface is the force due to the spring. The force exerted by the spring is towards the equilibrium position which is called the restoring force. The restoring force 𝐹 = – 𝑘𝑥 Here, 𝑘 is the force constant or stiffness factor of the spring. 1 Consider a frictionless surface block of mass 𝑚 is attached one end of the spring of negligible mass whose other end is fixed to a rigid wall. There are three forces on the mass: the weight, the normal force, and the force due to the spring. Two forces that act perpendicular to the surface are the weight and the normal force, which have equal magnitudes and opposite directions, and thus sum to zero. x
- 9. From Newton’s second law of motion 𝐹 = 𝑚𝑎 2 𝑎 = 𝑑𝑣 𝑑𝑡 = 𝑑 𝑑𝑡 𝑑𝑥 𝑑𝑡 = 𝑑2 𝑥 𝑑𝑡2 Substitute eqn. (2) in eqn. (1), we have 𝑚𝑎 = −𝑘𝑥 𝑎 = − 𝑘 𝑚 𝑥 𝑑2 𝑥 𝑑𝑡2 = −𝜔2 𝑥 𝑑2 𝑥 𝑑𝑡2 + 𝜔2 𝑥 = 0 𝐻𝑒𝑟𝑒 𝜔 = 𝑘 𝑚 3 4 𝜔 is called the angular frequency of the oscillator. The angular frequency depends only on the force constant and the mass, and not the amplitude. Equation (3) represents the equation of motion of the simple harmonic oscillator.
- 10. Solution (Displacement) of SHM 𝑥(𝑡) The restoring force of an SHM is directly proportional to the displacement of the block from its equilibrium position and is directed opposite to the direction of the displacement. Let the initial condition, 𝑥 = 𝐴 and 𝑣 = 0 at 𝑡 = 0, then the solution of eqn. (3) we get 𝑥 𝑡 = 𝐴 cos 𝜔𝑡 where 𝐴 is the maximum displacement which is called the amplitude of the motion. If 𝑇 is the time period for one complete oscillation, then 𝑥 𝑡 + 𝑇 = 𝑥 𝑡 𝐴 cos 𝜔 𝑡 + 𝑇 = 𝐴 cos 𝜔𝑡 𝜔𝑇 = 2𝜋 𝑇 = 2𝜋 𝜔 = 2𝜋 𝑚 𝑘 5 6 Time period
- 11. Frequency 𝑓 = 1 𝑇 = 𝜔 2𝜋 = 1 2𝜋 𝑘 𝑚 Angular frequency 𝜔 = 2𝜋 𝑓 = 𝑘 𝑚 7 8 Now, if we consider a sine function, the result will be the same. Further, taking a linear combination of sine and cosine functions is also a periodic function with period 𝑇. 𝑥 𝑡 = 𝐶 cos 𝜔𝑡 + 𝐷 sin 𝜔𝑡 where 𝐶 = 𝐴 cos 𝜙 and 𝐷 = 𝐴 sin 𝜙. 𝐶 and 𝐷 are initial conditions. Then the above equation becomes 𝑥 𝑡 = 𝐴 cos 𝜔𝑡 − 𝜙 In this equation 𝐴 (amplitude) and 𝜙 (phase angle or phase constant) are given by, 𝐴 = 𝐶2 + 𝐷2 𝑎𝑛𝑑 𝜙 = tan−1 𝐷 𝐶 Therefore, we can express any periodic function as a superposition of sine and cosine functions of different time periods with suitable coefficients. The period of the function is 2𝜋/𝜔. 9
- 12. Velocity, acceleration and total energy of SHM Using eqn. (9), the magnitude of the velocity of simple harmonic oscillator is 𝑣 = 𝑥 = 𝑑𝑥 𝑑𝑡 = 𝑑 𝑑𝑡 𝐴 cos 𝜔𝑡 − 𝜙 = −𝐴𝜔 sin 𝜔𝑡 − 𝜙 𝑣 = 𝐴𝜔 1 − 𝑥2 𝐴2 10 The cosine function oscillates between –1 and +1. The maximum velocity occurs at the equilibrium position (𝑥 = 0) when the mass is moving from 𝑥 = +𝐴. The maximum velocity in the negative direction is attained at the equilibrium position (𝑥 = 0) when the mass is moving from 𝑥 = −𝐴.
- 13. The acceleration of the particle is 𝑎 = 𝑥 = 𝑑2 𝑥 𝑑𝑡2 = 𝑑2 𝑑𝑡2 𝐴 𝑐𝑜𝑠 𝜔𝑡 − 𝜙 𝑎 = −𝐴𝜔2 cos 𝜔𝑡 − 𝜙 = −𝜔2 𝑥 In simple harmonic motion, the acceleration is directly proportional to the displacement but opposite in sign. The maximum acceleration 𝑎 = 𝐴𝜔2 occurs at the position (𝑥 = −𝐴), and the acceleration at the position (𝑥 = 𝐴) and is equal to −𝑎. 11 The total energy that a particle possesses while performing simple harmonic motion is energy in simple harmonic motion. To calculate the energy in simple harmonic motion, we need to calculate the kinetic and potential energy that the particle possesses.
- 14. Potential Energy(PE) of Particle Performing SHM Potential energy is the energy possessed by the particle when it is at rest. Consider a particle of mass 𝑚 performing simple harmonic motion at a distance 𝑥 from its mean position. The restoring force acting on the particle is 𝐹 = −𝑘𝑥 where k is the force constant. Now, the particle is given further infinitesimal displacement 𝑑𝑥 against the restoring force 𝐹. The work done 𝑑𝑤 during the displacement is The total work done is stored in the form of potential energy (PE) i.e., 𝑉 = −𝑑𝑤 = 𝑘𝑥 𝑑𝑥 = 1 2 𝑘𝑥2 = 1 2 𝑚𝜔2 𝑥2 𝑉 = 1 2 𝑚𝜔2 𝑥2 = 1 2 𝑚𝜔2 𝐴2 𝑐𝑜𝑠2 𝜔𝑡 − 𝜙 𝑑𝑤 = 𝐹 𝑑𝑥 = −𝑘𝑥 𝑑𝑥 12
- 15. Kinetic Energy (KE) in SHM Kinetic energy is the energy possessed by an object when it is in motion. The KE of a particle with mass 𝑚 performing simple harmonic motion is 𝐾𝐸 = 1 2 𝑚𝑥2 = 1 2 𝑚𝜔2 𝐴2 𝑠𝑖𝑛2 𝜔𝑡 − 𝜙 13 The total energy (TE) in SHM is the sum of its potential energy and kinetic energy. 𝑇𝐸 = 𝑃𝐸 + 𝐾𝐸 = 1 2 𝑚𝜔2 𝐴2 14 The total energy in the simple harmonic motion of a particle performing simple harmonic motion remains constant. Therefore, it is independent of displacement 𝑥. Thus, the total energy in the simple harmonic motion of a particle is: Directly proportional to its mass Directly proportional to the square of the frequency of oscillations and Directly proportional to the square of the amplitude of oscillation.
- 16. The law of conservation of energy states that energy can neither be created nor destroyed. The total energy in SHM will always be constant. However, kinetic energy and potential energy are interchangeable. The graph shows the kinetic and potential energy vs instantaneous displacement. At the mean position, the total energy in simple harmonic motion is purely kinetic and at the extreme position, the total energy in simple harmonic motion is purely potential energy. At other positions, kinetic and potential energies are interconvertible and their sum is equal to The nature of the graph is parabolic. 𝑇𝐸 = 𝑃𝐸 + 𝐾𝐸 = 1 2 𝑚𝜔2 𝐴2
- 17. Concepts of Simple Harmonic Motion (S.H.M) Amplitude: The maximum displacement of a particle from its equilibrium position or mean position is its amplitude, and its direction is always away from the equilibrium position. Period: The time taken by a particle to complete one oscillation is its period. Therefore, the period of S.H.M. is the least time after which the motion will repeat itself. Frequency: Frequency of S.H.M. is the number of oscillations that a particle performs per unit time. Phase: Phase of S.H.M. is its state of oscillation, and the magnitude and direction of displacement of particles represent the phase. 𝑇 = 2𝜋 𝜔 = 2𝜋 𝑚 𝑘 𝑓 = 1 𝑇 = 𝜔 2𝜋 = 1 2𝜋 𝑘 𝑚
- 18. Characteristics of SHM (i) In SHM, acceleration of the particle is directly proportional to its displacement, and directed towards the mean position. (ii) It can be represented by a single harmonic function of sine or cosine. (iii) Total energy of the particle executing SHM remains conserved. Directly proportional to its mass Directly proportional to the square of the frequency of oscillations and Directly proportional to the square of the amplitude of oscillation. (iv) It is a periodic motion.
- 19. Damped harmonic oscillator We know that in the ideal case the total energy of a harmonic oscillator remains constant. This implies that once such a system is set in motion it will continue to oscillate forever. Such oscillations are said to be free or undamped. Do you know of any physical system in the real world which experiences no damping? No. You must have observed that oscillations of a swing, a simple or torsional pendulum and a spring-mass system when left to themselves, die down gradually. This indicates that every oscillating system loses some energy as time elapses. Where does this energy go? Answer is, when a body oscillates in a medium it experiences resistance to its motion. This means that damping force comes into play.
- 20. Damping force can arise within the body itself, as well as due to the surrounding medium (air or liquid). The work done by the oscillating system against the damping forces leads to dissipation of energy of the system. That is, the energy of an oscillating body is used up in overcoming damping. A familiar example is that of brakes-we increase friction to reduce the speed of a vehicle in a short time. In general, damping causes a loss of energy. Therefore, we habitually try to minimize it.
- 21. A mass 𝑚 attached to a spring with a force constant 𝑘. The mass oscillates around the equilibrium position in a fluid with viscosity but the amplitude decreases for each oscillation. If damping is small, the period and frequency are constant and are nearly the same as for SHM, but the amplitude gradually decreases as shown. Equation of motion and solution of Damped harmonic oscillator This occurs because the non-conservative damping force removes energy from the system, usually in the form of thermal energy. Consider the forces acting on the mass. The net force is equal to the sum of restoring force 𝐹 = −𝑘𝑥 exerted by the spring and the damping force 𝐹𝑑 .
- 22. If the magnitude of the velocity is small i.e., the mass oscillates slowly, the damping force is proportional to the velocity and acts against the direction of motion. 𝛾 is called the damping coefficient. The net force on the mass is 𝐹𝑛𝑒𝑡 = 𝑚𝑎 = 𝑚 𝑑2 𝑥 𝑑𝑡2 = −𝑘𝑥 − 𝛾 𝑑𝑥 𝑑𝑡 𝐹𝑑 = −𝛾𝑣 = −𝛾 𝑑𝑥 𝑑𝑡 𝑑2 𝑥 𝑑𝑡2 = − 𝑘 𝑚 𝑥 − 𝛾 𝑚 𝑑𝑥 𝑑𝑡 = −𝜔2 𝑥 − 2𝑏 𝑑𝑥 𝑑𝑡 𝑑2 𝑥 𝑑𝑡2 + 2𝑏 𝑑𝑥 𝑑𝑡 + 𝜔2 𝑥 = 0 After rearranging terms, the equation of motion of a damped oscillator takes the form 1 2 𝜔 = 𝑘 𝑚 2𝑏 = 𝛾 𝑚 3 4 Note that a factor 2 has been introduced in the damping term as it helps to obtain a neat expression for the solution of this eqn. (2). Where
- 23. Eqn. (2) is a linear second order homogeneous differential equation with constant coefficients. If there were no damping, the second term in Eqn. (2) will be zero. The general solution of Eqn. (2) 𝑥 𝑡 = 𝐴𝑒𝛼𝑡 5 when 𝐴 and 𝛼 are unknown constants. Differentiating Eq.(5) twice with respect to time, we get 𝑑𝑥 𝑑𝑡 = 𝐴𝛼𝑒𝛼𝑡 𝑑2 𝑥 𝑑𝑡2 = 𝐴𝛼2 𝑒𝛼𝑡 Substituting these expressions in Eqn. (2), we get 𝛼2 + 2𝑏𝛼 + 𝜔2 𝐴𝑒𝛼𝑡 = 0 For this equation to hold at all times, we should either have 𝐴 = 0 which is trivial, or 𝛼2 + 2𝑏𝛼 + 𝜔2 = 0 This equation is quadratic in 𝛼. The two roots of this equation are 𝛼1 = −𝑏 + 𝑏2 − 𝜔2 𝛼2 = −𝑏 − 𝑏2 − 𝜔2
- 24. Thus, the two possible solutions of Eqn. (2) are 𝑥1 𝑡 = 𝐴1𝑒−𝑏𝑡 𝑒 𝑏2−𝜔2 𝑡 𝑥2 𝑡 = 𝐴2𝑒−𝑏𝑡 𝑒 − 𝑏2−𝜔2 𝑡 Since Eqn. (2) is linear, the principle of superposition is applicable. Hence, the general solution is obtained by the superposition of 𝑥1 and 𝑥2 : 𝑥 𝑡 = 𝑒−𝑏𝑡 𝐴1𝑒 𝑏2−𝜔2 𝑡 + 𝐴2𝑒 − 𝑏2−𝜔2 𝑡 The general solution of Eqn. (2) includes both exponential and harmonic terms. the quantity 𝑏2 − 𝜔2 can be positive, zero or negative depending on whether 𝑏 is greater than, equal to or less than 𝜔 respectively. These three possibilities are: 1. If 𝑏 > 𝜔, we say that the system is over (Heavy) damped 2. If 𝑏 = 𝜔, we have a critical damped system 3. If 𝑏 < 𝜔, we have an weak or light (under-damped) damped system Each of these conditions gives a different solution, which describes a particular behaviour. 6
- 25. Over (Heavy) Damping system When resistance to motion is very strong, the system is said to be heavily damped. Examples, vibrations of a pendulum in a viscous medium such as thick oil and motion of the coil of a dead beat galvanometer are heavily damped systems. a system is said to be heavily damped if 𝑏 > 𝜔. Then the quantity 𝑏2 − 𝜔2 is positive. If we put 𝛽 = 𝑏2 − 𝜔2 The general solution for damped oscillator given by Eqn. (6) reduces to 𝑥 𝑡 = 𝑒−𝑏𝑡 𝐴1𝑒𝛽 𝑡 + 𝐴2𝑒−𝛽 𝑡 This represents non-oscillatory behaviour. Such a motion is called dead-beat. The actual displacement will be determined by the initial conditions. 7
- 26. Let us suppose that to begin with the oscillator is in equilibrium position, i.e 𝑥 = 0 at 𝑡 = 0. Next we give it a sudden kick so that it acquires a velocity 𝑣 = 𝑣0 at 𝑡 = 0. Then Eq. (7) have 𝐴1 + 𝐴2 = 0 𝑖. 𝑒., 𝐴1 = −𝐴2 −𝑏 𝐴1 + 𝐴2 + 𝛽 𝐴1 − 𝐴2 = 𝑣0 𝐴1 = −𝐴2 = 𝑣0 2𝛽 On substituting these results in Eq. (7), we can write in compact form 𝑥 𝑡 = 𝑣0 2𝛽 𝑒−𝑏𝑡 𝑒𝛽 𝑡 − 𝑒−𝛽 𝑡 = 𝑣0 𝛽 𝑒−𝑏𝑡 sinh 𝛽𝑡 8 where sinh 𝛽𝑡 is hyperbolic sine function. From Eqn. (8) it is clear that 𝑥 (𝑡) will be determined by the interplay of an increasing hyperbolic function and a decaying exponential. These are shown in Figure (1). sinh 𝛽𝑡 𝑒−𝑏𝑡
- 27. Critical Damping system A system is critically damped if b is equal to the natural frequency 𝜔 of the system. This means that 𝑏2 − 𝜔2 = 0; so that Eqn. (6) reduces to 𝑥 𝑡 = 𝑒−𝑏𝑡 𝐴1 + 𝐴2 = 𝐴𝑒−𝑏𝑡 But the above equation has only one constant. Does this mean that it is not a complete solution? The reason is simple : the quadratic equation for 𝛼 has equal roots. So, the two terms in Eqn. (6) give the same time dependence and reduce to one term. Then the complete solution of Eqn. (2) 𝑥 𝑡 = 𝑃 + 𝑄𝑡 𝑒−𝑏𝑡 Where 𝑃 and 𝑄 are constants. 𝑃 has the dimensions of length and 𝑄 those of velocity. These can be determined easily from the initial conditions. Let us assume that the system is disturbed from its mean equilibrium position by a sudden impulse. 9
- 28. That is, at 𝑡 = 0, 𝑥(0) = 0 and 𝑑𝑥 𝑑𝑡 𝑡=0 = 𝑣0 This gives 𝑃 = 0 and 𝑄 = 𝑣0 , so that the complete solution is 𝑥 𝑡 = 𝑣0𝑡 𝑒−𝑏𝑡 10 At maximum displacement, 𝑑𝑥 𝑑𝑡 𝑥=𝑥𝑚𝑎𝑥 = 0 and 𝑑2𝑥 𝑑𝑡2 𝑥=𝑥𝑚𝑎𝑥 < 0. differentiating Eqn. (10) w.r.t time 𝑑𝑥 𝑑𝑡 𝑥=𝑥𝑚𝑎𝑥 = 𝑣0 𝑒−𝑏𝑡 − 𝑣0𝑏 𝑡 𝑒−𝑏𝑡 = 0 𝑣0 𝑒−𝑏𝑡 1 − 𝑏𝑡 = 0, 𝑡ℎ𝑒𝑛 𝑡 = 1 𝑏 𝑥𝑚𝑎𝑥 = 𝑣0 1 𝑏 𝑒−𝑏 1 𝑏 = 0.368 𝑣0 𝑏 = 0.736 𝑚𝑣0 𝛾 The displacement vs time graph of a critically damped system shown in figure
- 29. Weak or light Damping (Under-damping) system When 𝑏 < 𝜔 we refer to it as a case of weak damping. This implies that 𝑏2 − 𝜔2 is a negative quantity, i.e. 𝑏2 − 𝜔2 is imaginary. Let us rewrite it as 𝑏2 − 𝜔2 = −1 𝜔2 − 𝑏2 = 𝑖 𝜔2 − 𝑏2 = 𝑖𝜔𝑑 𝜔𝑑 = 𝜔2 − 𝑏2 = 𝑘 𝑚 − 𝛾2 4𝑚2 12 is a real positive quantity. Here note that for no damping (𝑏 = 0), 𝜔𝑑 reduces 𝜔, the natural frequency of the SHO. Using Eqn. (11), Eqn. (6) changes to We write the complex exponential in terms of sine and cosine functions. This gives 11 𝑥 𝑡 = 𝑒−𝑏𝑡 𝐴1𝑒𝑖𝜔𝑑 𝑡 + 𝐴2𝑒−𝑖𝜔𝑑 𝑡 𝑥 𝑡 = 𝑒−𝑏𝑡 𝐴1 cos 𝜔𝑑 𝑡 + 𝑖 sin 𝜔𝑑 𝑡 + 𝐴2 cos 𝜔𝑑 𝑡 − 𝑖 sin 𝜔𝑑 𝑡 𝑥 𝑡 = 𝑒−𝑏𝑡 𝐴1 + 𝐴2 cos 𝜔𝑑 𝑡 + 𝑖 𝐴1 − 𝐴2 sin 𝜔𝑑 𝑡 13
- 30. Let us now put 𝐴1 + 𝐴2 = 𝐴0 cos 𝜙 𝑖 𝐴1 − 𝐴2 = 𝐴0 sin 𝜙 14 where 𝐴0 and 𝜙 are arbitrary constants. These are given by solving Eqn. (14) 𝐴0 = 2 𝐴1𝐴2 cos 𝜙 = 𝐴1 + 𝐴2 𝐴0 = 𝐴1 + 𝐴2 2 𝐴1𝐴2 Using Eqn. (14), Eqn. (13) changes to Eqn. (15) represents the general solution of Eqn. (2) for a weakly damped oscillator (𝑏 < 𝜔). The damped oscillatory behavior described by Eqn. (15) is plotted in Figure for the particular case of 𝜙 = 0. Note that the amplitude decreases exponentially with time at a rate governed by 𝑏. So we con say that the motion of a weakly damped system is not simple harmonic motion. 𝑥 𝑡 = 𝐴0𝑒−𝑏𝑡 cos 𝜔𝑑 𝑡 − 𝜙 15 𝐴0𝑒−𝑏𝑡 −𝐴0𝑒−𝑏𝑡 𝐴0 cos 𝜔𝑑 𝑡 − 𝜙
- 31. We may conclude that damping results in decrease of amplitude and angular frequency. The period of oscillation is given by 𝑇 = 2𝜋 𝜔𝑑 = 2𝜋 𝜔2 − 𝑏2 = 2𝜋 𝑘 𝑚 − 𝛾2 4𝑚2 16 Energy decay in a damped harmonic oscillator The question now arises: How does damping influence the energy of a weakly damped oscillator? To answer this, the presence of damping the amplitude of oscillation decreases with time. This means that energy is dissipated in overcoming resistance to motion. The total energy of a harmonic oscillator is made up of kinetic and potential components. We can still use the same definition and write 𝐸𝑡 = 𝐾𝐸 + 𝑃𝐸 = 1 2 𝑚 𝑥2 + 1 2 𝑘 𝑥2 = 1 2 𝑚 𝑑𝑥 𝑑𝑡 2 + 1 2 𝑚𝜔2 𝑥2 17
- 32. where 𝒅𝒙 𝒅𝒕 denotes instantaneous velocity. For a weakly damped harmonic oscillator, the instantaneous displacement is given by Eqn. (15) : 𝑥 𝑡 = 𝐴0𝑒−𝑏𝑡 cos 𝜔𝑑 𝑡 − 𝜙 15 Differentiating Eqn. (15) with respect to time, and substitute instantaneous displacement 𝑥(𝑡) and velocity 𝒅𝒙 𝒅𝒕 in Eqn. (17) and simplified, the average energy of a weakly damped harmonic oscillator is < 𝐸 > = 1 2 𝑚 𝐴0 2 𝜔2 𝑒−2𝑏𝑡 = 𝐸0𝑒−2𝑏𝑡 Where 𝐸0 = 1 2 𝑚 𝐴0 2 𝜔2 is the total energy of the SHO or undamped oscillator. The average energy of a weakly damped oscillator decreases exponentially with time which is shown in figure. From Eqn. (18), observe that the rate of decay of energy depends on the value of 𝑏; larger the value of 𝑏, faster will be the decay. 𝐸0𝑒−2𝑏𝑡 18
- 33. Quality factor The quality factor (𝑄) represents the number of cycles completed by the oscillator before it "rings down“ or "runs out of energy (𝐸0𝑒−1 ) " (OR) The number of radians through which a damped system oscillates as its average energy decays to 𝐸0𝑒−1 is a measure of the quality factor (𝑄) The damping effect is by means of the rate of decay of average energy. The average energy of a weakly damped oscillator decays to 𝐸0𝑒−1 in time 𝑡 = 1 2𝑏 = 𝑚 𝛾 seconds. If 𝜔𝑑 is its angular frequency, then the oscillator will vibrate through 𝜔𝑑 𝑚 𝛾 radians. The number of radians through which a damped system oscillates as its average energy decays to 𝐸0𝑒−1 is a measure of the quality factor (𝑄) 𝑄 = 𝜔𝑑 2𝑏 = 𝜔𝑑𝑚 𝛾 19
- 34. Note that 𝑄 is only a number and has no dimensions. Damping parameter 𝛾 is small so that 𝑄 is very large. For a weakly damped mechanical oscillator, the quality factor can be expressed in term of the spring factor and damping constant. For weak damping, 𝜔𝑑 = 𝜔 = 𝑘 𝑚 𝑄 = 𝜔𝑑 2𝑏 = 𝜔𝑑𝑚 𝛾 = 𝑘 𝑚 𝑚 𝛾 = 𝑘𝑚 𝛾2 Then Quality factor 20 That is, the quality factor of a weakly damped oscillator is directly proportional to the square root of 𝑘 and inversely proportional to 𝛾. An undamped oscillator (SHO) (γ = 0) has an infinite quality factor
- 35. The quality factor is related to the fractional change in the frequency 𝜔 of an undamped oscillator. The frequency of damped oscillator is 𝜔𝑑 = 𝜔2 − 𝑏2 𝜔𝑑 2 𝜔2 = 1 − 𝑏2 𝜔2 = 1 − 1 4𝑄2 𝑄 = 𝑘𝑚 𝛾2 = 𝑘 𝑚2 𝑚 𝛾2 = 𝜔 2𝑏 𝜔𝑑 𝜔 = 1 − 1 4𝑄2 1/2 = 1 − 1 8𝑄2 Hence, the fractional change in 𝜔 is 1 8𝑄2. Note-1: Q is a measure of damping. It shows how many periods fit in one decay-time interval, or how many periods it takes for the oscillation to ring down, run out of energy. Note-2: Lightly damped oscillations are referred to as high Q, and heavier damped oscillations as low Q.
- 36. Mechanical Forced Oscillator
- 37. Electronic Oscillator