UNIT V: OSCILLATIONS

Transcription

1 UNIT V: OSCILLATIONS Introuction: Motion of boies can be broaly classifie into three categories: [] Translational motion [] Rotational motion [3] Vibrational / Oscillatory motion Translational motion: When the position of a bo varies linearly with time, such a motion is terme as translational motion. Example: A car moving on a straight roa, a ball moving on the groun. Rotational motion: When a bo as a whole oes not change its position linearly with time but rotates about its axis, this motion is sai to be rotational motion. Example: rotation of earth about its axis, rotation of a fly wheel on ball bearings. Vibrational /Oscillatory motion: When a bo executes back an forth motion which repeats over an again about a mean position, then the bo is sai to have Vibrational/oscillatory motion. If such motion repeats in a regular interval of time then it is calle Perioic motion or Harmonic motion an the bo executing such motion is calle Harmonic oscillator. In harmonic motion there is a linear relation between force acting on the bo an isplacement prouce. Example: bob of a penulum clock, motion of prongs of tuning fork, motion of balance wheel of a watch, the up an own motion of a mass attache to a spring. Note:. If there is no linear relation between the force an isplacement then the motion is calle unharmonic motion. Many systems are unharmonic in nature.. In some oscillatory systems the boies may be at rest but the physical properties of the system may unergo changes in oscillatory manner Examples: Variation of pressure in soun waves, variation of electric an magnetic fiels in electromagnetic waves. Parameters of an oscillatory system. Mean position: The position of the oscillating bo at rest.. Amplitue: The amplitue of an SHM is the maximum isplacement of the bo from its mean position. 3. Time Perio: The time interval uring which the oscillation repeats itself is calle the time perio. It is enote by T an its unit in secons. Perio = T= π Displacement Acceleration 4. Frequency: The number of oscillations that a bo completes in one secon is calle the frequency of perioic motion. It is the reciprocal of the time perio T an it is enote by n.

2 Frequency= n = T 5. Phase: It is the physical quantity that expresses the instantaneous position an irection of motion of an oscillating system. If the SHM is represente by y=a sin (ωt+ ), together with ω as the angular frequency. The quantity (ωt+ ) of the sine function is calle the total phase of the motion at time t an is the initial phase or epoch. All perioic motions are not vibratory or oscillatory. In this chapter we shall stu the simplest vibratory motion along one imension calle Simple Harmonic Motion (SHM) which usually occurs in mechanical systems. SIMPLE HARMONIC MOTION (SHM) A bo is sai to be unergoing Simple Harmonic Motion (SHM) when the acceleration of the bo is always proportional to its isplacement an is irecte towars its equilibrium or mean position. Simple harmonic motion can be broaly classifie in to two types, namely linear simple harmonic motion an angular simple harmonic motion. Linear Simple Harmonic Motion: If the bo executing SHM has a linear acceleration then the motion of the bo is linear simple harmonic motion. Examples: motion of simple penulum, the motion of a point mass tie with a spring etc., Angular Simple Harmonic Motion: If the bo executing SHM has an angular acceleration then the motion of the bo is angular simple harmonic motion. Examples: Oscillations of a torsional penulum A particle or a system which executes simple harmonic motion is calle Simple Harmonic Oscillator. Examples of simple harmonic motion are motion of the bob of a simple penulum, motion of a point mass fastene to spring, motion of prongs of tuning fork etc., UN DAMPED OR FREE VIBRATIONS If a bo oscillates without the influence of any external force then the oscillations are calle free oscillations or un ampe oscillations. In free oscillations the bo oscillates with its natural frequency an the amplitue remains constant (Fig.) Figure : Free Vibrations In practice it is not possible to eliminate friction completely.actually the amplitue of the vibrating bo ecreases to zero as a result of friction. Hence, practical examples for free oscillation/vibrations are those in which the friction in the system is negligibly small.

3 Examples of Simple harmonic oscillator a) Spring an Mass system The spring an mass system is an example for linear simple harmonic oscillator. Consier a block of mass m suspene from a rigi support through a mass less spring. The restoring force prouce in the spring obeys Hooke s Law. Accoring to Hooke s Law The restoring force prouce in a system is proportional to the isplacement. When the mass is isplace through a istance y an then release, it unergoes SHM. Then the restoring force (F) prouce is F = ky () The constant k is calle force constant or spring constant. It is a measure of the stiffness of the spring. The negative sign in equation () inicates that the restoring force is in the irection of its equilibrium position. Figure : A one-imensional simple harmonic oscillator. In equilibrium conition the linear restoring force (F) in magnitue is equal to the weight mg of the hanging mass, shown in figure. i.e, F = mg mg = ky mg anspring constant k = () y Now, if the mass is isplace own through a istance y from its equilibrium position an release then it executes oscillatory motion. y Velocity of the bo = v = an Acceleration of the bo = a = From Newton s secon law of motion for the block we can write

4 F = ma = m (3) from equation () substitue F= -ky in equation(3),we get -ky = m y k or + y = 0 m k m Substitue =ω, i.e., y y= (4) Whereωis the angular natural frequency. Equation (4) is the general ifferential equation for the free oscillator. The mathematical solution of the equation (4), y(t) represents the position as a function of time t. Let y(t) = Asin(ωt + ) be the solution. y(t) = Asin(ωt + ) (5) Where A is the amplitue an (ωt + ) is calle the phase, is the initial phase (i.e., phase at t=0). Perio of the oscillator (T) We have F = ma an also F = -ky. At equilibrium, magnitue of restoring force is equal to weight of hanging mass. i.e, ma = ky ky Acceleration = a = m The perio of the oscillator is Displacement y y m T = π =π π π Acceleration a ky k m k m We know that ω = an, rewrite perioexpression m ω k π ω k Time Perio T=, frequency n = ω T m Although the position an velocity of the oscillator are continuously changing, the total energy E remains constant an is given by E mv ky (6) The two terms in (6) are the kinetic an potential energies, respectively. VELOCITY AND ACCELERATION OF SHM We can fin velocity from the expression for isplacement, y =Asin(ωt+ ) Velocity in SHM The rate of change of isplacement is the velocity of the vibrating particle. By ifferentiating the above expression with time, we get

5 v = = Aωcos ωt+ v = Aω -sin (ωt+ ) y As sin(ωt+ )= A y v = Aω - A v=ω A -y This is the expression for velocity of the particle at any isplacement y. The maximum velocity is obtaine by substituting y =0, v max = A Since y = 0 correspons to its mean position, the particle has maximum velocity when it is at its mean position. At the maximum isplacement, i.e., at the extreme position y =A of the particle, the velocity is zero. Acceleration in SHM The rate of change of velocity is the acceleration of the vibrating particle. Differentiating velocity expression with respect to time t, we get acceleration a= = -Aω sin ωt+ y a= = -ω y The above equation gives acceleration of the oscillating particle at any isplacement. This equation is the stanar equation of SHM. Since y = 0 correspons to its mean position, the particle has minimum acceleration. At the maximum isplacement, i.e., at the extreme position y=a, maximum acceleration is A Graphical representation of isplacement, velocity an acceleration is as follows Let the isplacement of a particle executing SHM be, y = Asin ωt+ velocity v= = Aωcos ωt+ an Ac cleration a= =-Aω sinωt+ y Table : Displacement, Velocity an Acceleration for various values of time t when φ=0 Displacement Velocity Acceleration Time Ωt Asin ωt A cos ωt A sin ωt t = Aω 0 T t = 4 +A 0 A T t = 0 - Aω 0 t = 3T 4 3 -A 0 A

6 t = T 0 + Aω 0 The variation of isplacement, velocity an acceleration with time t is shown graphically: b) Spring Constant in Series an Parallel combination SPRINGS IN SERIES: Consier two springs A an B of force constants K an K be connecte in series an a mass m attache to the lower en of the bottom most spring as shown in the figure 3. When mass m is pulle own a little an release. Let y be the isplacement of mass m at any instant of time. Let y an y be the extensions of two springs of A an B respectively. Then total extension in the spring is y = y + y In series the restoring force in each spring is same, so, F=-K y =-K y F isplacement y =- an y =- K K Total extension of the spring is y = y Where K eff K +K,K eff F y F F F Keff K K is effective force constant of the spring KK K K K K K K +K

7 Figure 3: Springs in Series SPRINGS IN PARALLEL: Consier two springs A an B of force constants K an K be connecte in parallel an a mass m is attache at the lower en as shown in the figure 4. Let mass m be pulle a little an release. Let y be the isplacement of mass m from equilibrium position. Let F an F be the restoring forces evelope in springs A an B respectively. Then net restoring force is, F = F + F Where F = -K y, F =- K y Total restoring force F = (F + F ) WhereK eff -Ky = - (K + K ) y = K eff y iseffectiveforceconstant of spring So, time perio of the system is given by T = π m K = π m K + K eff Figure 4: Springs in Parallel c) Torsional Penulum Simple harmonic motion can also be angular. In this case, the restoring torque require for proucing SHM is irectly proportional to the angular isplacement an is irecte towars its mean position. A heavy metal isc suspene by means of a wire from a rigi support constitutes a Torsional Penulum (Fig. 5).When the wire is twiste by an angle θ, a restoring torque/couple acts on it, tening it to return to its mean position. Restoring torque or couple (τ) = C θ..() Figure 5: Torsional Penulum Where, C is the torsional constant. It is equal to the moment of the couple require to prouce unit angular isplacement. Its unit is Nmra. The negative sign

8 inicates that the torque is acting in the opposite irection to the angular isplacement. The angular acceleration prouce by the restoring couple in the wire, α = an I be the moment of inertia of the metal isc about the axis of the wire. Accoring to Newton s secon law, the equation of torque or couple acting on it is given by τ = I α = I () Therefore, on equating () an () we have, I = -Cθ We get angular acceleration, C = θ, I C θ Substituting =ω, we have =-ω θ I The above relation shows that the angular acceleration is proportional to angular isplacement an is always irecte towars its mean position. Therefore, the motion of the isc is always simple harmonic motion (SHM). 0 (3) This is the ifferential equation of angular simple harmonic motion of the Torsional Penulum. Therefore, the time perio an frequency of oscillator is given by relation, Perio π I C T= = π Where ω= ω C I an Frequency n C T I Note: For a given wire of length l, raius r an rigiity moulus η Torsionalconstant of the wire C r l 4 DAMPED VIBRATIONS: Free vibrations are oscillations in which the friction/resistance consiere is zero or negligible. Therefore, the bo will keep on vibrating inefinitely with respect to time. In real sense if a bo set into vibrations, its amplitue will be continuously ecreasing ue to friction/resistance an so the vibrations will ie after some time. Such vibrations are calle ampe vibrations (Fig. 6).

9 Figure 6: Dampe vibrations DIFFERENTIAL EQUATION OF DAMPED VIBRATIONS AND SOLUTION A block of mass m connecte to the free en of a spring is partially immerse in a liqui (Fig. 7) an is subjecte to small vibrations. The amping force encountere is more when the block moves in the liqui an hence the amplitue of vibration ecreases with time an ultimately stops vibrating; this is ue to energy loss from viscous forces. Let y to be the isplacement of the bo from the equilibrium state at any instant of time t, then / is the instantaneous velocity. Figure 7: Vertical spring an mass system The two forces acting on the bo at this instant are: i. A restoring force which is proportional to isplacement an acts in the opposite irection, it may be written as F restoring = ky ii. A frictional or amping force is irectly proportional to the velocity an is oppositely irecte to the motion, it may be written as F r amping The net force on an oscillator subjecte to a linear amping force that is linear in velocity is simply the sum is thus given by

10 F= F restoring + F amping y r k = -ky -r But by Newton'slaw of motion, F = m wheremis themassof thebo an is the accleration of the bo. y Then, m =-ky-r or + + y= () m m k m Let r m =b an, then the above equation takes the form, y y=0 b () This is the ifferential equation of secon orer. In orer to solve this equation, we assume its solution as αt y = Ae (3) Where A an α are arbitrary constants. Differentiating equation (3) with respect to time t, we have y = Aαe an = Aα e α t α t By substituting these values in equation (), we have t t t A e ba e A e =0 Ae t or b 0 For the above equation to be satisfie, either y=0, or b 0 Since y=0, correspons to a trivial solution, one has to consier the solution b 0 The stanar solution of the above quaratic equation has two roots, it is given by, b 4b 4 b b Therefore, the general solution of equation () is given by ( b b ) t ( b b ) t y C e D e (4) Where C an D are constants, the actual solution epens upon whether b b., b an

11 Case : Heavy amping ( b In this case, ) b is real an less than b, therefore in equation (4) both the exponents are negative. It means that the isplacement y of the particle ecreases continuously with time. That is, the particle when once isplace returns to its equilibrium position quite slowly without performing any oscillations (Fig.8). Such a motion is calle overampe or aperioic motion. This type of motion is shown by a penulum moving in thick oil or by a ea beat moving coil galvanometer. Figure 8: Heavy amping or overampe motion Case : Critical amping ( b ) By substituting b in equation (4) the solution oes not satisfy equation ().Hence we consier the case when then can be written as b is not zero but is a very small quantity β. The equation (4) y C e De ( b ) t ( b ) t bt t t y e ( C e De ) sinceβissmall, wecan approximate, t t e t an e t,on the basis of exponentialseries expansion bt y e C ( t) D( t) bt y e ( C D) t ( C D) which is the prouct of terms bt As t is present in e an also in the term t ( C D), both of them contribute to the variation of y with respect to time. But by the virtue of having t in the exponent, the term e bt preominantly contributes to the equation. Only for small values of t, the term bt t ( C D) contributes in a magnitue comparable to that of e. Therefore, though y ecreases throughout with increase of t, the ecrement is slow in the beginning, an then ecreases rapily to approach the value zero. i.e., the bo attains equilibrium position (Fig.9). Such amping motion of a bo is calle critical amping. This type of motion is exhibite by many pointer instruments such as voltmeter, ammeter...etc in which the pointer moves to the correct position an comes to rest without any oscillation.

12 Figure 9: Critical amping motion. Case 3: Low amping ( b ) This is the actual case of ampe harmonic oscillator. In this case us write b-ω =i ω -b =iβ where β = ω -b an i= -. Then,equation (4) becomes b is imaginary. Let y = Ce (-b + iβ )t (-b - iβ )t -bt iβt -iβt y = e (Ce +De ) -bt y = Rewriting e (C(cosβ C+D=Asin t+isinβ t)+d(cosβ an i(c-d)=acos t-isinβt)), Where A an are constants. -bt y = e (C+D)cosβt+i(C-D) sinβt) -bt y = e Asincosβt+Acossinβt +De -bt y= Ae sin(β t+ ) bt This equation y Ae sin( t ) represents the ampe harmonic oscillations. The bt oscillations are not simple harmonic because the amplitue ( Ae ) is not constant an ecreases with time (t). However, the ecay of amplitue epens upon the amping factor b. This motion is known as uner ampe motion (Fig.0). The motion of penulum in air an the motion of ballistic coil galvanometer are few of the examples of this case. Figure 0: Uner ampe motion The time perio of ampe harmonic oscillator is given by π π π T= = β ω -b k r - m 4m The frequency of ampe harmonic oscillator is given by k r n= T m 4m FORCED VIBRATIONS:

13 In the case of ampe vibrations, the amplitue of vibrations ecrease with the time exponentially ue to issipation of energy an the bo eventually comes to a rest. When a bo experiences vibrations ue to the influence of an external riving force the bo can continue its vibration without coming to a rest. Such vibrations are calle force vibrations. For example: When a tuning fork is struck on a rubber pa an its stem is place on a table, the table is set in vibrations with the frequency of the fork. These oscillations of the table are the force oscillations/vibrations. So force vibrations can also be efine as the vibrations in which the bo vibrates with frequency other than natural frequency of the bo, an they are ue to applie external perioic force. DIFFERENTIAL EQUATION OF FORCED VIBRATIONS AN D SOLUTION Suppose a particle of mass m is connecte to a spring. When it is isplace an release it starts oscillating about a mean position. The particle is riven by external perioic force ( Fosin ). The oscillations experiences ifferent kins of forces viz, ) A restoring force proportional to the isplacement but oppositely irecte, is given by F restoring = -ky, where k is known as force constant. ) A frictional or amping force proportional to velocity but oppositely irecte, is given by F amping = - r, where r is the frictional force/unit velocity. 3) The applie external perioic force is represente by ( Fosin ), where F o is the maximum value of the force an ω is the angular frequency of the riving force. The total force acting on the particle is given by, F = F + F +F net external restore amp = Fo sinω t - r - ky y r k Fosinω t + + y= m m () m By Newton's secon law of motion F= m. y Hence,m =F osinω t - r - ky The ifferential equation of force vibrations is given by r k F m m m o Substitute = b, = ω an = f, then equation () becomes y +b +ω y=f sinω- Where b is the amping constant, o () F is the amplitue of the external riving force an ω is the angular frequencyof theexternalforce. The solution of the above equation () is given by

14 y= Asin(ω t ) (3) Where, A is the amplitue of the force vibrations. By ifferentiating equation (3) twice with respect to time t, we get y Acos( ) an sin( ) A By substituting the values of y, / an y/ inequation (), we get Asin( t ) b A cos ( t ) Asin ( t ) f sin ( t ) or A( )sin( t ) b Acos ( t ) f sin ( t )cos f cos ( t )sin (4) If equation (4) hols goo for all values of t, then the coefficients of sin (ω t- φ) an cos(ω t- φ) must be equal on both sies, then A( ) f cos (5) an b A f sin (6) By squaring an aing equations (5) an (6), we have ( ) 4 A b A f A f b ( ) 4 (7) By iviing equation (6) by (5), we get tan, Phase tan (8) b b Substituting the value of A in equation (3) we get

15 y f sin( ) t (9) ( ) 4b This is the solution of the ifferential equation of the Force Harmonic Oscillator. From equations (7) an (8), it is clear that the amplitue an phase of the force oscillations epen upon ( ), i.e., these epen upon the riving frequency (ω ) an the natural frequency (ω) of the oscillator. We shall stu the behavior of amplitue an phase in three ifferent stages of frequencies i.e, low frequency, resonant frequency an high frequency. Case I: When riving frequency is low i.e., ( ). In this case, amplitue of the Vibrations is given by f A 4b As, an 4b 0 0 F0 f m F0 F0 A f an ω ω k k m m bω an tan tan (0) 0 - ω ω This shows that the amplitue of the vibration is inepenent of frequency of the riving force an is epenent on the magnitue of the riving force an the force-constant (k). In such a case the force an isplacement are always in phase. Case II: When i.e., frequency of the riving force is equal to the natural frequency of the bo. This frequency is calle resonant frequency. In this case, amplitue of vibrations is given by f A ( ) 4b f Fo m Fo A b ( r m) r b b an tan tan tan ( ) 0 Uner this situation, the amplitue of the vibrations becomes maximum an is inversely proportional to the amping coefficient. For small amping, the amplitue is large an for large amping, the amplitue is small. The isplacement lags behin the force by a phase π/. k m

16 Case III: When ( ) i.e., the frequency of force is greater than the natural frequency of the bo. In this case, amplitue of the vibrations is given by f A 4b 4 since, f ω is very large, 4b F0 m F0 m A an tan - bω ω ω tan 4 b tan ( 0) This shows that amplitue epens on the mass an continuously ecreases as the riving frequency ω is increase an phase ifference towars π. RESONANCE: If we bring a vibrating tuning fork near another stationary tuning fork of the same natural frequency as that of vibrating tuning fork, we fin that stationary tuning fork also starts vibrating. This phenomenon is known as Resonance. Resonance is a phenomenon in which a bo vibrates with its natural frequency with maximum amplitue uner the influence of an external vibration with the same frequency. Theory of resonant vibrations: (a) Conition of amplitue resonance. In case of force vibrations, the expression for amplitue A an phase is given by, f A b an tan ( ) 4b The amplitue expression shows the variation with the frequency of the riving force ω. For a particular value of ω, the amplitue becomes maximum. The phenomenon of amplitue becoming a maximum is known as amplitue resonance. The amplitue is maximum when ( ) 4b is minimum. If the amping is small i.e., b is small, the conition of f maximum amplitue reuce to A max. b (b) Sharpness of the resonance. We have seen that the amplitue of the force oscillations is maximum when the frequency of the applie force is at resonant frequency. If the frequency changes from this value, the amplitue falls. When the fall in amplitue for a small change from the resonance conition is very large, the resonance is sai to be sharp an if the fall in amplitue is small, the resonance is terme as flat. Thus the term sharpness of resonance can be efine as the rate of fall in amplitue, with respect to the change in forcing frequency on either sie of the resonant frequency.

17 Figure : Effect of amping on sharpness of resonance Figure, shows the variation of amplitue with forcing frequency for ifferent amounts of amping. Curve () shows the variation of amplitue when there is no amping i.e., b=0. In this case the amplitue is infinite at. This case is never realize in practice ue to friction/issipation forces, as a slight amping factor is always present. Curves () an (3) show the variation of amplitue with respect to low an high amping. It can be seen that the resonant peak moves towars the left as the amping factor is increase. It is also observe that the value of amplitue, which is ifferent for ifferent values of b (amping), iminishes as the value of b increases. This inicates that the smaller is the amping, sharper is the resonance or large is the amping, flatter is the resonance. EXAMPLE FOR ELECTRICAL RESONANCE: LC AND LCR CIRCUIT LC CIRCUIT: The electric version of the Harmonic oscillator is the LC circuit mae from an inuctor an a capacitor, An LC circuit is also calle a tank circuit, tune circuit or resonant circuit, is an electric circuit built with a capacitor C an an inuctor L connecte together as shown in figure. These circuits are use for proucing signals at a particular frequency or accepting a signal from a more composite signal at a particular frequency. LC circuits are basic electronics components in various electronic evices, especially in raio equipment use in circuits like tuners, filters, frequency mixers an oscillators. The main function of an LC circuit is generally to oscillate with minimum amping.

18 Figure : LC circuit The circuit contains inuctor L an charge capacitor C as shown in the figure. Suppose a charge conenser of capacity C an initial charge q 0 is ischarge through inuctance L of negligible resistance. At any instant when the charge of conenser is q, the potential q i ifference across it will be an the inuce emf ue to inuctance is L, then the equation C of emf is given by i q L + = () C But i= rate of flow of charge = q q L + =0 C q q or L + =0 C q q or + =0 LC LC This is the general ifferential Putting equation ω = of oscillations, we have of LC circuit. Let its solution be q +ω = () q = q 0 sin(ωt+φ) q q Where, q 0 an φ are constants. This is the equation of ischarge of the conenser through the inuctance. This equation represents simple harmonic oscillations. The frequency of the oscillations is given by ω n= = π π LC Time perio of the oscillationsis given by T= π LC n LCR CIRCUIT: An L-C-R circuit fe by an alternating emf is a classic example for a force harmonic oscillator. Consier an electric circuit containing an inuctance L, capacitance C an resistance R in series as shown in the figure 3. An alternating emf has been applie to a circuit is represente by E sin t. o

19 Figure 3: Series LCR circuit. Let q be the charge on the capacitor at any instant an I be the current in the circuit at any instant. The potential ifference across the capacitor is q C, the back emf ue to self inuctance in the inuctor is I L an, the potential rop across the resistor is IR. The sum of voltages across the three LCR elements illustrate must be equal the voltage supplie by the source element. Hence the voltage equation at any instant is given by, V +V +V = V (t) L R C S I q L +IR+ = Eosinω C Differentiating, we get I I q L +R + =ω Eocos C t q I I I but = I, L +R + = ω Eocosω C o + + I= cosω I R I ωe L LC L I R I ωeo π + + I= sinω+ () L LC L This is the ifferential equation of the force oscillations in the electrical circuit. It is similar to equation of motion of a mechanical oscillator riven by an external force. y F o +b +ω y= sinω t m () The explicit an precise connection with the mechanical oscillation equation is given below:

20 Force oscillations (refer previous section) Starting with equation from force vibrations, we have LCR Circuit application The equivalent equation for LCR circuit is given by y F o + b + ω y = sinω t m Amplitue of mechanical vibrations is given by A F m o b ( ) 4 o + + I= sin ω + I R I ω E π L LC L Amplitue of current I o is given by I0 Eo L ( ) 4b R Substitute for an 4 b, we get LC L Io LC Eo L R L In the enominator,multiply an ive the term LC an by L,we get Io Io Eo L L R LC L L Eo L R L C L L by

21 Eo Io L L R L C substitute = X C(Capacitive reactance) Cω an Lω =X L(Inuctive reactance) Io Eo E o C L X X R X X R L C The solution of the equation () for the current at any instant in the circuit is of the form Eo I sin( ) sin( ) (3) Io R L C Electrical Impeance: The ratio of amplitues of alternating emf an current in ac circuit is Eo Eo calle the electrical impeance of the circuit. ie.., Z = Io I Z Eo Eo The amplitue of the current is I o = = Z R + ωl- ωc o Impeance of the circuit Z = R + ωl- (4). ωc The quantity ωl- is the net reactance of the circuit which is the ifference between ωc the inuctive reactance (X L)=ω Lan the capacitive reactance (X C)=.Equation (3) ω C. shows that the current I lags in phase with the applie emf L given by C X L X C tan tan R R o E sin t by an angle an is The following three cases arise:. When X L >X C, is positive, that is, the current lags behin the emf. Circuit is inuctive.

22 . When X L <X C, is negative, that is, the current leas behin the emf. Circuit is capacitive. 3. When X L =X C, is zero, that is, the current is in phase with the emf. Circuit is resistive. Electrical Resonance: Accoring to equation (4), the current have its maximum amplitue when X L XC 0 or L 0 or L or C C LC Where ω is the angular frequency of the applie emf, while ω = LC is the (angular) natural frequency of the circuit. Hence, the maximum amplitue of the current oscillations occurs when the frequency of the applie emf is exactly equal to the natural (un-ampe) frequency of the electrical circuit. This is the conition of electrical resonance. Sharpness of resonance an Banwih: When an alternating emf is applie to an LCR circuit, electrical oscillations occur in the circuit with the frequency ω is equal to the applie emf. The amplitue of these oscillations (current amplitue) in the circuit is given by I o R E o L C Eo =, Where Z is theimpeanceof thecircuit. Z At resonance, when the frequency ω of the applie emf is equal to the natural frequency of the circuit, the current amplitue I o is maximum an is equal to E o /R. Thus at LC resonance the impeance Z of the circuit is R. At other values of ω, the current amplitue I o is smaller an the impeance Z is larger than R. E o / E o R ω ω ω Figure 4: Graphical plot of current vs frequency. The variation of the current amplitue I o with respect to applie emf frequency ω is shown in the figure 4. I o attains maximum value (E o /R) when ω has resonant value ω an ecreases as

23 ω changes from ω. The rapiity with which the current falls from its resonant value (E o /R) with change in applie frequency is known as the sharpness of resonance. It is measure by the ratio of the resonant frequency ω to the ifference of two frequencies ω an ω taken at of the resonant (ω) value. i.e., Sharpness of resonance (Q) =, ω an ω are known as the half power frequencies. The ifference of half power frequencies, ω - ω is known as ban-wih. The smaller is the banwih, the sharper is the resonance. SI Sample Questions (one mark each) CO No. What is simple harmonic motion?. Write the general equation representing SHM. 3. List any two characteristics of SHM. 4. A particle executes a S.H.M. of perio 0 secons an amplitue of meter. 3 Calculate its maximum velocity. 5. Hyrogen atom has a mass of.68x0-7 kg, when attache to a certain 3 massive molecule it oscillates as a classical oscillator with a frequency0 4 cycles per secon an with amplitue of 0-0 m. Calculate the acceleration of the oscillator. 6. A bo executes S.H.M such that its velocity at the mean position is 4cm/s an its amplitue is cm. Calculate its angular velocity What is free vibration? 8. What is ampe vibration? 9. Give any two examples for ampe vibration. 0. What is force vibration?. Given two vibrating boies what is the conition for obtaining resonance?. Explain why a loae bus is more comfortable than an empty bus? 3. What is a simple harmonic oscillator? 4. What is torsional oscillation? 5. If two springs are connecte in parallel, what is its equivalent spring constant? 6. If two springs are connecte in series, what is its equivalent spring constant? 7. Displacement of a particle of mass 0g executes SHM given by an its isplacement at t=0 is 3cm where the amplitue 3 isn5cm. Calculate the initial phase of the particle. 8. Name the two forces acting on a system executing ampe vibration. 9. How critical amping is beneficiary in automobiles? 0. What is restoring force? Descriptive Questions.. Every SHM is perioic motion but every perioic motion nee not be SHM.

24 Why? Support your answer with an example... Distinguish between linear an angular harmonic oscillator? 3.. Setup the ifferential equation for SHM. 4.. Define the terms (i) time perio (ii) frequency (iii) phase an (iii) angular frequency of oscillations 5.. What is the phase ifference between (i) velocity an acceleration (ii) acceleration an isplacement of a particle executing SHM? 6.. Show graphically the variation of isplacement, velocity an acceleration of a particle executing SHM. 7.. Explain the oscillations of a mass attache to a horizontal spring. Hence euce an expression for its time perio. 8.. Derive an expression for the time perio of a bo when it executes angular SHM 9.. What is amping? On what factors the amping epens? 0.. What are ampe vibrations? Establish the ifferential equation of motion for a ampe harmonic oscillator an obtain an expression for isplacement. Discuss the case of heavy amping, critical amping an low amping... What o you mean by force harmonic vibrations? Discuss the vibrations of a system executing simple harmonic motion when subjecte to an external force... What is riven harmonic oscillator? How oes it iffer from simple an ampe harmonic oscillator? 3.. What is resonance? Explain the sharpness of resonance. 4.. Illustrate an example to show that resonance is isastrous sometimes. SI Problems No. A particle executes SHM of perio 3.4 secon an amplitue 5cm. Calculate its maximum velocity an maximum acceleration. Solution: The maximum velocity at y =0 in v = ω A -y, v max = A 0.raian T 3.4 vmax cm / sec 0.0 m / s At the maximum isplacement, i.e., at the extreme position x=a, maximum acceleration is - A a m / s. A circular plate of mass 4kg an iameter 0.0 metre is suspene by a wire which passes through its centre. Fin the perio of angular oscillations for small isplacement if the torque require per unit twist of the wire is 4 x 0-3 N- m/raian. Solution: MR 4(0.05) Moment of Inertia I 0.005kgm CO 3 3

25 I Time perio is given by T 3.4 =7.0s. 3 C 4x0 3. A mass of 6kg stretches a spring 0.3m from its equilibrium position. The mass is remove an another bo of mass kg is hange from the same spring. What woul be the perio of motion if the spring is now stretche an release? Solution: F mg m F=ky, k= = = =96N/m an T=π = 3.4 =0.45s y y 0.3 k A vibrating system of natural frequency 500 cycles/sec, is force to vibrate with a perioic force/unit mass of amplitue 00 x 0-5 N/kg in the presence of a amping/ unit mass of 0.0 x 0-3 ra/s. Calculate the maximum amplitue of vibration of the system. Given :Natural frequency = 500 cycles/sec, Amplitue of the force / unit mass, Fo/m=00 x 0-5 N/kg, Damping coefficient, r/m = 0.0 x 0-3 ra/s Solution: Maximum amplitue of vibration f Fo m A b ( r m) A m Themaximumamplitueof vibration of the system is 0.038meter 5. A circuit has an inuctance of henry an resistance 00Ω. An A. C. supply of 50 cycles is applie to it. Calculate the reactance an impeance offere by the circuit. Solution: The inuctive reactance is X L =ω L =πnl Here ω =πn=π x 50= 00π ra/sec an L= Henry. X L =ω L=πnL =π x 50x =00Ω. Z R X The impeance is L 6. A series LCR circuit has L=mH, C=0.µF an R=0Ω.Calculate the resonant frequency of the circuit. Solution: The resonant angular frequency of the circuit is given by LC Here L=mH an C=0.µF 5 0 ra / s x0 3 3 Reference Books Oscillations an Waves, Satya Prakash, Pragati Prakshan, thir eition, 005. Engineering Physics

26 Hitenra K Malik an A K Singh, Tata McGraw Hill Eucation Private Limite, 009, ISBN: A Text book of Engineering Physics Dr. M N Avahanulu, Dr. P. G. Kshirsagar, S. Chan & Company Private limite. Revise eition Engineering Physics R K Gaur an S L Gupta, Dhanpat Rai Publications, Revise eition 0.