OSCILLATIONS

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Denis Booker
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1 OSCIAIONS Important Points:. Simple Harmonic Motion: a) he acceleration is directly proportional to the displacement of the body from the fixed point and it is always directed towards the fixed point in the path of the body. aα x (Or) a = kx Acceleration (a) = F = k x. m m a = ω x Where ω = k m (Or) d x + ω x = 0 dt. Every harmonic is periodic. But every periodic is not harmonic. 3. Amplitude: he maximum displacement of a vibratin particle from its mean position is called amplitude (A). It is a vector quantity. 4. ime Period: he time taken for one oscillation is called time period π = Where anular velocity ω 5. Frequency: he number of oscillations completed by a particle in one second is called frequency. Frequency f = cycles/sec.

2 6. Phase: Phase is that which denotes the state of vibration 7. Displacement: Displacement of a particle executin simple harmonic motion is iven by y = Asin( ωt ± φ) Where φ = Initial phase ω = Anular frequency A = Amplitude 8. Velocity: Velocity of the particle executin simple harmonic motion is iven by V = Aω cosωt (Or) V = ω A y Vmax = Aω 9. Acceleration: Acceleration of a particle executin simple harmonic motion is iven by a = Aω sin( ωt ± φ) a = ω y amax = Aω For simple harmonic motion is iven by = π y a Where y = displacement and a = acceleration

3 0. Enery of the particle in SHM: a) b) ω ω ω PE = m y = m A Sin t KE = mv = mω ( A y ) = ω cos ω m A t c) otal enery = mω A = π n ma d) At the mean position KE is maximum and PE is minimum. e) At the extreme position KE is minimum and PE is maximum.. Simple Pendulum: ime period of simple pendulum executin S.H.M. is = π l Where l = lenth of the simple pendulum = acceleration due to ravity. Seconds Pendulum: A pendulum with time period seconds is called second pendulum. Its lenth on earth is 00 cm. 3. oaded Sprin: i) If a mass M is suspended vertically from a sprin and if the sprin elonates by x then sprin constant is K F = Or x M K = x

4 ii) ime period of the loaded sprin is M = π or = π K x iii) If two sprins of force constants k and k are joined in series, the effective force kk constant k = k + k iv) If two sprins of force constants k and k are joined in parallel, the effective force constant k = k + k. v) When a sprin of force constant k is cut into n equal parts, the sprin constant of each part is nk vi) If a uniform sprin of sprin constant K is cut into two pieces of lenths in the ratio l : l, then the force constants of the two sprins will be k k( l + l ) k( l + l ) = and k = l l

5 Very Short Answer Questions. Give two examples of periodic motion which are not oscillatory? A. ) Motion of the seconds hand in a watch ) Revolution earth around the sun.. he displacement in S.H.M. is iven by y = a sin (0t +4). What is the displacement when it is increased by π / ω? A. In time equal to time period, the particle comes to same position. So displacement of the particle remains same even after time is increased by π ω. 3. A irl is swinin seated in a swin. What is the effect on the frequency of oscillation if the stands? A: When the irl stands up, her centre of mass moves up and lenth of swin decreases. Hence frequency of oscillation increases, since v l 4. he bob of a simple pendulum is a hollow sphere filled with water. How will the period of oscillation chane, if the water beins to drain out of the hollow sphere? A. If water beins to drain out of the hollow sphere, its centre of mass beins to shift below the centre of the sphere. Hence the lenth of the pendulum increases and time period also increases. When the entire water is drained out of the sphere, the centre of mass shifts to centre of sphere and the time period attains its initial value. 5. he bob of a simple pendulum is made of wood. What will be the effect on the time period if the wooden bob is replaced by an identical bob of aluminum? A: Since the sizes of metal bob and wooden bob are equal, the position of centre of mass does not chane. herefore the lenth of the pendulum remains same so, time period remains constant.

6 6. Will a pendulum clock ain or lose time when taken to the top of a mountain? A: On the top of mountain, value of is less. Period of oscillation oscillation increases and clock loses time. l, there period of 7. A pendulum clock ives correct time at the equator. Will it ain or lose time if it is taken to the poles? If so, why? A. ime period = π l he value of is maximum at the poles and hence time period of the pendulum clock decreases and hence ains time. 8. What fraction of the total enery is K.E when the displacement is one half of a amplitude of particle executin S.H.M. A: Kinetic enery otal enery A mω A 4 3 A = = y mω A 4 = Fraction of total enery in the kinetic enery 9. What happens to the enery of a simple harmonic oscillator if its amplitude is doubled? 3 = 4 A. Enery of simple harmonic oscillator E = mω A E α A If the amplitude is doubled, enery becomes four times. 0. Can the pendulum clocks be used in an artificial satellite? A. No. In an artificial satellite, the bodies are weihtless. Hence the pendulum clocks cannot be used in an artificial satellite.

7 Short Answer Questions. Define simple harmonic motion. Give two examples? A. Definition: A body is said to be in simple harmonic motion, if the acceleration (a) is directly proportional to its displacement (y) from the fixed point and is always directed towards the fixed point in its path. a y Example: ) Oscillations of a simple pendulum. ) Vibrations of a tunin fork. Present raphically the variations of displacement, velocity and acceleration with time for a particle in S.H.M A: Consider a particle is in SHM with amplitude A. he displacement of the particle is iven by y = Asin( ωt + φ) () Velocity of the particle at any instant can be obtained differentiatin displacement equation with respect to time dv V = = aω cos( ωt + φ) () dt o et acceleration of the particle aain differentiate the velocity equation with respect to time dv a = = Aω sin( ωt + φ) (3) dt GRAPHICA REPRESENAION: i) Displacement time raph

8 ii) Velocity time raph iii) Acceleration time raph 3. What is phase? Discuss the phase relations between displacement, velocity and acceleration in simple harmonic motion? A: Phase: he physical quantity which ives the information about the position and direction of motion of particle in SHM at any instant from mean position is called phase. In eneral, the displacement of a particle in SHM can be written as Y = Asin( ωt ± φ), where ( ωt ± φ) is called phase, expressed in radian. When the particle starts from mean position, dy π Y = Asinωt velocityv = = Aω cosωt = Aω sin ωt +. he phase difference between dt π displacement and velocity is. dv Acceleration a = = Aω sinωt = Aω sin( ωt + π ). he phase difference between displacement dt π and acceleration is π and phase difference between velocity and acceleration is. 4. Obtain an equation for frequency of oscillation of a sprin of force constant K to which a mass m is attached. A. Consider a mass less sprin of force constant K is loaded with mass m. et it be pulled a little down and left to make vertical oscillations. et y be the displacement of the mass.

Restorin force F = K y k From Newton s second law, ma = - ky Or a =. y m Or aα y Hence the oscillations of a loaded sprin are simple harmonic. Acceleration of the particle in S.H.M at any time is iven by a = ω y, k ω =.

9 Restorin force F = K y k From Newton s second law, ma = - ky Or a =. y m Or aα y Hence the oscillations of a loaded sprin are simple harmonic. Acceleration of the particle in S.H.M at any time is iven by a = ω y, k ω =. m But, ω n = π n = π k m 5. Derive the expression for the kinetic enery and potential enery of simple harmonic oscillator? A. Kinetic Enery: he velocity of the particle executin simple harmonic motion at any position V = ω A y he kinetic enery of the particle K. E m A y = mv = ω ( ) Also, the velocity of the particle executin simple harmonic motion at any position is iven by, v = Aω cosωt K. E = mv = ma ω Cos ω t

10 Potential Enery: Consider a particle of mass m executin SHM. If F is the force at the displacement y Restorin force F = ma = mω y Work is done in displacin the body throuh a small displacement dy is iven by W = F dy = mω ydy Now the total work done in displacin the body from 0 to y is iven by y W = dw = mω ydy = mω y 0 his work done is stored as potential enery P. E = mω y = mω A sin ωt 6. How does the enery of a simple pendulum vary as it moves from one extreme position to the other durin its oscillation? A. Consider a simple pendulum of lenth l and let m be the mass of the bob. et A and B are extreme positions and O be the mean position. et A be its amplitude and y be the displacement of the bob. At the Extreme Positions A or B (y = A): PE = mω y m A = = 0 = ω And KE mω ( A y ) otal enery = PE + KE = mω A + 0 = mω A --- ()

11 At the mean position O ( y = 0 ): P. E m y = ω = 0 KE = mω ( A y ) = mω A otal enery = PE + KE = 0 + mω A = mω A --- () At any position P (OP = y): P. E = mω y KE = mω A y ( ) otal enery = PE + KE ( ) = mω y + mω A y = mω A --- (3) Hence from equations (), () & (3) it is clear that the total enery of a simple pendulum is constant at any point on its path. 7. Derive the expression for displacement, velocity and acceleration of a particle executes SHM A: Consider a particle movin alon the circumference of a circle of radius A and centre O with a uniform speed V or with an anular velocity ' ω '. Suppose at time t = 0, the particle is at point A such that XOA = φ. At any time t, suppose the particle reaches the point P such that AOP = ωt. Draw PN YY

12 Clearly, displacement of projection N from centre O at any instant t is y = ON. In riht anle trianle ONP, PON = ωt + φ ON y = cos( ωt + φ) or = cos( ω + φ) OP A his equation ives displacement of a particle in S.H.M at any instant. o find velocity, differentiate the displacement with time. dy V Aω ωt φ ω A x dt = = sin( + ) = [ y = Acos( t + )] ω φ he neative sin shows that the velocity of N is directed towards left. o find acceleration, differentiate the velocity with time dv dt ω ω φ ω a = = A sin( t + ) = x

13 on Answer Questions. Define Simple Harmonic Motion? Show that the projection of uniform circular motion on any diameter is simple harmonic. A. Definition: A body is said to be in S.H.M, if the acceleration is directly proportional to its displacement from the fixed point and is always directed towards a fixed point in its path. Expression: Consider a particle P movin on the circumference of a circle of radius A with uniform anular velocity ω. et O be the centre of the circle. et PN be the perpendicular drawn to the diameter YY from P. As P moves on the circumference of the circle, N moves on the diameter about the O. et the position of N at any time t so that XOP = θ he displacement of N with respect to the fixed point O in the path is iven by ON = Y = Asinθ (Or) ( ) Y = Asinωt θ = ωt... () he rate of chane of displacement is called velocity.

14 dy v = = Aω cosωt dt he rate of chane of velocity is called Acceleration. dv dt ω ω a = = A Sin t Or a = ω y ( y = Asinωt) Acceleration a = ω y Hence the acceleration of N is directly proportional to the displacement in manitude but in opposite direction, always towards the fixed point O in the path. Hence the motion of N is simple harmonic.. Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period. What is second s pendulum? A. Simple Pendulum: et l be the lenth and m be the mass of the bob of a simple pendulum. et the bob be iven a small anular displacement θ and released..et OB = y = Displacement from the mean position. y θ =...() l S θ l O y B m sinθ θ m m cos θ

15 At the extreme positions B the tanential acceleration a = he torque about the point of suspension τ = l ( m sin θ ) sinθ he restorin torque τ = Iα Where I is the moment of inertia and τ is the anular acceleration ml Iα = l ( m sin θ ) orα = θ I (When is very small,sinθ θ ) Comparin this with the equation for SHM, ω = ml I Hence the bob of the pendulum executes SHM. But, π I = = π and ω ml I = ml = π l Second s Pendulum: A simple pendulum whose time period is two seconds is called a second s pendulum. 3. Derive the equation for the kinetic enery and potential enery of a simple harmonic oscillator and show that the total enery of a particle in simple harmonic motion is constant at any point on its path A: i) Expression of K.E (K): he displacement of a simple harmonic oscillator is iven by x = Asinωt Velocity of SHO is, dx V = = Aω cosωt = ω A x dt he K.E of SHO is iven by, ω ω K = mv = ma cos t

16 (Or) K = mω ( A x ) ii) Expression of P.E (U) : When the displacement of SHO (particle) from its equilibrium position is x, the restorin force actin on it is F = -Kx If we displace the particle further throuh a small distance dx, then work done aainst the restorin force is iven by dw = -Edx = +Kx dx he total work done in movin the SHO from mean position (x = 0) to displacement x is iven by x x x W = dw = Kxdx = K = Kx a 0 his work done aainst the restorin force is stored as P.E of SHO Hence, P.E of SHO = Kx = mω x (Or) P.E of SHO = ω cos ( ω + φ) U m A t he total enery of oscillator, E = K + U = mω ( A x ) + mω x If the particle is at, (left extreme position) x = -A, K = 0, U m A = ω, then E K U m A = + = ω () If the particle is at (mean position) x = 0, K m A = ω, U = 0 hen E K U m A = + + ω () If the particle is at (riht extreme position) x = A, K = 0, U m A = ω then E K U m A = + = ω (3)

17 Hence the total mechanical enery (E) remains constant at any point he variation of K.E (K), P.E (U) and.e (E) with displacement x is shown in fiure. he raphs for K and U are parabolic while that for E is a straiht line parallel to displacement axis. At x = 0, the enery is all kinetic and for x = ± A, the enery is all potential.

18 Problems. he bob of a pendulum is made of a hollow brass sphere. What happens to the time period of the pendulum, if the bob is filled with water completely? Why? Sol: = π Where = lenth of the pendulum When s hollow brass sphere is completely filled with water, centre of mass does not chane. Hence lenth of the pendulum remains same herefore, time period of hollow brass sphere filled with water completely has no chane.. wo identical sprins of force constant K are joined one at the end of the other (in series). Find the effective force constant of the combination. F F F K A. otal elonation x = x + x = + Keff = K K K eff Effective force constant of combination K = eff K 3. What are physical quantities havin maximum value at the mean position in SHM? A. Velocity and kinetic enery. 4. A particle executes SHM such that, the maximum velocity durin the oscillation is numerically equal to half the maximum acceleration. What is the time period? A: Vmax = amax ButVmax = Aω ; a A A A max = ω ω = ω ω = ime period π π = = = π sec ω

19 5. A mass of k attached to a sprin of force constant What is the time taken? 60Nm makes 00 oscillations. A. m = k, k = 60Nm m = π = = K 7 60 ime for 00 oscillations = = 55.3s 6. A simple pendulum in a stationary life has time period. What would be the effect on the time period when the lift i) Move Up with uniform velocity ii) Moves Down with uniform velocity iii) Moves Up with uniform acceleration a iv) Moves Down with uniform acceleration a v) Beins o fall freely under ravity? A: ime period of a simple pendulum in a stationary lift; = π () ime period of a simple pendulum in a lift movin up with uniform acceleration is up = π () + a ime period of a simple pendulum in a lift movin down with uniform acceleration is down = π (3) a i) Moves up with uniform velocity, acceleration of lift at = 0 hen op = π = [ ()] No chane in time period ii) Moves down with uniform velocity a = 0 hen down = π = [ ()] No chane in time period

20 iii) () () + a up = up < he lift moves up with uniform acceleration a, time period decreases iv) (3) () a down = down > he lift moves down with uniform acceleration a, time period increases v) When the lift falls freely under ravity a = = down infinity = 0 = 7. A particle executin SHM has amplitude of 4cm and its acceleration at a distance of cm from the mean position is cms. What will be its velocity when it is at a distance of cm from its mean position? A. A = 4cm, a = 3cms, x = cm, x = cm ; V =? But, a x = ω 3 = ω ω = 3 V = ω A x = 3 4 = 3 8. A simple harmonic oscillator has a time period of s.what will be the chane in the phase 0.5 s after leavin the mean position? A. = s, t = 0.5 s Phase chane φ = ωt = π t π π φ = 0.5 = rad 4

21 9. A body describes simple harmonic motion with an amplitude of 5cm and a period of 0.s. Find the acceleration and velocity of the body when the displacement is a) 5cm b) 3cm c) 0 cm A: A = 5cm, = 0.cm a) If y = 5, 4π a = ω y = 5 = 5π ms 0.04 ω = π V ω A y ω = = = b) If y = 3, 4π a = ω y = 3 = 3π ms 0.04 π V = ω A y = π V = 4 = 40π cms = 0.4π ms 0. c) If y = 0cm, a = ω y = ω (0) = 0 V ω A y π π = = = V = 0.5π ms 0. he mass and radius of a planet are double that of the earth. If the time period of a simple pendulum on the earth is, Find the time period on the planet? GM A: E = π = where E = R E E E p GM p GM E E = π But p = R = 4R = p p E p = π = π E E = p

22 . Calculate the chane in the lenth of a simple pendulum of lenth m, when its period of oscillation chanes from s to.5s? 3 Sol: = m ; = s, =.5s = s = =?? 9 9 = = = Chane in lenth, = = 6 6 Chane in lenth is decreased by 7 6 m. A freely fallin body takes seconds to reach the round on a planet, when it is dropped from a heiht of 8m. If the period of a simple pendulum is π seconds on the planet, calculate the lenth of the pendulum? h = t 8 = A. For a freely fallin body, ( ) = 4ms 4 π l = = = m 4π 4π 3. he period of a simple pendulum is found to increase by 50% when the lenth of the pendulum is increased by 0.4m. Calculate the initial lenth and the initial period of oscillation at a place where = 9.8ms 3 A: = 50% = = = 0.6m and = 9.8ms Initial time period = π = 4 π ()

23 Final time period = π = 4 π () 4π 9 4π 0.6 () () = ( l l ) = π π 0.6 = = = =.93 =.39s Initial lenth π 0.6 4π 4π = =.4 = = 0.48m 5 4. A clock reulated by a second s pendulum keeps correct time. Durin summer the lenth of the pendulum increases to.0m. How much will the clock ain or lose in one day? A: he correct time of a clock, = s.0 ime period of a clock durin summer, = π = π =.073s 9.8 Increase in time period for each oscillation = 0.073s he clock loses time for each oscillation = 0.073s No of oscillations made by pendulum keepin correct time is a day s s = = = 43, 00s oss of time by clock in a day = x 43,00 = 80s 5. he time period of a body suspended from a sprin is. What will be the new time period if the sprin is cut into two equal parts and the mass is suspended i) From one part ii) Simultaneously from both the parts? A: he time period of a body of mass m suspended from a sprin of sprin constant K is m = π () K l If the sprin of lenth l is cut into two equal parts, lenth of each part l =

24 et the sprin constant of each part is K he sprin constant elonation lenth of the sprin, K l l K = l = l, K = K m i) With one part of the sprin, time period = π () K () K K () K K = = = ii) he same mass is suspended simultaneously from both the parts, time period m = π () K (3) () K K = But K = K + K = K = K = 4K K = = 4K

SIMPLE HARMONIC MOTION PREVIOUS EAMCET QUESTIONS ENGINEERING. the mass of the particle is 2 gms, the kinetic energy of the particle when t =

SIMPLE HARMONIC MOTION PREVIOUS EAMCET QUESTIONS ENGINEERING. the mass of the particle is 2 gms, the kinetic energy of the particle when t = SIMPLE HRMONIC MOION PREVIOUS EMCE QUESIONS ENGINEERING. he displacement of a particle executin SHM is iven by :y = 5 sin 4t +. If is the time period and 3 the mass of the particle is ms, the kinetic enery