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

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1 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 of the particle when t = 4 is iven by π ns : 4 [009 E] ) 0.4 Joules ) 0.5 Joules 3) 3 Joules 4) 0.3 Joules π Sol: he iven equation is y = 5sin 4t+ 3 () comparin with the equation y = sin ( ωt+ φ ) From the equation, π π π = = = s ω 4 π π When t =, y = 5sin ω = 4rads π = 5cos =.5m 3 KE.. = mω y = 0 3 ( 4 ) ( 5 ) (.5 ) = 0.3 J. particle is executin simple harmonic motion with an amplitude and time period. he displacement of the particle after period from its initial position is : [008 E] ) ) 4 3) 8 4) Zero ns : 4 Sol: fter a time period of or interal multiples of the particle comes to mean position. Hence displacement is zero. 3. he manitude of imum acceleration is times that of imum velocity of a simple harmonic oscillator. he time period of the oscillator in seconds is (007 E) ) 4 ) 3) 4) 0.5 ns : Sol: a = ω.() v = ω..() a = πv [iven ] = ( ) ω π ω ω = π = π π s ω = π =

2 4. he time period of a simple pendulum is. When the lenth is increased by 0cm, the period is. When the lenth is decreased by 0cm, its period is. hen relation between,, is ) = + ) = 3) = + 4) = Sol: he time period of a simple pendulum is l = π () When the lenth increases by 0cm then l + 0 = π () When the lenth decreases by 0cm then π l 0 = (3) Squarin () & (3) and addin l 0 l = π l = 4π = + = (004 E) 5. When a body of mass.0 k is suspended from a certain liht sprin hanin vertically, its lenth increases by 5cm. By suspendin.0 k block to the sprin and if the block is pulled throuh 0cm and released, the imum velocity of it in m/s is (cceleration due to ravity = 0 m/s ) (003E) ) 0.5 ) 3) 4) 4 Sol: Force = m = kx where k is force constant m k = = 00Nm x From law of conservation of enery kx = mv ( ) kx 00 0 V = = = m V = ms 6. body of mass m is suspended to an ideal sprin of force constant k. he expected chane in the position of the body due to an additional force F actin vertically downwards is : (005 E)

3 ) 3F k ) F k 3) 5F k 4) 4F k ns : Sol: If F is the force actin and K is the force constant. Let the chane in position of the body due to addition force is x. So F = kx F x = k 7. n object is attached to the bottom of a liht vertical sprin and set vibratin. he imum speed of the object is 5 cm/sec and the period is 68 milli seconds. he amplitude of the motion in centimeters is (003 E) ) 3.0 ).0 3).5 4).0 Sol: Given V = 5cms = 5 0 ms = s = 68 ms π V = = ω = = 5 0 m 3 =.5 cm 8. body executes S.H.M. under the action of a force F with a time period 4/5 seconds. If the force is chaned to F it executes S.H.M. with time period 3/5 seconds. If both the forces F and F act simultaneous in the same direction on the body. Its time period in seconds is (00 E) ) /5 ) 4/5 3) 5/4 4) 5/ π 4π mr Sol: Force F = mrω = mr = s the two forces are actin simultaneously Resultant force = F = F+ F = + [ Since m & r are constants) = = + = = s 6 9 5

4 9. If the displacement (x) and velocity (v) of a particle executin S.H.M. are related throuh the expression 4v = 5 - x then its time period is (00 E) ) π ) π 3) 4 π 4) 6 π Sol: Give expression is 4V = 5 x () Dividin equation () by 4 5 x V = V = x () Comparin () with V = ω x π ω = = = 4π s 0. wo particles P and Q start from oriin and execute S.H.M. alon x-axis with same amplitude but with periods 3 seconds and 6 seconds respectively. he ratio of the velocities of P and Q when they are at mean position is (00 E) ) : ) : 3) : 3 4) 3 : ns : Sol: Velocity of a particle executin S.H.M. V = ω y π π But ω = V = y s V V 6 V = = V : V = : 3. body is executin S.H.M. at a displacement x its P.E. is E and at a displacement y its P.E is E. he P.E at displacement (x+y) is ) E = E E ) E = E + E 3) E = E+ E 4) E = E E (00 E) Sol: he potential enery stored in a body E = kx K E = kx E =. x.() K E = ky E =. y.() Potential enery (E) at displacement (x+y) = ( ) k x+ y

5 From (), () and (3) k E = ( x+ y)...(3) k k k ( x + y) = ( x) + ( y) E = E + E. body of mass k executin S.H.M., its displacement y cm at t seconds is iven by y=6sin (00t + π /4). Its imum kinetic enery is (000 E) ) 6J ) 8J 3) 4J 4) 36J ns : Sol: Comparin the iven equation with y = sin ( ωt+ φ) π We et = 6 0 m, ω = 00 rads -, φ = 4 rad ( ) ( ) ( ) KE = mω = = 8 J 3. particle executin S.H.M. has an amplitude of 6 cm. Its acceleration at a distance of cm from the mean position is 8cm/s. he imum speed of particle is (000E) ) 8cm/s ) cm/s 3) 6 cm/s 4)4cm/s ns : a = ω y 8= ω Sol: ( ) ω = 4 ω = rads V = ω = 6 = cms MEDICL 4. simple pendulum is executin SHM with a period of 6sec between two extreme positions B and C about a point O. If the lenth of the arc BC is 0cm. how lon will the pendulum take the move from position C to a position D towards O exactly midway between C and O [009 M] ) 0.5sec ) sec 3).5 sec 4) 3 sec ns : Sol: ime period = s mplitude = 5 cm π he equation of SHM is y = sin t π = sin t π π sin = sin 6 6 t t = s 5. irl swins on a cradle in sittin position. If she stands, the time period of cradle [008 M]

6 ) decreases ) increases 3) remains constant 4) first increases then it remains constant ns : Sol: s the lenth of the pendulum decreases time period also decreases 6. he displacement of a particle of mass 3 m executin simple harmonic motion is iven by y = 3sin(0.t) in SI units. he kinetic enery of the particle at a point which is at a distance equal to of its amplitude from its mean position is (007 M) ) x 0 3 J ) 5 x 0 3 J 3) 0.48 x 0 3 J 4) 0.4 x 0 3 J ns : Sol: Given equation is y=3sin(0.t) comparin with standard equation y = sin ( ωt) 3 = 3 m, ω =0., m= 3 0 k KE. = mω ( x ) = ( 0. ) ( 3 ) 3 = J π 7. he simple harmonic motion of a particle is represented by the equation x = 4cos 88t+ 4. he ns : frequency (in Hz) and the initial displacement (in m) of the particle are (006 M) ) 4, ) 6, 3) 4,3 4) 6,3 Sol: Comparin the iven equation with x= cos( ωt+ φ ) i) ω t = 88t ω = 88 π n= 88 n= 4Hz ii) Initial displacement [t=0] π 4 x = 4cos = = 4 8. body executin S.H.M. has a imum velocity of and a imum acceleration of 4. Its amplitude in metres is (005 M) ) ) ) 0.5 J 4) 0.5 ns : 4 Sol: V ω =, a = ω V a ω () ω = = = = 0.5m 4 9. he equation of motion of particle executin SHM is a+ 6π x = 0. In this equation a is the linear acceleration in m/s, of the particle at a displacement x in meters. he time period of SHM, in seconds is (004 E) ) /4 ) / 3) 4) ns :

7 Sol: Given that 6π x 0 a+ = a= 6π x -ve sin indicates that acceleration and displacement are opposite in directions displacement ime period = π acceleration = π 6 = s π 0. he time period of a particle in simple harmonic motion is 8 seconds. t t = 0 it is at the mean position. he ratio of the distances travelled by it is the first and second seconds is: (003 M) ) ) 3) 4) 3 Sol: he equation of a particle executin S.H.M is y = sinωt distance traveled in s π y = sin = = S 8 π Distance traveled in seconds = y = sin = 8 S = Distance traveled in nd second = y - y = = S S = = /. he mass and diameter of a planet are two times those of earth. If a seconds pendulum is taken to it, the ns : 4 Sol: time period of the pendulum in seconds is (00 M) ) ) / 3) 4) We known that GM = R p M p R E = E M E R p

8 M E R E = = ME RE We know that time period of simple pendulum = l/ π = P E E p P = = s P. If two bodies of same mass are executin S.H.M. with frequencies in the ratio : and amplitudes in the ratio :3 then the ratio of their total eneries is (00M) ) : 3 ) : 9 3) : 4 4) : 6 ns : mω = ( ) m π n E n = = = E n 3 9 Sol: E = otal enery of a body executin SHM = 3. he time period of a liht loaded sprin is 3.5 seconds. On chanin the load by k, the period decreases by 0.5 seconds. he initial load on the sprin is (00 M) ) 3 (0/3) k ) 4 (0/3) k 3) 5 (0/3) k 4) 6 (0/3) k ns : m π k m = m 3.5 m = 3 m 0 On simplifyin m= 3 k 3 Sol: he time period of a mass loaded sprin = = 4. particle is executin simple harmonic motion with an amplitude of 4cm. t the mean position the velocity of the particle is 0 cm/s. he distance of the particle from the mean position when its speed becomes 5cm/s is (000 M) ) 3 cm ) 5 cm 3) 3cm 4) 5cm Sol: mplitude = 4 cm Maximum velocity V = 0cm/sec But V = aω

9 V 0 ω = = =.5rads a 4 Velocity V = ω a y 5 5= 6 y y = 3cm 5. particle executes simple harmonic motion with a period of s and amplitude m. he shortest time it takes to reach point ) ) 4 Sol: ime period = sec m from its mean position in seconds is (000 M) mplitude = metre Displacement = 3) 8 metre 4) 6

10 We know y = asinω t π = sin t 0 π sin 45 = sin t t = sec 8

OSCILLATIONS

OSCILLATIONS 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