Simple Harmonic Motion I Sem

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Phoebe Strickland
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1 Simple Hrmonic Motion I Sem Sllus: Differentil eqution of liner SHM. Energ of prticle, potentil energ nd kinetic energ (derivtion), Composition of two rectngulr SHM s hving sme periods, Lissjous figures. Prolems. (3+1=4 hrs.) The to nd fro motion of prticle out fixed point (men position) in which the ccelertion is proportionl to displcement nd directed towrds the fixed point is clled simple hrmonic motion (SHM). It cn e shown tht the projection of uniform circulr motion long n of its dimeters is lws simple hrmonic motion. The displcement of prticle executing SHM t n instnt of time, t is given the eqution: = A sinωt. Definitions: 1. Displcement (): The distnce of the prticle from the men position t n given instnt of time is clled displcement. 2. Amplitude (A): The mximum displcement on either side of the men position is clled mplitude. 3. Velocit (v): The rte of displcement is clled velocit. We know tht the displcement of prticle executing SHM is given = A sinωt. Then the velocit will e v = d dt = d dt A sinωt = Aω cosωt = Aω 1 sin 2 ωt = ω A 2 A 2 sin 2 ωt v = ω A 2 2 From the ove eqution, it follows tht the velocit is mximum t the men position ( = 0) nd is given v mx = Aω. It lso follows tht the velocit is minimum nd is equl to zero t the extreme positions where = A. 1 P g e

2 Simple Hrmonic Motion I Sem 4. Accelertion (): The rte of chnge of velocit is clled ccelertion. It is given = dv dt = d dt = ω 2 Asinωt Aω cosωt ccelertion, = ω 2 From the ove eqution it follows tht the ccelertion of prticle executing SHM is zero t the men position ( = 0) nd is mximum t the extreme positions ( = A). The mximum vlue is given mx = ω 2 A. 5. Time Period (T): The time tken the prticle executing SHM to complete one oscilltion is clled its time period or periodic time or simpl period. 6. Frequenc (f): The frequenc of prticle executing SHM is the numer of oscilltions completed the prticle in one second. It should e noticed tht the frequenc is the reciprocl of the time period. ie., f = 1 T 7. Phse (φ): The stte of virtion of prticle executing SHM is clled phse. It gives the position nd direction of motion of the prticle t the given instnt of time. It is mesured the frction of the time period tht hs elpsed since the prticle hs lst crossed its men position in given direction. Differentil eqution of liner SHM: Consider prticle of mss, m executing SHM out fixed point. We know tht the restoring force, F cting on the prticle is proportionl to the displcement, nd is lws directed towrds the men position. i. e., F = k. (1) where k is constnt of proportionlit clled the force constnt. But, ccording to the Newton s second lw of motion, F = m = m d2 dt 2 2 P g e

3 Sustituting this vlue in eqution (1), we get Simple Hrmonic Motion I Sem m d2 dt 2 = k i. e., d 2 dt 2 = k m = ω2 where ω = k m represents the ngulr velocit of the prticle. d 2 dt 2 + ω2 = 0 This is the differentil eqution representing simple hrmonic motion. Energ: The prticle executing SHM possesses oth the potentil energ nd kinetic energ. Expression for potentil energ: Consider prticle of mss m executing simple hrmonic motion. Let e the displcement of the prticle t given instnt of time, t. Let it e displced through smll distnce, d. If F is the restoring force cting on the prticle t distnce from the men position, then the work done in displcing the prticle through d is given dw = F d Hence, the totl work done in displcing the prticle through distnce is given But, we know tht W = dw 0 = F d. (1) 0 F = m = mω 2 neglecting the negtive sign tht hs onl the phsicl significnce. Sustituting for F in eqution (1), we get W = mω 2 d = 1 2 mω P g e

Simple Hrmonic Motion I Sem This work done is stored in the prticle in the form of potentil energ, V nd hence the potentil energ of prticle executing SHM is given V = 1 2 mω2 2.

4 Simple Hrmonic Motion I Sem This work done is stored in the prticle in the form of potentil energ, V nd hence the potentil energ of prticle executing SHM is given V = 1 2 mω2 2.. (2) Expression for kinetic energ: The kinetic energ, T of prticle of mss, m moving with velocit, v is given the expression T = 1 2 mv2 (3) We know tht the velocit of prticle executing SHM is given v = ω A 2 2. Sustituting this vlue in eqution (3), we get the expression for kinetic energ T = 1 2 mω2 A 2 2 (4) Totl energ: The totl energ, E of prticle executing SHM is the sum of its potentil energ nd kinetic energ, given E = T + V = 1 2 mω mω2 A 2 2 E = 1 2 mω2 A 2 From the ove expression, it follows tht the totl energ of prticle is independent of the displcement nd hence is constnt. Grphicl Representtion: 4 P g e

5 Simple Hrmonic Motion I Sem Composition of two rectngulr SHM s of the sme frequenc: Composition refers to determintion of the resultnt. The determintion of the resultnt motion of prticle sujected to two mutull perpendiculr SHM s of the sme frequenc is clled composition of two rectngulr SHM s of the sme frequenc. The resultnt pth trced the prticle sujected to two mutull perpendiculr SHM s of the sme frequenc is clled Lissjous figure. Consider two mutull perpendiculr SHM s of the sme frequenc represented x = sinωt. 1 And = sin ωt + φ. (2) where nd re the respective mplitudes nd φ is the phse difference etween the two SHM s. From eqution (1) it follows tht sinωt = x nd cosωt = 1 From eqution (2) it follows tht = sinωt. cosφ + cosωt. sinφ = sinωt. cosφ + cosωt. sinφ Sustituting for sinωt nd cosωt in the ove eqution, we get 2 = x. cosφ + 1 x2 2. sinφ i. e., x. cosφ = 1 x2 2. sinφ Squring oth the sides, Rerrnging the terms nd simplifing, we get x. cosφ 2 = 1 x2 2. sin2 φ x2 2. cos2 φ 2x cosφ = sin2 φ x2 2 sin2 φ x cosφ = sin2 φ (3) This is the eqution tht represents the resultnt of two rectngulr SHM s of the sme frequenc. 5 P g e

6 Simple Hrmonic Motion I Sem Eqution (3) represents n olique ellipse nd hence it follows tht the pth trced the resultnt of two rectngulr SHM s of the sme frequenc is n ellipse tht cn e enclosed in rectngle of sides 2 nd 2 s shown in figure (1). Specil Cses: Figure (1) 1. φ = 0. i.e., when there is no phse difference etween the two SHM s. In this cse, sinφ = 0 nd cosφ = 1. Sustituting these vlues in eqution (3), we get x = 0 x 2 = 0 i. e., ± x = 0 ± = ± x The eqution represents stright line hving positive slope pssing through the origin nd ling in the first nd the third qudrnts s shown in figure (2) Figure (2) 2. φ = π. In this cse sinφ = 1 nd cosφ = 1. Sustituting these vlues in eqution (3), we get x = 1 2 The ove eqution represents n olique ellipse s shown in figure (3) 3. φ = π. In this cse sinφ = 1 nd cosφ = 0. Sustituting these vlues in eqution (3), we 2 get Figure (3) = 1 This represents smmetric ellipse s shown in figure (4). 6 P g e Figure (4)

7 Simple Hrmonic Motion I Sem Further, if the mplitudes re equl, i.e., =, then we get = 1 x2 + 2 = 2 This represents circle s shown in figure (5). Figure (5) 4. φ = 3π. In this cse sinφ = 1 nd cosφ = 1. Sustituting these vlues in eqution (3), we get x = 1 2 This represents n olique ellipse s shown in figure (6). Figure (6) 5. φ = π. In this cse sinφ = 0 nd cosφ = 1. Sustituting these vlues in eqution (3), we get x = x + 2 = 0 i. e., ± x + = 0 ± = x Figure (7) This represents stright line with negtive slope pssing through the origin nd ling in the 2 nd nd the 4 th qudrnts s shown in figure (7). The sme figures would repet in the reverse direction when the phse difference φ vries from π to 2π. All these figures from (1) to (7) re the pth trced the resultnt of the two rectngulr SHM s of the sme frequenc under vrious phse differences nd re clled Lissjous Figures. 7 P g e

8 Simple Hrmonic Motion I Sem Question Bnk 2 mrk questions: 1. If prticle is executing SHM with frequenc of 0.4Hz nd mplitude 0.2m, clculte its mximum ccelertion. 2. The displcement of prticle executing SHM is given = sin8πt. Clculte its frequenc nd mplitude. 3. A prticle is executing n SHM with period of 4s nd mplitude 2cm. Wht is its mximum speed? 4. A prticle is executing n SHM of mplitude 2cm nd period 2s. Clculte its velocit t the men position. 5. Wht re Lissjous Figures? 6. Wht is the nture of the Lissjous figure when two SHM s of the sme period nd sme mplitude hving phse difference of π 2 cting t right ngles superpose? 7. Wht is the nture of the Lissjous figure when two rectngulr SHM s of the sme period nd virting in phse superpose? 8. Wht is the phse difference etween two SHM s of the sme period cting t right ngles if the Lissjous Figure is n olique ellipse in the 1 st nd the 3 rd qudrnts when the superpose? 5 Mrk Questions: 1. Set up differentil eqution for simple hrmonic motion. 2. For prticle executing n SHM, mplitude is 15cm nd period is 2π s. Find its velocit when the displcement is 8cm from the men position. 3. Derive n expression for the potentil energ of prticle executing SHM. 4. Write note on Lissjous Figures. 10 Mrk Questions: 1. Derive n eqution for the composition of two rectngulr SHM s hving the sme period nd explin n two cses. 2. Derive expressions for the potentil energ, kinetic energ nd hence the totl energ of prticle executing SHM. Give the grphicl representtion. 8 P g e

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