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SHM · Lecture 19 of 26

Superposition of SHMs: adding with phasors, and perpendicular SHMs

By Lalit, IIT Madras

Superposition of SHMs in JEE: two SHMs on one line added with phasors into R = root(A² + B² + 2AB cos φ), then SHMs at right angles.

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A chalkboard phasor diagram showing arrows A and B at angle φ with their resultant R at angle δ, beside the boxed formulas R = √(A² + B² + 2AB cos φ) and x = R sin(ωt + δ).

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What this lecture covers

Under superposition, displacements add. Two SHMs on one line with the same ω give another SHM with that ω, shown with the sine of a sum and then with phasors, whose components give R = root(A² + B² + 2AB cos φ). Amplitudes 4 and 3 at 90 degrees make 5, and 3 and 5 at 60 degrees make 7. Three SHMs add the same way, but unequal omegas do not give SHM. Perpendicular SHMs trace a straight line or an ellipse.

In this lecture

  1. Play from The principle of superposition
  2. Play from Two SHMs on one line, same ω
  3. Play from Expanding the sin of a sum
  4. Play from The phasor of an SHM
  5. Play from Adding with phasors, with examples
  6. Play from Three SHMs on one line
  7. Play from Unequal omegas: the sum is not SHM
  8. Play from Perpendicular SHMs: line or ellipse
  9. Play from The ellipse, the circle and direction

Check yourself

  1. Asked at play from

    Question 1: Two SHMs act on a particle along the same line, and both have the same angular frequency ω. What kind of motion results when they are added?

    1. An SHM with the same ω
    2. An SHM with double the ω
    3. Not an SHM
    Show the answerHide the answer

    Answer: An SHM with the same ω

  2. Asked at play from

    Question 2: Adding two SHMs on a line needs the trigonometric identity for the sine of a sum. How does sin(ωt + φ) expand?

    1. sin ωt cos φ + cos ωt sin φ
    2. sin ωt sin φ + cos ωt cos φ
    3. sin ωt + sin φ
    Show the answerHide the answer

    Answer: sin ωt cos φ + cos ωt sin φ

  3. Asked at play from

    Question 3: An arrow of length B makes an angle φ with a given line. What is the length of its component along that line?

    1. B cos φ
    2. B sin φ
    3. B tan φ
    Show the answerHide the answer

    Answer: B cos φ

  4. Asked at play from

    Question 4: Two SHMs on a line are x₁ = 4 sin ωt and x₂ = 3 sin(ωt + 90°), so their phasor arrows of lengths 4 and 3 are at a right angle. What is the amplitude R of the resultant SHM?

    1. 7
    2. 5
    3. 1
    Show the answerHide the answer

    Answer: 5

  5. Asked at play from

    Question 5: Two SHMs on a line are x₁ = 3 sin ωt and x₂ = 5 sin(ωt + 60°). Using cos 60° = 1/2, what is the amplitude R of the resultant SHM?

    1. 8
    2. 7
    3. √34
    Show the answerHide the answer

    Answer: 7

  6. Asked at play from

    Question 6: A particle is acted on by two SHMs along one line with different angular frequencies ω₁ and ω₂, giving x = A sin ω₁t + B sin ω₂t. Recalling that SHM means acceleration = −(one constant) × x, is the sum an SHM, and if so with what angular frequency?

    1. Yes, with ω₁ + ω₂
    2. Yes, with the average ω
    3. No, it is not an SHM
    Show the answerHide the answer

    Answer: No, it is not an SHM

  7. Asked at play from

    Question 7: For the conic x²/9 + y²/4 = 1, what are the values of x on putting y = 0?

    1. 3 or −3
    2. 9 or −9
    3. 2 or −2
    Show the answerHide the answer

    Answer: 3 or −3

  8. Asked at play from

    Question 8: A particle moves on an ellipse with semi-axes A along x and B along y, starting from the top (x = 0, y = B) at t = 0; a little later x has increased and y has decreased. In which direction does the particle go round the ellipse?

    1. Clockwise
    2. Anticlockwise
    Show the answerHide the answer

    Answer: Clockwise

  9. Asked at play from

    Question 9: A particle has x = 3 sin ωt and y = −3 cos ωt. What is the shape of its path, and in which direction does it go round?

    1. Circle, anticlockwise
    2. Circle, clockwise
    3. Ellipse, anticlockwise
    Show the answerHide the answer

    Answer: Circle, anticlockwise

Read the lectureIn Roman Hinglish · about 2,050 words

Play from The principle of superposition

Aaj hum superposition padhenge. Particle ek line par hai, aur O uski mean position hai. Us par do ya zyada SHM ek saath lagein, toh uski final motion kya hogi?

Ye idea aage waves aur AC current mein bhi aayega. Uske O se displacement ko hum x kahenge, jo meter mein hai. Right mein usey plus lete hain aur left mein minus.

Principle seedha hai. Pehla cause akela lage, toh kisi moment par particle ka displacement x₁ hota hai. Doosra cause akela lage, toh usi moment par displacement x₂ hota hai. Dono saath lagein, toh hum dono ko jodte hain, yaani x = x₁ + x₂. Displacements sign ke saath judte hain.

Ek example lo. Paani par ek cork hai, aur us par do ripples aati hain. Unmein se pehli akeli usey 3 centimeter uthaaye aur doosri 2 centimeter, toh dono saath 5 centimeter uthaati hain.

Ab sign wala example lo. x₁ = +3 centimeter hai aur x₂ = −2 centimeter hai, toh hamein x = +1 centimeter milta hai. Displacements judte hain, par amplitudes tabhi judti hain jab peaks ek saath aayein, warna sum ka peak chhota rehta hai.

Play from Two SHMs on one line, same ω

Pehla case sabse important hai, jismein dono SHM ek hi line par hain aur dono ka ω same hai. ω angular frequency hai, jo radian per second mein hota hai aur hamesha positive rehta hai. Ab andaaza lagao, dono ko jodne par kaisi motion banegi?

Sahi. Algebra aur phasors dono se dikhayenge ki sum phir SHM hai, aur ω wahi rehta hai.

Pehle SHM ko hum likhte hain x₁ = A sin ωt, jismein A amplitude hai, jo meter mein hai aur positive rehta hai, aur t time hai, jo seconds mein hai. Doosre ko likhte hain x₂ = B sin(ωt + φ), jismein B uska amplitude hai aur φ uska phase constant hai, jo radian mein hai. Pehle ka phase constant zero hai, isliye φ hi batata hai ki doosra kitna aage hai.

Play from Expanding the sin of a sum

Inhein jodne ke liye trigonometry wale chapter ki ek identity chahiye. Toh batao, sin(ωt + φ) kaise khulta hai?

Sahi. Sin(ωt + φ) = sin ωt cos φ + cos ωt sin φ. Aage cos ko sine form mein badalne mein bhi yahi kaam aayega.

Toh doosre term ko kholte hain, aur sin ωt aur cos ωt wale terms saath rakhte hain. Dono mein ωt wahi hai. Aisa combination phir usi ω ka ek hi sine banta hai, bas uska amplitude aur phase constant naye hote hain. Yaani resultant phir usi ω ka ek SHM hai.

Play from The phasor of an SHM

Par trig har baar aasaan nahi padti, khaaskar teen ya chaar SHM mein. Isliye hum phasor use karenge, jismein phase kaam aayega. Phasor ek vector hai jo ek SHM ko represent karta hai, yaani ye phasor circle ka wahi arrow hai. Ye sirf maths ki convenience hai, aur iska koi physical matlab nahi hai. Jo bhi quantity sine ki tarah badle, uska phasor ban sakta hai.

Phasor aise banta hai. Vector ki do cheezein hoti hain, length aur direction. SHM ko hum likhte hain x = A sin(ωt + φ), toh uske phasor ki length A hai. Direction top wali reference line se clockwise ωt + φ hai, yaani phase. Uska shadow diameter par x deta hai.

Ab hum us SHM ka phasor banayenge jise likhte hain x = −A sin ωt. Length negative nahi hoti, toh minus ka matlab hai ulti direction, yaani angle mein 180° aur jud jaate hain. Isliye −A sin ωt = A sin(ωt + 180°). Ye arrows force ya velocity nahi hain, bas SHM ko dikhane ka tareeka hain.

In arrows ko jodne ke liye vectors wale chapter ki ek baat chahiye. Length B ka ek arrow kisi line se φ angle par hai. Us line ke along uska part kitna hota hai?

Sahi. Angle ke paas wala part cos leta hai aur saamne wala sin. Toh along part hai B cos φ, aur across part hai B sin φ.

Play from Adding with phasors, with examples

Wahi do SHM jodte hain. Pehle ka arrow A hai, jo reference line par hai, aur doosre ka B hai, jo φ aage hai. In dono ka vector sum hamara teesra arrow hai. Ab in teeno ko saath ghumao, toh inke shadows x₁, x₂ aur x bante hain. Har moment par sum ka shadow x₁ + x₂ ke barabar rehta hai, isliye sum arrow hi resultant SHM ka phasor hai.

t = 0 par lautkar doosre arrow ko todte hain, along B cos φ aur across B sin φ. Toh sum ka along part hai A + B cos φ, across part hai B sin φ, aur uski length hai R, yaani naya amplitude. Pythagoras aur sin² + cos² = 1 se hamein milta hai R = √(A² + B² + 2AB cos φ).

Sum arrow reference se jitna aage hai, wo angle δ hai. Tan δ = across/along, yaani B sin φ/(A + B cos φ). Resultant ko hum likhte hain x = R sin(ωt + δ), jismein amplitude R hai, phase constant δ hai, aur ω wahi rehta hai.

Do special cases dekhte hain. φ zero ho, toh arrows ek direction mein hain aur R = A + B. φ 180° ho, toh arrows ulti direction mein hain aur R = A − B ka size. Baaki φ par, jaise 90°, R in dono ke beech rehta hai. Formula ratna zaroori nahi hai, tareeka yaad rakho, yaani arrows banao, jodo, aur length aur angle padho.

Pehla example lo, jismein hum do SHM jodte hain. Inhein likhte hain x₁ = 4 sin ωt aur x₂ = 3 sin(ωt + 90°). Arrows 4 aur 3 hain aur right angle par hain, toh sum arrow unke triangle ka hypotenuse hai. R kitna hoga?

Sahi. Cos 90° zero hota hai, toh R² = 4² + 3², yaani 16 + 9, yaani 25. Isliye R hai 5, 7 nahi.

Angle ke liye hamein milta hai tan δ = 3/4, yaani δ lagbhag 37° hai. Toh resultant ko hum likhte hain x = 5 sin(ωt + 37°).

Ab formula se ek aur example lo. Is baar hum likhte hain x₁ = 3 sin ωt aur x₂ = 5 sin(ωt + 60°). δ ke liye hamein milta hai tan δ = 5√3/11, par pehle R chahiye. Cos 60° = 1/2. R kitna hoga?

Sahi. R² = 9 + 25 + 2 × 3 × 5 × 1/2, yaani 34 + 15, yaani 49. Toh R hai 7, 8 nahi.

Play from Three SHMs on one line

Teen SHM hon toh? Logic wahi rehta hai. Sab ek line par hon aur ω same ho, toh resultant usi ω ka ek SHM hai.

Ek bhi ω alag ho, toh wo SHM nahi rehta. Isiliye hum ω pehle dekhte hain. Principle wahi hai, yaani x = x₁ + x₂ + x₃.

Ab ye teen SHM lo, x₁ = 8 sin ωt, x₂ = 5 cos ωt, aur x₃ = −3 sin ωt. Inhein sine form mein likhte hain. Cos ωt = sin(ωt + 90°), aur −sin ωt = sin(ωt + 180°). Toh 8 wala arrow reference par hai, 5 wala 90° aage, aur 3 wala 180° aage hai.

Ek line wale arrows pehle jodte hain. 8 upar hai aur 3 neeche hai, toh net 5 upar bachta hai. Phir 5 aur 5 right angle par hain, toh Pythagoras se hamein milta hai R = 5√2. Dono barabar hain, isliye angle 45° hai. Toh hum likhte hain x = 5√2 sin(ωt + 45°).

Ye arrows jodne ka ek tareeka tha, aur order koi bhi ho resultant wahi aata hai. Doosra tareeka components ka hai, jo tilted arrows mein aasaan padta hai. Question mein diya hai x₁ = A sin ωt, x₂ = A sin(ωt + 60°), aur x₃ = A sin(ωt + 90°). Hamein amplitude nikaalna hai.

Har arrow ko along aur across todte hain. Along mein hamein milta hai A + A cos 60° + A cos 90°, yaani 3A/2. Across mein hamein milta hai A sin 60° + A sin 90°, yaani A × (1 + √3/2).

Dono ke squares jodne par hamein milta hai A² × (4 + √3). Toh R = A√(4 + √3). Angle bhi nikal sakta hai, par question amplitude poochta hai.

Aise questions JEE mein aate hain. Isliye vectors jodna pakka rakho, kyunki components ka tareeka har angle par chalta hai.

Play from Unequal omegas: the sum is not SHM

Agla case ω alag-alag hone ka hai. Hum jodte hain x = A sin ω₁t + B sin ω₂t. SHM ki pehchaan yaad karo, yaani acceleration = −(ek hi constant) × x. Toh kya ye sum SHM hai?

Sahi. x ko do baar differentiate karo, toh pehle term se aata hai −ω₁² A sin ω₁t aur doosre se aata hai −ω₂² B sin ω₂t. Dono ke constants alag hain, isliye acceleration −(ek constant) × x nahi ban sakta.

Par superposition ab bhi chalta hai. Kisi moment par ek SHM se 3 centimeter aur doosre se 2 centimeter ho, toh saath mein 5 centimeter hota hai. Bas motion SHM nahi rehti. ω aur 2ω jodo, ya ω aur 3ω, toh shape sine se alag aati hai. Ek line par same ω ho, tabhi SHM milta hai.

Play from Perpendicular SHMs: line or ellipse

Ab perpendicular directions lete hain. y particle ka vertical displacement hai, jo meter mein hai. Upar usey plus lete hain aur neeche minus. Hum likhte hain x = A sin ωt aur y = B sin(ωt + φ). Path chahiye, toh φ ke special values lekar t eliminate karte hain.

φ zero lo. Tab hamein milta hai x = A sin ωt aur y = B sin ωt, yaani y = (B/A) × x. Ye origin se guzarti seedhi line hai. Particle is line par point (A, B) se point (−A, −B) tak aage-peeche chalta hai. Ye SHM hai, aur iska amplitude hai √(A² + B²).

Ab φ = π rakhte hain. Tab hamein milta hai y = −B sin ωt, yaani y = −(B/A) × x. Ye doosri taraf jhuki line hai, aur particle point (A, −B) aur point (−A, B) ke beech chalta hai. Ye bhi SHM hai, aur uska amplitude wahi hai.

Ab φ = 90° rakhte hain. Tab hamein milta hai y = B cos ωt. Isliye x/A = sin ωt aur y/B = cos ωt. Dono ka square karke jodo, aur sin² + cos² = 1 lagao, toh hamein milta hai x²/A² + y²/B² = 1.

Play from The ellipse, the circle and direction

Ye conic sections wale chapter ki equation hai. Ek example lo, x²/9 + y²/4 = 1. Ismein y zero rakho, toh x kitna hoga?

Sahi. y zero par x² = 9, toh x = +3 ya −3. Yahi ellipse hai, jismein x ki sabse badi value 3 hai aur y ki sabse badi value 2 hai.

Toh ye ellipse hai. x sabse bada A hai, jahan y zero hota hai, aur y sabse bada B hai, jahan x zero hota hai. t zero par particle ka x zero hai aur y = B, yaani wo top par hai. Thodi der baad x badhta hai aur y ghatta hai, yaani particle right ki taraf neeche jaata hai. Ye clockwise hai ya anticlockwise?

Sahi. Top par right ki taraf jaata point clockwise ghoomta hai.

Ellipse wali motion SHM nahi hai. Superposition se hamein line milti hai, ellipse milti hai, aur φ plus ya minus 90° par A = B ho toh circle milta hai. Ab ek question lo, jismein diya hai x = 3 sin ωt aur y = −3 cos ωt. Path aur direction batao.

Sahi. Square karke jodo, toh x² + y² = 9 sin² + 9 cos², yaani 9, toh radius 3 ka circle banta hai. t zero par particle neeche hai, aur thodi der baad x plus ho jaata hai, yaani wo neeche se right ki taraf jaata hai, aur ye anticlockwise hai.

Direction ka shortcut suno. x ko A sin ωt likho, toh t zero par x zero hai aur wo right mein ja raha hai. Us waqt y plus ho, toh particle upar wale aadhe mein right jaata hai, yaani clockwise. y minus ho, toh neeche wale aadhe mein jaata hai, yaani anticlockwise.

Ab poora summary dekhte hain. Displacements judte hain. Ek line par same ω ho, toh phir SHM milta hai. ω alag ho, toh SHM nahi milta. Perpendicular mein same ω par line ya ellipse banti hai.

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