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SHM · Lecture 22 of 26

Forced oscillations and resonance

By Lalit, IIT Madras

Forced oscillations and resonance for JEE SHM: resonance at ω = ω₀, and the steady amplitude of a driven spring-block.

Watch lecture 22

A chalkboard titled 'Amplitude of the steady oscillation' shows a block of mass m on a spring fixed to a wall with a force F0 cos ωt acting on it and the amplitude A marked, beside the formula A = (F0/m) / √((ω² − ω0²)² + (bω/m)²) and a note that for small damping with ω far from ω0 the b term is dropped.

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What this lecture covers

Left alone, a swing slowly loses amplitude; an outside push can return the energy. The sign of the push's work decides the outcome: against the motion it removes energy, with the motion it adds, and five joules in for five lost keeps the amplitude steady. Resonance is a push in sync, ω = ω₀, doing positive work throughout. Driven by F₀ cos ω t, a spring-block settles to an amplitude proportional to 1/|ω² − ω₀²| for small damping, giving the resonance curve.

In this lecture

  1. Play from Forced oscillation: a swing left alone
  2. Play from The sign of work: pushing a swing
  3. Play from Push gives more than the air takes
  4. Play from In sync: ω and ω zero
  5. Play from Resonance: ω equals ω zero
  6. Play from A spring-block with an outside force
  7. Play from Amplitude of a forced oscillation
  8. Play from Small damping: drop the damping term
  9. Play from The resonance curve
  10. Play from Amplitude and energy peak at resonance

Check yourself

  1. Asked at play from

    Question 1: A child on a swing is pulled to one side and released, with air resistance acting and nobody pushing afterwards. What happens to the amplitude of the swing?

    1. It slowly falls
    2. It stays the same
    3. It grows
    Show the answerHide the answer

    Answer: It slowly falls

  2. Asked at play from

    Question 2: When a force acts on a body in the same direction as the body's displacement, what is the sign of the work done by that force?

    1. Positive
    2. Negative
    3. Zero
    Show the answerHide the answer

    Answer: Positive

  3. Asked at play from

    Question 3: A child on a swing is to gain energy from every push we give. How should the child be moving relative to us at the moment of each push?

    1. Moving away
    2. Coming towards us
    Show the answerHide the answer

    Answer: Moving away

  4. Asked at play from

    Question 4: On a swing, air resistance removes 5 joule of energy every cycle and each push supplies 5 joule per cycle. What happens to the amplitude of the swing?

    1. It stays the same
    2. It grows
    3. It falls
    Show the answerHide the answer

    Answer: It stays the same

  5. Asked at play from

    Question 5: A swing is pushed repeatedly from outside. The push repeats with angular frequency ω, and a second angular frequency ω₀ (in rad/s, always positive) is also in play. Whose angular frequency is ω₀?

    1. The swing's own
    2. The push's
    Show the answerHide the answer

    Answer: The swing's own

  6. Asked at play from

    Question 6: A swing is pushed once every time it completes a cycle. How is the push's angular frequency ω related to the swing's own angular frequency ω₀?

    1. ω = ω₀
    2. ω = 2ω₀
    3. ω = ω₀/2
    Show the answerHide the answer

    Answer: ω = ω₀

  7. Asked at play from

    Question 7: For a system that is driven by an external push, what does resonance mean?

    1. Push ω = own ω₀
    2. A very big push
    3. Zero damping
    Show the answerHide the answer

    Answer: Push ω = own ω₀

  8. Asked at play from

    Question 8: A block of mass m on a spring of stiffness k, with natural angular frequency ω₀ = √(k/m), is driven by F₀ cos ωt with small damping and ω far from ω₀, so the damping term is dropped and A = (F₀/m)/√((ω² − ω₀²)²), with F₀ and m fixed. A is directly proportional to which quantity?

    1. 1/|ω² − ω₀²|
    2. |ω² − ω₀²|
    3. 1/(ω² − ω₀²)²
    Show the answerHide the answer

    Answer: 1/|ω² − ω₀²|

  9. Asked at play from

    Question 9: A block on a spring has natural angular frequency ω₀ = 10 rad/s and very small damping, and it is driven by an external force once at an angular frequency of 12 rad/s and once at 20 rad/s. In which case is the steady amplitude larger?

    1. ω = 12 rad/s
    2. ω = 20 rad/s
    3. Both the same
    Show the answerHide the answer

    Answer: ω = 12 rad/s

Read the lectureIn Roman Hinglish · about 1,700 words

Play from Forced oscillation: a swing left alone

Pichhli lecture mein humne damped oscillation dekha. Resistance har cycle mein energy le jaati hai, isliye amplitude ghatta jaata hai, aur oscillation aakhir mein ruk jaati hai. Aaj sawal ulta hai, ki use chalta kaise rakhein. Iske liye baahar se ek external force lagana padega, jo oscillation ki help kare. Naam mein hi ye hai, forced oscillation, yaani force lagakar chalai gayi oscillation.

Ek chhota example lo, swing par baitha ek child, jise hum lagaatar swing karte dekhna chahte hain. Pehle use ek side kheenchkar bas chhod do, aur hum kuch na karein. Uske amplitude ka kya hoga?

Sahi. Air ki resistance har cycle mein thodi energy le jaati hai, toh amplitude dheere-dheere ghatta hai, aur child aakhir mein neeche latka reh jaata hai.

Ye wahi damping hai. Air ki resistance energy le jaati hai, isliye swing rukti hai. Ab socho, agar koi external source har cycle mein theek utni hi energy waapas de de jitni air le gayi, tab kya hoga? Tab child swing karta reh sakta hai. Poora topic isi ek idea par chalta hai.

Play from The sign of work: pushing a swing

Push ka asar work se tay hota hai. Uske liye work, energy and power wale chapter ki ek baat yaad karo. Jab force aur displacement same direction mein hon, toh us force ka work kya hoga?

Sahi. Same direction mein work positive hota hai, aur positive work body ki energy badhaata hai. Force displacement ke opposite ho, toh work negative, aur energy ghatti hai.

Ab push ki timing dekho. Maan lo humne galat samay par push kiya. Tab child hamari taraf aa raha tha, aur humne use ulta dhakka de diya.

Push ke dauran child hamari taraf chal raha hai aur push uske opposite, toh uska work negative hota hai. Yaani energy badhne ki jagah aur nikal gayi, aur humne damping badha di. Aisa nahi karna, kyunki hamein system ki energy banaye rakhni hai aur badhani hai.

Toh sahi timing kya hai? Child ki energy badhani hai, toh push tab lagayein jab child hamse door ja raha ho, ya tab jab wo hamari taraf aa raha ho?

Sahi. Tab push usi direction mein hai jismein child chal raha hai, toh work positive, aur har push se child ki energy badhti hai. Kisi bhi side se push karo, rule ek hi hai. Push us direction mein lagao, jismein swing chal rahi hai.

Play from Push gives more than the air takes

Ab energy ka hisaab dekho. Maan lo har cycle mein push ka work, air jitni energy le jaati hai usse zyada hai. Tab har cycle mein kuch energy bach jaati hai, system ki energy badhti jaati hai, aur amplitude bhi badhta hai. Child har baar aur ooncha jaata hai.

Ab maan lo har cycle mein air 5 joule energy le jaati hai, aur hamara push bhi har cycle mein 5 joule deta hai. Tab amplitude ka kya hoga?

Sahi. Jitni energy gayi, utni waapas aa gayi, toh net change zero hai. Energy wahi rehti hai, amplitude wahi rehta hai, aur swing chalti rehti hai.

Aur push bilkul band kar den, toh sirf air ka negative work bachta hai. Har cycle mein energy ghatti hai, amplitude ghatta hai, aur swing aakhir mein ruk jaati hai. Toh teen hi cases hain. Push zyada de toh amplitude badhta hai, barabar de toh wahi rehta hai, aur kam de ya kuch na de toh ghatta hai.

Play from In sync: ω and ω zero

Jo oscillation baahar ke force ki wajah se chalti rehti hai, use forced oscillation kehte hain. Naam force se isliye hai ki wo force hata lo toh swing thodi der jhoolkar ruk jaayegi. Motion ko zinda rakhne wala wahi baahari force hai.

Ek condition aur hai. External agent aur system dono in sync hone chahiye, tabhi har push system ki energy badhaata hai. Sync na ho, toh kabhi push sahi direction mein lagega aur kabhi ulta, jaisa abhi dekha.

In sync kab honge, ye dekhne ke liye do angular frequencies alag rakhni hongi. Swing ki apni angular frequency ko ab hum ω0 likhenge, jo radian per second mein hoti hai aur hamesha positive. Aur push kitni tezi se repeat hota hai, wo ω hai. Dhyan do ki pichhli lectures mein plain ω hi block ki apni angular frequency tha, par yahan ω push ka hai. Batao, ω0 kiska hai?

Sahi. ω0 swing ki apni angular frequency hai, jo wo bina kisi push ke bhi rakhti hai. ω us push ki hai, jo baahar se lag raha hai.

Swing jab bhi apni ek cycle poori kare, tabhi push lagao, toh push ki period swing ki apni period ke barabar hogi. Tab ω aur ω0 ka relation kya hoga?

Sahi. Same period matlab same angular frequency, kyunki ω = 2π/time period. Toh push ki ω, swing ki ω0 ke barabar hoti hai.

Play from Resonance: ω equals ω zero

Jab push aur system ki period exactly match karti hai, toh wo dono in sync hain, aur isi condition ko resonance kehte hain. Resonance par ω = ω0 hota hai. Swing ko har cycle mein ek baar push karo, toh wo resonance mein hai. Har cycle mein do baar push karo, toh unmein se ek ulta padega, aur wo resonance nahi hai.

Ye sawal exam mein baar-baar aata hai. Resonance ka matlab kya hai?

Sahi. Resonance ka matlab hai ki external agent ki angular frequency system ki natural angular frequency ke barabar hai, yaani ω = ω0.

Resonance par external agent system par har samay positive work karta rehta hai, kyunki push hamesha velocity ki direction mein lagta hai, yaani jis taraf body chal rahi hai. Uska work resistance ke loss se zyada ho, toh amplitude badhta hai. Barabar ho, toh amplitude constant rehta hai, aur forced oscillation chalti rehti hai. In sync na ho, toh push ka kuch hissa energy jodta hai aur kuch nikaal leta hai, isliye amplitude chhota rehta hai.

Play from A spring-block with an outside force

Ab yahi kahani spring-block par dekho. Block ka mass m hai, spring ka constant k hai, aur block mean position ke dono taraf aage-peeche chalta hai, damping ke saath. Is par baahar se ek external force bhi lag raha hai, jo aage-peeche badalta rehta hai.

Block ko mean position se x door rakho, aur right ko plus lo. Us par do force hum pehle se jaante hain. Spring ka force −kx hai, aur resistance ka force −b × velocity. Yahan b damping constant hai, jo kilogram per second mein hota hai aur hamesha positive. Aur minus sign batata hai ki resistance velocity ke opposite hai.

Teesra force wo external force hai jo hum lagaate hain, yaani F0 cos ωt. F0 uski sabse badi value hai, newton mein. ω wahi hai, yaani push kitni tezi se repeat hota hai, aur t time hai, seconds mein. Cos ki wajah se ye force kabhi right mein lagta hai aur kabhi left mein.

Teeno ko jodo. Net force = mass × acceleration hota hai, aur acceleration x ka double derivative hai. Toh hamein milta hai m × x ka double derivative + b × x ka derivative + kx = F0 cos ωt. Damped oscillation ki equation wahi hai, bas uske right side ka zero hat gaya aur external force aa gaya.

Play from Amplitude of a forced oscillation

Is equation ko hum solve nahi karenge, kyunki wo kaafi mushkil hai, aur hamein equation se nahi, uske solution se matlab hai. Thodi der baad block steady oscillation mein aa jaata hai, aur uska amplitude A ek fixed value leta hai. Us A ki form hamein yaad rakhni hai.

Answer ye hai, A = (F0/m) / √((ω² − ω0²)² + (bω/m)²). Yahan ω push ka hai, aur ω0 block ki apni angular frequency hai, jo √(k/m) hoti hai. Steady oscillation bhi push ki ω par chalti hai, apni ω0 par nahi.

Play from Small damping: drop the damping term

Ab is formula se exam jaisa ek sawal dekhte hain. Small damping ke liye, forced oscillation ka amplitude kiske directly proportional hota hai? Ise formula se khud nikaalte hain.

Small damping ka matlab hai ki b bahut chhota hai. Sawal mein yahi diya hai. Toh jab tak ω, ω0 se door hai, denominator mein b wala term pehle term ke saamne lagbhag zero hai, aur hum use hata sakte hain.

Tab root ke andar sirf (ω² − ω0²)² bacha hai. Kisi bhi quantity ke square ka root uska size hota hai, sign ke bina, aur amplitude positive hi hota hai. Toh A kiske directly proportional hai?

Sahi. A = (F0/m) / (ω² − ω0²) ka size. F0/m fixed hai, toh A, 1/us size ke proportional hai.

Ab ω = ω0 rakho. Denominator ka pehla term zero ho jaata hai, aur small damping mein doosra bhi bahut chhota hai, toh denominator zero ke paas aa jaata hai aur amplitude bahut bada ho jaata hai. Jo b wala term humne hataya tha, wahi yahan amplitude ko infinite hone se rokta hai. Theek resonance par hamein milta hai A = F0/(bω0).

Play from The resonance curve

Ise graph par dekho. Horizontal axis par push ki ω hai, jis par ω0 likha hai, aur vertical axis par amplitude. ω badhao, toh amplitude ω0 ke paas pahunchkar ek peak banata hai. Damping kam ho, toh peak ooncha aur patla hota hai, aur damping zyada ho, toh peak neecha aur chauda.

Block ki ω0 10 radian per second hai, aur damping bahut kam hai. Push ki ω 12 radian per second ho ya 20 radian per second, amplitude kismein zyada hoga?

Sahi. ω² − ω0² nikalo. 12 par wo 144 − 100, yaani 44 hai, aur 20 par 400 − 100, yaani 300. Denominator chhota ho toh amplitude bada hota hai, isliye 12 wala lagbhag 7 times bada hai.

Play from Amplitude and energy peak at resonance

Toh jab external agent aur system resonate karte hain, yaani ω = ω0, tab amplitude bhi sabse zyada hota hai aur energy bhi. Dono jaise friends ban gaye hain. Wo ek hi rhythm mein chalte hain, aur har push ka poora asar aata hai.

Chalo, poora summary dekhte hain. External force se oscillation chalti rehti hai, yahi forced oscillation hai. Push tabhi energy jodta hai, jab wo usi direction mein ho jismein body chal rahi hai.

Resonance par, yaani ω = ω0, push hamesha in sync rehta hai aur amplitude sabse zyada hota hai. Equation mein damped oscillation wali equation ke right side par F0 cos ωt aa jaata hai. Small damping mein, ω0 se door, A, 1/(ω² − ω0²) ke size ke proportional hai. Agli lecture mein hum damping ke teen cases dekhenge, underdamped, overdamped aur critically damped.

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