Velocity and acceleration in SHM — v = ω√(A² − x²): Question bank

1.6.4 · D5 · Physics › Oscillations & Waves › Velocity and acceleration in SHM — v = ω√(A² − x²)

Traps se pehle, woh chaar symbols jinpar yeh sab tika hai:

Woh do tools jinpar hum poore waqt depend karte hain:

True or false — justify

True or false: extremes par (

x = \pm A

) particle sabse tezi se move kar raha hota hai kyunki usne sabse zyada distance travel ki hai.

False —

x = \pm A

par root

A^{2} - x^{2} = 0

hai, isliye

v = 0

; particle momentarily ruka hua hota hai jab woh reverse karta hai. Travel ki gayi distance ka instantaneous speed se koi lena-dena nahi.

True or false: centre par (

x = 0

) acceleration maximum hoti hai.

False —

a = - ω^{2} x

, aur

x = 0

par yeh

a = 0

deta hai. Centre woh jagah hai jahan speed peak karti hai lekin acceleration zero ho jaati hai; yeh dono anti-phase mein hain.

True or false: velocity aur acceleration apne maxima ek hi jagah reach karte hain.

False —

v_{m a x} = ω A

centre par,

∣ a ∣_{m a x} = ω^{2} A

edges par. Yeh opposite locations par peak karte hain, jo exactly

90°

180°

phase ki kahaani hai.

True or false: SHM mein acceleration constant hoti hai, bilkul freely falling body ki tarah.

False — free fall mein

a = g

(fixed) hota hai, lekin yahan

a = - ω^{2} x

position ke saath badalta hai: centre par zero, edges par sabse bada. Sirf

a \propto - x

hi SHM define karta hai, Simple Harmonic Motion — definition a = −ω²x ke anusaar.

True or false:

v = ω A^{2} - x^{2}

pehle se hi motion ki direction bata deta hai.

False — bare root sirf magnitude (speed) deta hai. Physical velocity

\pm

carry karta hai kyunki har point se ek baar baahri aur ek baar andar aate hue, opposite sign ke saath guzra jaata hai.

True or false: amplitude

A

ko double karne se maximum speed double ho jaati hai lekin period unchanged rehta hai.

True —

v_{m a x} = ω A

A

ke saath scale karta hai, lekin

ω

(aur isliye

T = 2 π / ω

)

k

aur

m

se set hota hai, na ki aap ise kitna kheenchte ho. Dekho Angular frequency ω and time period T.

True or false:

x = A /2

par speed exactly

v_{m a x}

ki aadhi hoti hai.

False —

v = ω A^{2} - (A /2)^{2} = ω A 3 /2 \approx 0.87 v_{m a x}

. Speed beech mein zyada rehti hai aur sirf edge ke paas tezi se girती hai; half-speed

x = 3 A /2 \approx 0.87 A

par hoti hai.

True or false: restoring acceleration hi object ko oscillate karti hai drift away hone ki bajaye.

True —

a = - ω^{2} x

mein minus sign hamesha particle ko equilibrium ki taraf wapas drive karta hai, isliye har displacement reverse hoti hai; minus hatao aur motion exponentially bhag jaati hai instead of cycle karne ke.

Spot the error

v = ω A^{2} - x^{2}

, isliye

x = 2 A

par speed

ω A^{2} - 4 A^{2} = ω A - 3

hai." — error kahan hai?

Particle kabhi

x = 2 A

tak pahuncha hi nahi: SHM poori tarah

- A \leq x \leq A

range mein rehta hai kyunki

x = A sin (ω t + ϕ)

aur

sin

kabhi

1

se zyada nahi ho sakta. Isliye

x = 2 A

physical domain ke baahir hai; root imaginary isliye ho jaata hai kyunki yeh ek flag hai ki tumne ek aisi displacement ke baare mein poocha jo motion produce nahi kar sakti, na ki isliye ki koi real speed complex hai.

"Kyunki

a = - ω^{2} x

aur

x

negative ho sakta hai, acceleration kabhi kabhi positive hoti hai — isliye yeh hamesha restoring nahi hoti." — flaw pakdo.

Arithmetic sahi hai lekin conclusion nahi: jab

x < 0

a = - ω^{2} x > 0

+ x

direction mein point karta hai — jo, hamare sign convention ke anusaar, centre ki taraf wapas hai. "Restoring" ka matlab hai "

x

ke opposite sign," aur minus sign dono taraf yeh guarantee karta hai.

v (x)

paane ke liye humne

v = A ω cos (ω t + ϕ)

x

ke saath respect mein differentiate kiya." — kya galat hua?

Humne

x

ke saath differentiate nahi kiya; humne

sin^{2} + cos^{2} = 1

use karke time eliminate kiya Pythagorean identity sin² + cos² = 1 se.

v

ko differentiate karne se acceleration milta hai,

v (x)

relation nahi.

"Energy deta hai

\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = E

, isliye

v = 2 E / m - (k / m) x^{2}

, jo

ω A^{2} - x^{2}

se alag formula hai." — step by step reconcile karo.

Yeh dono identical hain. Pehle, stiffness–frequency link

k = m ω^{2}

hai (defining SHM force

F = - k x = ma = - m ω^{2} x

), isliye

k / m = ω^{2}

. Doosra, total energy turning point

x = A

par apni all-potential value ke barabar hai jahan

v = 0

E = \frac{1}{2} k A^{2}

, isliye

2 E / m = (k / m) A^{2} = ω^{2} A^{2}

. Dono substitute karo:

ω^{2} A^{2} - ω^{2} x^{2} = ω A^{2} - x^{2}

. Same statement, energy language mein (Energy in SHM — kinetic and potential).

"Velocity displacement se

90°

aage hai, isliye acceleration velocity se

90°

aage hona chahiye aur displacement se

45°

aage." — galti pakdo.

Aakhri number galat hai:

a

v

90°

aage hai, aur

90° + 90° = 180°

, isliye

a

x

180°

aage hai (anti-phase),

45°

nahi. Phases add hote hain, average nahi — dekho Phase and phase difference.

"Reference circle par, shadow ki speed circle ke top par sabse tez hoti hai." — ise correct karo.

Projected (shadow) speed sabse tez hoti hai jab point horizontal diameter cross karta hai (shadow centre par,

x = 0

), aur extreme left/right par zero hoti hai (shadow

\pm A

par). Dekho Reference circle (projection of uniform circular motion).

Why questions

Speed edges ke paas itni tezi se kyun girti hai rather than linearly fall karne ke?

Kyunki

v = ω A^{2} - x^{2}

: square root

x = 0

ke paas flat rehta hai aur

x = \pm A

ke paas tezi se girta hai, jahan do nearly-equal squares cancel ho jaate hain. Geometrically "bachli jagah" ek semicircle ki height ki tarah sihriti hai, straight line ki tarah nahi.

Velocity form

t

ke bina itni useful kyun hai?

Yeh seedha jawab deta hai "yahan kitni tezi se?" bina pehle woh time solve kiye jab particle us point par pahuncha — tum ek inverse-sine step aur poora phase bookkeeping skip kar dete ho.

Acceleration formula

a = - ω^{2} x

mein koi amplitude

A

kyun nahi hai, jabki

∣ a ∣_{m a x}

mein hai?

a

sirf is par depend karta hai ki tum kahan ho (

x

) aur oscillator kitna stiff hai (

ω^{2}

); amplitude sirf us largest

x

ke through enter karta hai jo motion reach karta hai, deta hai

∣ a ∣_{m a x} = ω^{2} A

x = \pm A

par.

Velocity displacement se

90°

aage kyun honi chahiye, peeche nahi?

Kyunki

v = \frac{d x}{d t}

sin

ko differentiate karne se

cos

milta hai, jo quarter-cycle pehle peak karta hai. Rate of change hamesha quantity se aage rehta hai, isliye

v

x

se aage hai, kabhi peeche nahi.

a = - ω^{2} x

mein

\pm

kyun nahi aata jabki

v

mein aata hai?

Acceleration poori tarah position se fix hoti hai (ek value of

x

→ ek

a

), isliye koi ambiguity nahi. Velocity nahi: same

x

ko do opposite directions ke saath visit kiya jaata hai, isliye sign choice aata hai.

v = \pm ω A^{2} - x^{2}

mein kaun sa sign outbound hai aur kaun sa inbound?

"Outbound" ko centre se door jaana maano.

+ x

side par, door jaana matlab

x

badhna hai, isliye

v > 0

(woh

+

root); wapas aana matlab

v < 0

(woh

-

root).

- x

side par yeh flip ho jaata hai: outbound hai

x

ghataana, isliye

v < 0

, aur inbound hai

v > 0

. Magnitude dono taraf same hoti hai — sirf direction (sign) badalta hai.

Swing ke middle ko "no force ka point" kehna galat kyun hai even though

a = 0

wahan?

a = 0

ka matlab us instant par zero net restoring force hai, lekin particle sabse tezi se move kar raha hota hai aur maximum momentum carry karta hai seedha iske through — chalte rehne ke liye koi force zaruri nahi, aur ek restoring force us moment reappear hoti hai jab yeh centre se hata.

Edge cases

Exactly

x = 0

par

v

aur

a

kya hain?

v = ω A

(maximum speed) aur

a = 0

(koi restoring acceleration nahi). Sabse tez lekin momentarily un-accelerated — yeh centre se coast karta hua guzarta hai.

Exactly

x = \pm A

par

v

aur

a

kya hain?

v = 0

(turning point par ruka hua) aur

∣ a ∣ = ω^{2} A

(maximum restoring acceleration) centre ki taraf point karta hua. Speed aur acceleration centre ke versus apne roles swap kar lete hain.

Degenerate case

A = 0

mein poori picture ka kya hota hai?

Tab

x = 0

har time ke liye,

v = ω 0 = 0

, aur

a = 0

: bilkul koi oscillation nahi. Ek zero-amplitude "oscillator" bas equilibrium par baitha rehta hai.

Kya hota hai jab

ω \to 0

(spring infinitely soft ho jaati hai)

A

fixed rakhte hue?

v_{m a x} = ω A \to 0

aur

∣ a ∣_{m a x} = ω^{2} A \to 0

: motion infinitely slow ho jaati hai aur period

T = 2 π / ω \to \infty

. Limit mein kisi finite time mein kuch nahi hilta.

Kya

v = ω A^{2} - x^{2}

us waqt valid hai jab particle

x = A

par turn around karta hai?

Haan — yeh correctly

v = 0

deta hai wahan. Turning point woh ek jagah hai jahan speed aur "root ke neeche bachli jagah" dono vanish ho jaate hain, yahi exactly woh reason hai ki particle reverse karta hai.

Connections

Velocity and Acceleration in SHM — v = ω√(A² − x²)
Simple Harmonic Motion — definition a = −ω²x
Energy in SHM — kinetic and potential
Phase and phase difference
Reference circle (projection of uniform circular motion)
Angular frequency ω and time period T
Pythagorean identity sin² + cos² = 1