SHM differential equation — solution - x = A cos(ωt + φ): Question bank

1.6.2 · D5 · Physics › Oscillations & Waves › SHM differential equation — solution - x = A cos(ωt + φ)

Shuru karne se pehle, vocabulary ki ek reminder taaki koi symbol neeche tumse chhupi na rahe:

$x$ = centre se displacement (metres), $+$ ek taraf measure kiya, $-$ doosri taraf.
$ω$ = angular frequency (rad/s), ek fixed number jo physics mein baka hua hai ( $k / m$ ek spring ke liye).
$A$ = amplitude, sabse door tak $x$ kabhi pahunchta hai.
$ϕ$ = phase, woh "head start" jo tumhe batata hai ki clock wiggle mein kahan se shuru hua.
$\overset{x}{¨}$ = acceleration ( $x$ ka doosra time-derivative); defining law hai $\overset{x}{¨} = - ω^{2} x$ .

True or false — justify

Har extra oscillation cycle exactly utna hi time leta hai jitna pichla wala liya.

True — period

T = 2 π / ω

sirf

ω

par depend karta hai, aur ideal SHM mein

ω

kabhi nahi badalta, isliye cycles ek clockwork ki tarah identical hain.

Amplitude double karne se period half ho jaata hai.

False —

T = 2 π / ω

mein

A

bilkul bhi nahi hai, isliye period untouched rehta hai. Bada swing faster hota hai (bada

v_{m a x} = A ω

), extra distance utne hi time mein cover karta hai.

Bilkul centre par (

x = 0

) acceleration zero hoti hai.

True —

a = - ω^{2} x

, aur

x = 0

a = 0

ho jaata hai. Restoring pull ghar par zero hai kyunki tumhare paas koi jagah nahi hai jahan se wापस kheecha ja sake.

Bilkul centre par velocity zero hoti hai.

False — centre woh jagah hai jahan speed maximum hoti hai (

v_{m a x} = A ω

). Zero speed sirf turning points

x = \pm A

par hoti hai.

x = A sin (ω t + ϕ)

fundamentally alag motion describe karta hai

x = A cos (ω t + ϕ)

se.

False — sine bas cosine ek phase se shifted hai (

sin θ = cos (θ - π /2)

), isliye yeh wahi SHM hai cycle mein alag point se shuru ki gayi; shift ko

ϕ

mein absorb kar lo.

Agar do identical springs mein masses

m

aur

2 m

hain, toh bhaari wala faster oscillate karta hai.

False —

ω = k / m

, isliye zyada mass ka matlab hai chhota

ω

aur lamba period; bhaari mass zyada sluggish hai.

SHM ke liye acceleration versus displacement ka graph origin se guzarti ek straight line hai.

True —

a = - ω^{2} x

x

mein linear hai slope

- ω^{2}

ke saath; negative slope ke saath origin se guzarti ek straight line SHM ki pehchaan hai.

Velocity aur displacement ek hi instant mein apne maxima tak pahunchte hain.

False — woh quarter-cycle apart hain:

v

centre par peak karta hai (

x = 0

) jabki

x

ends par peak karta hai jahan

v = 0

hoti hai. Woh

9 0^{\circ}

out of phase hain — dekho Phase and phase difference.

SHM oscillator ki total energy centre aur extreme ke beech move karte waqt badlati hai.

False — energy bas kinetic aur potential ke beech trade karti hai; total constant rehti hai, jo exactly woh conserved quantity hai jo derivation ke Step 1 ne uncovaar ki. Dekho Energy in SHM.

Spot the error

"Kyunki release par

v_{0} = 0

hai, isliye phase

ϕ = 0

hona chahiye."

Incomplete —

v_{0} = 0

sirf itna batata hai ki start ek turning point hai (

cos ϕ = \pm 1

). Tumhe abhi bhi

x_{0}

ka sign check karna hoga:

x_{0} > 0

ϕ = 0

milta hai, lekin

x_{0} < 0

ϕ = π

milta hai.

"Maine

f = 2

Hz seedha

cos (f t)

mein plug kar diya."

Wrong variable — argument angular frequency

ω = 2 π f

use karta hai,

f

nahi.

f

use karne se rotation

2 π

ke factor se undercount ho jaata hai, isliye motion bahut slow lagegi.

A = x_{0}^{2} + v_{0}^{2}

initial conditions se."

Units clash — tum length

^{2}

ko velocity

^{2}

se add nahi kar sakte. Velocity ko pehle

ω

se divide karna hoga:

A = x_{0}^{2} + v_{0}^{2} / ω^{2}

, taaki dono terms length squared hon.

"Solution hai

x = A cos (ω t)

, bas — ek equation, ek answer."

Ek constant miss hai —

\overset{x}{¨} = - ω^{2} x

second-order hai aur do arbitrary constants maangta hai.

ϕ

drop karne se yeh information kho jaati hai ki tum cycle mein kahan se shuru hue the.

"Max acceleration aur max speed ek hi jagah hote hain."

Nahi —

a_{m a x} = A ω^{2}

ends par hota hai (

x = \pm A

, sabse strong pull), jabki

v_{m a x} = A ω

centre par hota hai (

x = 0

, zero pull). Woh opposite locations par hain.

"Kyunki

tan ϕ = - v_{0} / (ω x_{0})

hai, bas

arctan

le lo aur ho gaya."

Itna kaafi nahi —

tan

har

18 0^{\circ}

par repeat karta hai, isliye akela

arctan

ϕ

aur

ϕ + π

mein fark nahi bata sakta. Tumhe

x_{0}

ka sign (

cos ϕ

fix karta hai) ya

v_{0}

(

sin ϕ

fix karta hai) use karke sahi quadrant choose karna hoga.

6 0^{\circ}

tak swing kiya gaya ek pendulum

x = A cos (ω t + ϕ)

ω = g / L

ke saath follow karta hai."

Sirf small angles par — yeh neat formula assume karta hai ki restoring force displacement mein linear hai, jo ek small-angle approximation hai.

6 0^{\circ}

par motion sahi SHM nahi rahi. Dekho Simple pendulum.

Why questions

SHM solution zaroor ek sinusoid kyun honi chahiye, na ki, maano, ek parabola?

Kyunki law

\overset{x}{¨} = - ω^{2} x

kehta hai ki doosra derivative function ki ek flipped, scaled copy hai — aur cosine/sine hi akele aisi functions hain jisme yeh self-referential property hai; ek parabola ka doosra derivative ek constant hota hai, khud se proportional nahi.

Amplitude compute karte waqt hum

v_{0}

ω

se kyun divide karte hain?

Ek speed ko length mein convert karne ke liye taaki ise position se compare kiya ja sake: kyunki

v_{m a x} = A ω

, kisi bhi velocity ko

ω

se divide karne se equivalent displacement (metres) milta hai, tumhe ise ek square root ke neeche add karne deta hai.

\overset{x}{¨} = - ω^{2} x

mein minus sign itna important kyun hai?

Minus sign hi acceleration ko ghar ki taraf point karta hai; plus sign ke saath (

\overset{x}{¨} = + ω^{2} x

) solution growing exponentials hote jo infinity ki taraf ud jaate — koi oscillation hi nahi hoti.

Ek SHM system ko exactly do initial conditions kyun chahiye?

Kyunki equation mein ek second derivative hai, ise do baar integrate karne se do unknown constants aate hain; tum unhe do facts se pin karte ho — kahan shuru kiya (

x_{0}

) aur kitni tez (

v_{0}

) —

A

aur

ϕ

ke roop mein encode karke.

Max speed centre par kyun hai jab wahan force zero hai?

Particle ne poori inward journey restoring force se accelerate hoke ki hai; jab tak force centre par zero tak girta hai, saari woh pushing maximum speed mein accumulate ho chuki hai — force zero, lekin velocity peak.

Hum SHM ko kisi cheez ki circle mein ghoomne ki shadow kyun imagine kar sakte hain?

Steady rate

ω

par rotate karte circle par ek point ka horizontal projection

A cos (ω t + ϕ)

hota hai — literally humara solution — isliye SHM hai hi uniform circular motion jo edge-on dekha gaya. Dekho Reference circle and SHM as projection.

Restoring force ka displacement ke proportional hona (sirf oppose karna nahi) ek constant period kyun guarantee karta hai?

Sirf strict proportionality hi acceleration

= - ω^{2} x

ko ek constant

ω

ke saath banati hai; woh ek

ω

ek fixed period set karta hai chahe tum kitna bhi pull karo, isliye SHM isochronous hai.

Edge cases

Agar tum mass ko exactly

x = 0

par zero velocity ke saath rakh do toh motion kya hogi?

Woh hamesha ke liye wahan raha —

x_{0} = 0

aur

v_{0} = 0

ke saath amplitude

A = 0

hai, isliye koi oscillation nahi; equilibrium ek valid (trivial) solution hai.

Period kya hogi jab spring constant

k \to 0

ω = k / m \to 0

, isliye

T = 2 π / ω \to \infty

— ek infinitely floppy spring infinitely lamba period deta hai, yani koi return force nahi, koi oscillation nahi.

Agar tum maximum displacement par sabse bada possible outward push ke saath shuru karo, toh kya hoga?

Tum nahi kar sakte —

x = A

by definition woh turning point hai jahan speed zero hoti hai; wahan koi bhi outward velocity tumhe

A

se aage le jaayegi, jo contradict karta hai ki

A

maximum hai. Amplitude simply badh jaati hai jitni bhi energy tum inject karo ushe accommodate karne ke liye.

Turning points

x = \pm A

par velocity aur acceleration kya hain?

Velocity zero hai (momentarily at rest) jabki acceleration apne maximum magnitude

A ω^{2}

par hai jo centre ki taraf point kar rahi hai — standstill ka instant sabse strong pull ka instant hai.

Ideal (undamped) SHM ke liye solution

t \to \infty

hone par kya predict karta hai?

Yeh unchanged amplitude

A

ke saath hamesha ke liye oscillate karta rehta hai — koi energy lost nahi hoti. Real systems friction ki wajah se time ke saath shrink hoti hain; uske liye ek extra term chahiye, jo Damped oscillations deta hai.

Agar restoring force sirf stretch se aaye (ek rubber band jo compress hone par slack ho jaata hai), toh kya motion abhi bhi SHM hai?

Nahi — ek band jo sirf stretch hone par pull karta hai woh ek single linear

- k x

force nahi deta sab

x

values ke across, isliye

\overset{x}{¨} = - ω^{2} x

slack side par fail karta hai aur motion pure SHM nahi hai. Compare karo Hooke's Law and the spring force se, jo ek two-sided linear spring assume karta hai.