Simple harmonic motion — definition, restoring force F = −kx: Question bank

1.6.1 · D5 · Physics › Oscillations & Waves › Simple harmonic motion — definition, restoring force F = −kx

Yaad rakhne wali cheezein: $F = - k x$ (force displacement ke opposite hoti hai), $a = - ω^{2} x$ , $ω = k / m$ , $T = 2 π m / k$ , $v = ω A^{2} - x^{2}$ , $E = \frac{1}{2} k A^{2}$ .

True ya false — justify karo

Neeche har claim ya thoda subtle tarike se sahi hai ya thoda subtle tarike se galat. Reason, verdict se zyada important hai.

Saari SHM oscillation hai, aur saari oscillation SHM hai.

False. SHM ek special oscillation hai jisme

F \propto - x

zaroori hai; ek bouncing ball ya large-angle pendulum oscillate karta hai lekin uski force

x

mein linear nahi hai, isliye constant period nahi hota.

Amplitude double karne se period double ho jaata hai.

False.

T = 2 π m / k

mein amplitude hai hi nahi — zyada khichaav se badi force bhi milti hai aur bada top speed bhi, jo exactly cancel ho jaate hain (isochronism). Dekho Energy in SHM.

Amplitude double karne se maximum speed double ho jaati hai.

True.

v_{m a x} = A ω

linearly

A

ke saath badhta hai;

ω

k / m

se fix hai, usme do guna amplitude se do guna peak speed milti hai.

Amplitude double karne se total energy chaar guna ho jaati hai.

True.

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

, isliye

A \to 2 A

karne se energy

4

guna ho jaati hai.

F = - k x

mein minus sign ka matlab hai ki force ek negative number hai.

False. Ye ek direction rule hai: force displacement ke opposite direction mein point karti hai.

x < 0

par,

F = - k (neg) > 0

, yaani positive, phir bhi ghar ki taraf point karti hai.

Turning points

x = \pm A

par acceleration zero hota hai kyunki mass momentarily rest mein hota hai.

False. Wahan velocity zero hoti hai, lekin

a = - ω^{2} x

maximum hota hai

x = \pm A

par — wahi peak force exactly motion ko reverse karti hai.

x = 0

par mass ko zero force hai lekin ye "kuch nahi ho raha" wale sense mein equilibrium nahi hai.

True-ish.

F = 0

x = 0

isliye koi instantaneous push nahi, phir bhi mass top speed se guzarta hai — rukta nahi kyunki usme momentum hai.

Usi spring par heavier mass hamesha slower oscillate karega.

True.

T = 2 π m / k

m

ke saath badhta hai; zyada inertia same restoring force ke saath zyada sluggishly respond karti hai.

Usi mass par stiffer spring (bada

k

) faster oscillate karega.

True.

ω = k / m

k

ke saath badhta hai; same displacement ke liye stronger pull-back ka matlab hai quicker cycles.

Agar aapko

ω

pata hai, toh aap period jaante ho chahe amplitude ya phase kuch bhi ho.

True.

T = 2 π / ω

sirf

ω = k / m

par depend karta hai; amplitude

A

aur phase

ϕ

set karte hain kitna bada aur kab, kabhi kitna lamba nahi.

Error dhundho

Har snippet mein ek galat step hai. Use naam do aur theek karo.

F = - k x^{3}

SHM hai kyunki force phir bhi displacement ko oppose karti hai."

Error: SHM ko force proportional (linear)

x

mein chahiye.

F \propto - x^{3}

displacement oppose karta hai aur oscillate karta hai, lekin

T

amplitude par depend karta hai, isliye ye SHM nahi hai.

F = ma

aur

F = - k x

se seedha

a = - k x

milta hai."

Error: mass drop kar diya.

ma = - k x \Rightarrow a = - (k / m) x = - ω^{2} x

;

m

bhoolne se wrong units aur wrong

ω

milta hai.

U = \frac{1}{2} k x^{2}

isliye potential energy

x = 0

par sabse badi hoti hai."

Error:

\frac{1}{2} k x^{2}

sabse chhota (zero)

x = 0

par hota hai aur sabse bada

x = \pm A

par. Equilibrium energy valley ka bottom hai, top nahi.

"Speed

x = \pm A

par greatest hoti hai kyunki wahan displacement sabse zyada hai."

Error:

v = ω A^{2} - x^{2}

x = \pm A

par zero ho jaata hai aur

x = 0

par peak karta hai. Bada displacement matlab bada potential, badi speed nahi.

ω = k / m

se, bada amplitude bada

ω

deta hai."

Error:

ω

mein sirf

k

aur

m

hain, koi

A

nahi. Amplitude ka angular frequency par koi asar nahi hota.

x = A cos (ω t + ϕ)

SHM nahi ho sakta kyunki ye

A

se start hota hai,

0

se nahi."

Error: starting point

ϕ

se set hota hai, SHM hai ya nahi isse nahi. Koi bhi

ϕ

phir bhi

a = - ω^{2} x

satisfy karta hai; cosine aur sine same motion hai bas phase mein shift hai.

F = - k x

(vertical spring) mein constant gravitational pull add karna SHM destroy kar deta hai."

Error: gravity sirf equilibrium ko ek naye point

x_{0} = m g / k

par shift karti hai. Us naye centre ke baare mein

F = - k (x - x_{0})

— phir bhi linear, phir bhi SHM, same

ω

Why questions

Mechanism explain karo, sirf rule restate mat karo.

Kyon same motion (ek sine wave) springs, pendulums, aur crystal mein atoms ke liye dikhti hai?

Kisi bhi stable equilibrium ke paas potential ek parabola hota hai

U \approx \frac{1}{2} k x^{2}

, isliye

F = - d U / d x = - k x

sab ke liye — shared linear restoring force shared motion force karti hai. Dekho Taylor Series.

SHM solution sine ya cosine hi kyun hona chahiye, kuch aur kyun nahi?

Equation

a = - ω^{2} x

poochh rahi hai "kaun sa function do derivatives ke baad minus-itself return karta hai?" — sirf

sin

aur

cos

(aur unke combinations) karte hain, isliye ye forced hai, guessed nahi.

Energy conservation hume time solve kiye bina speed kyun dhundhne deta hai?

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

v

ko directly position

x

se link karta hai; rearrange karne par

v = ω A^{2} - x^{2}

milta hai bina kisi clock ke. Poora derivation Energy in SHM mein.

Bada initial pull oscillation ko jaldi "khatam" kyun nahi karta?

SHM frictionless hai; energy

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

conserved hai, isliye amplitude kabhi decay nahi karta. Sirf added dissipation (ek

- b v

term) ise shrink karta hai — dekho Damped Oscillations.

x = 0

fastest point kyun hai jabki wahan kuch push nahi ho raha?

Spring ki saari potential energy

x = 0

par kinetic energy mein convert ho jaati hai; zero force ka matlab us instant mein speed ka koi change nahi, lekin mass pehle se apni maximum speed carry kar raha hota hai.

Uniform circular motion ka projection SHM kyun trace karta hai?

Ek point jo circle mein constant

ω

se ghoom raha hai, uski ek axis par chhaya

x = A cos (ω t + ϕ)

follow karti hai — same equation SHM ki. Dekho Uniform Circular Motion.

Hum

F = - k x

mein

k > 0

kyun insist karte hain?

k > 0

ke saath force back khichti hai equilibrium ki taraf (stable, oscillates). Agar

k < 0

hota to force door push karta — point unstable hota aur object bhaag jaata, koi oscillation nahi.

Edge cases

Boundary aur degenerate situations jo standard formulas quietly assume karte hain.

Agar amplitude

A = 0

ho toh kya hoga?

Mass hamesha ke liye equilibrium par baith jaata hai:

x (t) = 0

v = 0

F = 0

E = 0

. Ye trivial (degenerate) solution hai — technically zero motion wali SHM.

Agar

k \to 0

ho (spring floppy ho jaaye) toh?

ω = k / m \to 0

aur

T \to \infty

: koi restoring force nahi, isliye koi oscillation nahi — mass bas freely drift karta hai. SHM boundary par break down ho jaata hai.

Agar

m \to 0

ho toh?

ω = k / m \to \infty

aur

T \to 0

: infinitely light mass infinitely fast oscillate karta — physically ek signal ki inertia hi timescale set karti hai.

Ek real spring ko bahut door kheenchne par

F = - k x

(aur isliye SHM) kyun fail ho sakta hai?

Real springs sirf ek elastic range ke andar Hooke's Law

F = - k x

follow karte hain; bahut zyada khichaav se force linear nahi rehti (stiffen ya permanently deform ho jaati hai), isliye motion SHM nahi rehti. Dekho Hooke's Law.

Pendulum ke liye "SHM" description quietly kab break hoti hai?

Sirf small angles par hi

F \approx - (m g / L) x

(linear) hota hai. Bade swings par true force

\propto sin θ

nonlinear hai, period amplitude ke saath badhta hai, aur ye SHM rehna band ho jaata hai. Dekho Simple Pendulum.

Agar mass exactly rest mein equilibrium par start kare (

x = 0

v = 0

), toh baad mein kya motion hogi?

Kuch nahi hilta: zero displacement se zero force aur zero speed, isliye

E = 0

aur ye wahin reh jaata hai. Motion ke liye ya toh initial displacement chahiye ya initial velocity.

Kya restoring force displacement mein linear hai, $F = - k x$ with $k > 0$ ? ::: Agar haan to ye SHM hai (constant period, koi amplitude dependence nahi); agar force nonlinear hai (jaise $- k x^{3}$ ya large-angle $sin θ$ ) to ye oscillate kar sakta hai lekin SHM nahi hai.

Connections

Parent: SHM definition
Energy in SHM — speed aur amplitude traps ke peeche energy reasoning
Hooke's Law — elastic-limit edge case
Simple Pendulum — jahan small-angle approximation khatam hoti hai
Damped Oscillations — kyun real amplitudes decay karte hain
Uniform Circular Motion — projection view
Taylor Series — kyun har valley bottom ek parabola hai

Question bank — Simple harmonic motion — definition, restoring force F = −kx

True ya false — justify karo

Error dhundho

Why questions

Edge cases

Quick Recall

Connections