WHY ask this? Intuitively gravity adds a constant downward pull, so you'd guess slower or faster oscillation. Let's derive and forecast-then-verify.
Hang the mass. Two forces: spring +k direction and gravity mg down. Let y point downward, measured from the natural length.
At equilibrium the mass sits at extension y0 where forces balance:
ky0=mg⇒y0=kmg
This y0 is the static extension — purely a shift of the rest point.
Now displace to general y. Net downward force:
F=mg−ky
Newton: my¨=mg−ky. Substitute the new coordinate x=y−y0 (displacement from the new equilibrium), so y¨=x¨:
mx¨=mg−k(x+y0)=mg−kx−ky0==0(mg−ky0)−kx⇒mx¨=−kx
Bonus trick: since y0=mg/k, we get k=mg/y0, so
T=2πkm=2πgy0
Measure the static stretch y0 → get the period without knowing k or m!
HOW to derive ω from energy: differentiate E w.r.t. time (energy constant so dE/dt=0):
dtdE=mvv˙+kxv=v(mx¨+kx)=0
Since v=0 generally, mx¨+kx=0 — same SHM equation. ✓ Two roads, one answer.
Imagine a toy car tied to a stretchy rubber band on a smooth table. Pull it and let go: the band yanks it back, it zooms past the middle, stretches the other way, gets yanked back again — back and forth forever (if nothing slows it). The stiffer the band, the quicker the wiggle; the heavier the car, the lazier the wiggle. Now hang the same band from the ceiling with a weight: gravity just drops the weight to a new resting spot, but if you bounce it, it bounces at the exact same speed as on the table — gravity only moved the "home base," it didn't change the bounce.
Dekho, spring-mass system SHM ka sabse simple example hai. Jab tum mass ko kheencho, spring usko wapas pull karti hai — yeh hai restoring force, aur Hooke's law kehta hai F=−kx. Yani jitna zyada displace karoge, utna zyada force wapas kheechega. Newton ka law lagao: mx¨=−kx, isse seedha mil jaata hai ω=k/m aur T=2πm/k. Yaad rakho: stiff spring (bada k) = fast wiggle, heavy mass (bada m) = slow wiggle.
Ab vertical case ka twist samjho. Spring ko ceiling se latkao, gravity mass ko thoda neeche kheench degi — yeh static stretch hai y0=mg/k. Bahut log sochte hain ki gravity period change kar degi, par aisa nahi hota! Jab tum displacement equilibrium se naapte ho (natural length se nahi), to gravity ka term cancel ho jaata hai, kyunki ky0=mg. Bachi hui equation wahi mx¨=−kx hai. Matlab period BILKUL same: T=2πm/k. Gravity sirf "home base" shift karti hai, beat nahi badalti.
Ek shaandar shortcut: kyunki k=mg/y0, to T=2πy0/g. Iska matlab agar tum sirf static stretch y0 measure kar lo, to bina m ya k jaane period nikal sakte ho. Exam mein yeh trick bahut kaam aata hai.
Energy angle bhi samajh lo: total energy 21kA2 constant rehti hai, equilibrium par puri KE (max speed vmax=Aω), aur extreme points par puri PE. Yeh do raaste — Newton aur energy — dono same ω dete hain, to confidence high rahega.
