For particle undergoing SHM,
V=ωA2−x2⇒V2=ω2(A2−x2)
V1=ωA2−x12⇒V12=ω2(A2−x12) ...(i)
V2=ωA2−x22⇒V22=ω2(A2−x22) ...(ii)
V12−V22=ω2(x22−x12)
ω=x22−x12V12−V22
T=2πV12−V22x22−x12
NEET UG 2015 — Physics Waves & Oscillations
A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are V1 and V2 respectively. Its time period is:
Held on 30 Apr 2015 · Verified 9 Jul 2026.
2πV12+V22x12+x22
2πV12−V22x22−x12
2πx12+x22V12+V22
2πx12−x22V12−V22
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