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NEET 2015
Exam
NEET
Year
2015
Subject
Physics
Chapter
Oscillations

NEET 2015 · Physics · Oscillations

A particle is executing SHM along a straight line. Its velocities...

A particle is executing SHM along a straight line. Its velocities at distances x1{x}_{1} and x2{x}_{2} from the mean position are V1{V}_{1} and V2{V}_{2} respectively. Its time period is:

Options

  1. A2πx12+x22V12+V222\pi\sqrt{\dfrac{{x}_{1}^{2} + {x}_{2}^{2}}{{V}_{1}^{2} + {V}_{2}^{2}}}
  2. B2πx22−x12V12−V222\pi\sqrt{\dfrac{{x}_{2}^{2} - {x}_{1}^{2}}{{V}_{1}^{2} - {V}_{2}^{2}}}
  3. C2πV12+V22x12+x222\pi\sqrt{\dfrac{{V}_{1}^{2} + {V}_{2}^{2}}{{x}_{1}^{2} + {x}_{2}^{2}}}
  4. D2πV12−V22x12−x222\pi\sqrt{\dfrac{{V}_{1}^{2} - {V}_{2}^{2}}{{x}_{1}^{2} - {x}_{2}^{2}}}

Correct answer

B. 2πx22−x12V12−V222\pi\sqrt{\dfrac{{x}_{2}^{2} - {x}_{1}^{2}}{{V}_{1}^{2} - {V}_{2}^{2}}}

Step-by-step solution

For particle undergoing SHM,

V=ωA2−x2⇒V2=ω2(A2−x2)V = \omega\sqrt{{A}^{2} - {x}^{2}} ⇒ {V}^{2} = {\omega}^{2}\left({A}^{2} - {x}^{2}\right)

V1=ωA2−x12⇒V12=ω2(A2−x12){V}_{1} = \omega\sqrt{{A}^{2} - {x}_{1}^{2}} ⇒ {V}_{1}^{2} = {\omega}^{2}\left({A}^{2} - {x}_{1}^{2}\right) ...(i)

V2=ωA2−x22⇒V22=ω2(A2−x22){V}_{2} = \omega\sqrt{{A}^{2} - {x}_{2}^{2}} ⇒ {V}_{2}^{2} = {\omega}^{2}\left({A}^{2} - {x}_{2}^{2}\right) ...(ii)

V12−V22=ω2(x22−x12){V}_{1}^{2} - {V}_{2}^{2} = {\omega}^{2}\left({x}_{2}^{2} - {x}_{1}^{2}\right)

ω=V12−V22x22−x12\omega = \sqrt{\dfrac{{V}_{1}^{2} - {V}_{2}^{2}}{{x}_{2}^{2} - {x}_{1}^{2}}}

T=2πx22−x12V12−V22T = 2\pi\sqrt{\dfrac{{x}_{2}^{2} - {x}_{1}^{2}}{{V}_{1}^{2} - {V}_{2}^{2}}}
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