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Standard XII

Physics

Relating Circular Motion to SHM

A particle mo...

Question

A particle moving along x axis has acceleration f, at t time. Given
f=F(1-t/T) where F and T are constants. The particle at t=0 has zero velocity. When f=0 the possible velocity is

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Solution

$d e a r s t u d e n t g i v e n t h a t f = F (1 - \frac{t}{T}) a n d a = \frac{d v}{d t} = f \frac{d v}{d t} = F - \frac{F t}{T} o n i n t e g r a t i n g v = F t - \frac{F t^{2}}{2 T} + c g i v e n t h a t a t t = 0, v = 0 t h i s g i v e s c = 0 s o, v = F t - \frac{F t^{2}}{2 T} n o w a t f = 0, f = F (1 - \frac{t}{T}) g i v e s t = T t h e n f r o m v = F t - \frac{F t^{2}}{2 T}; a n d t = T w e g e t v = F T / 2 R e g a r d s$

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A particle moving along x axis has acceleration f, at t time. Given f=F(1-t/T) where F and T are constants. The particle at t=0 has zero velocity. When f=0 the possible velocity is

dear studentgiven thatf=F(1-tT)and a=dvdt=fdvdt=F-FtTon integratingv=Ft-Ft22T+cgiven that at t=0, v=0this gives c=0so,v=Ft-Ft22Tnow at f=0,f=F(1-tT) gives t=Tthen fromv=Ft-Ft22T; and t=Twe get v=FT/2Regards

A particle moving along x axis has acceleration f, at t time. Given
f=F(1-t/T) where F and T are constants. The particle at t=0 has zero velocity. When f=0 the possible velocity is