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  •  According to the giving kinematic relation of the particle as below -  <img alt="\mathrm{V(\theta)=\frac{e^{i \theta}-e^{-i \theta}}{2 i}}" src="https://entrancecorner.oncodecogs.co

 According to the giving kinematic relation of the particle as below -  \mathrm{V(\theta)=\frac{e^{i \theta}-e^{-i \theta}}{2 i}}   where  i  is  complex number. Acceleration is proportional to-
 

Option: 1

\cos \theta


Option: 2

\sin \theta


Option: 3

e^{i \theta}


Option: 4

i \sin \theta


Answers (1)

best_answer

Velocity of the particle is given by-
   \mathrm{ V(\theta)=\frac{e^{i \theta}-e^{-i \theta}}{2 i}}..................(A)
Now,
but we know that
\mathrm{ \begin{aligned} & e^{i \theta}=\cos \theta+i \sin \theta \\ & e^{-i \theta}=\cos \theta-i \sin \theta \end{aligned}}
\mathrm{ Add ()1) \&(2) \, \, \, \, \, \, } \\ { we \ get, } \\ { \, \, \, \, \, \sin \theta=\frac{e^{i \theta}-e^{-i \theta}}{2 i}}
Value of \mathrm{\sin \theta } is same as value of \mathrm{v(t) } in equation (A) -
\mathrm{ v(\theta)=\sin \theta}
Acceleration will become 
\mathrm{ \frac{d V(\theta)}{d t}=\frac{d}{d \theta}(\sin \theta) =-\cos \theta \frac{d v(\theta)}{d \theta} \propto \cos \theta}

 

Posted by

Suraj Bhandari

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