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Give examples of a one-dimensional motion where

a) the particle moving along positive x-direction comes to rest periodically and moves forward

b) the particle moving along positive x-direction comes to rest periodically and moves backward

Answers (1)

(a) Let us consider a motion where

x(t) = \omega t - \sin \; \omega t

Thus, v=\frac{dx}{dt}

              = \omega - \omega\; \cos \; \omega t

& a=\frac{dv}{dt}

        =\omega ^{2}\; \sin \; \omega t

      x(t) = 0, v=0 \; and \; a=0; at \; \omega t = 0

      x(t) = \pi >0, v= \omega -\omega \; \cos \pi = 2\; \omega >0 \; and\; a=0; at \; \omega t = \pi

     x(t) = 2\pi >0, v=0 \; and\; a=0; at\; \omega t = 2\pi .

(ii) Let us consider a function of motion where,

x(t) = -a \sin \; \omega t

x(t) = -a \; \sin 0 = 0; at\; t=0

x(t) = -a \sin \frac{2\pi }{T}.\frac{T}{4}= -a\sin \frac{\pi }{2}=-a;at\; t=\frac{T}{4}

x(t) = -a \sin \frac{2\pi }{T}.\frac{3T}{4}=-a\; \sin \frac{3\pi}{2}

           = -a \sin (\pi +\frac{\pi}{2}) = -a ( -\sin \frac{\pi}{2}) = a ; at\; t = \frac{3T}{4}

x(t) = -a \sin \frac{2\pi}{T}.\frac{T}{2} = -a \sin \pi = 0; at\; \; t = \frac{T}{2}

x(t) = -a \sin \frac{2\pi}{T}.T = -a\; \sin 2\pi = +0; at\; \; t = T

Thus the displacement of the particle is in negative direction and it comes to rest periodically.

Thus,

-a \sin \; \omega t is a periodic function.

V=v=\frac{dx(t)}{dt}

               =\frac{d}{dt}.(-a\; \sin\; \omega t)

               =-a\; \omega \cos\; \omega t

Now, v=a\omega \; \cos\; 0^{o}=-\omega a;at\; t=0

v = -\omega\; a \; \cos \frac{2\pi}{T}.\frac{T}{4}=-\omega\; a\; \cos\; \frac{\pi}{2}=0;at\; t=\frac{T}{4}

v = -\omega\; a \; \cos \frac{2\pi}{T}.\frac{T}{4}=-a\; \omega\; \cos\; \pi=+\omega\; a;at\; t=\frac{T}{2}

v = -\omega\; a \; \cos \frac{2\pi}{T}.\frac{3T}{4}=-\omega\;a \cos\; (\pi+\frac{\pi}{2})at\; t=\frac{3T}{4}

v = -a\; \omega\; \cos \frac{2\pi}{T}.T=-\omega\;a \cos\; 2\pi=-\omega\;a\; at\; t=T

 After zero displacement, velocity changes periodically.

Thus, x(t) = -a \sin \omega\; t is the function required.

(i) Now, let us consider a function

x(t) = a \sin \omega\: t.

x(0) = 0

x(\frac{T}{4}) = a\; \sin\; \frac{2\pi}{T}.\frac{T}{4}=a\; \sin\frac{\pi}{2}=a

x(\frac{T}{2}) = a\; \sin\; \frac{2\pi}{T}.\frac{T}{2}=a\; \sin\; \pi=a

x(\frac{3T}{2}) = a\; \sin\; \frac{2\pi}{T}.\frac{3T}{2}=a\; \sin\;\left ( \pi+\frac{\pi}{2} \right )=-a

x(T) = a\; \sin\; \frac{2\pi}{T}.T=a\; \sin\;2\pi=0

Thus, the particle moves in a positive direction, periodically with zero displacements.

Hence x(t) = a \sin \omega t is the required function.

 

Posted by

infoexpert22

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