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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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