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The motion of a particle of mass m is given by x = 0 for t < 0 s, x( t ) = A sin 4p t for 0 < t <(1/4) s (A > o), and x = 0 for t >(1/4) s. Which of the following statements is true?
(a) The force at t = (1/8) s on the particle is $-16\pi ^2 A m.$
(b) The particle is acted upon by on impulse of magnitude $4\pi ^2 A m.$ at t = 0 s and t = (1/4) s.
(c) The particle is not acted upon by any force.
(d) The particle is not acted upon by a constant force.
(e) There is no impulse acting on the particle.

Answers (1)

The correct answer is: -

(a) The force at $t =\frac{1}{8}$ s on the particle is $-16\pi ^2 A m.$

(b) The particle is acted upon by on impulse of magnitude $4 \pi ^2 A m.$ at t = 0s and $\left (t =\frac{1}{4} \right )s$

(d) The particle is not acted upon by a constant force.

Explanation: Here, mass = m

Thus, x(t) = 0, for t<0

$x(t) = A \sin4\pi t, for 0 < t < \left ( \frac{1}{4} \right ) s$

and

$x(t) =0, for t > \left ( \frac{1}{4} \right ) s$

Now,

For

$0< t < \left ( \frac{1}{4} \right ) s, x(t)=A\sin 4\pi t$

$v =\frac{ dx}{dt }= 4A\pi \cos4\pi t$

$a =\frac{ dv}{dt}= -16 \pi ^2A \sin 4\pi t$

$F(t)=ma(t)= -16 \pi ^2A m\sin 4\pi t$

Now, it’s clear that force is a function of time, hence opt (d) is verified.

(a) At $t=\left (\frac{1}{8} \right )s$

$a = -16\pi A\sin \frac{\pi}{2}$

$a(t) = -16\pi ^{2}A$

thus, force at $t =\frac{1}{8}$

$F = ma = m(-16\pi ^2A)$

Hence, $F = -16\pi ^{2}AmN.$

Hence, opt (a)

(b) Impulse = Change in momentum between the values t = 0s and $\left (t =\frac{1}{4} \right )s$

Now we already know from above that, F(t) varies from 0 at t = 0s to the max. value at $F (t)= -16\pi ^{2}AmN.$ at $\left (t =\frac{1}{4} \right )s$ & by the eq. $\overrightarrow{I}=\overrightarrow{F}t$ at $\left (t =\frac{1}{4} \right )s$ .

$|I|=16\pi^2Am\times\frac{1}{4}=4\pi^2Am$

Hence opt (b).

Posted by

infoexpert24

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