Ncert Exercises & Solutions

The motion of a particle of mass $m$ is given by $x=0$ for $t<0 ~\text s$ , $x(t)=A~\text {sin}~4\pi t$ for $0<t<(1/4) \text s ~(A>0)$ , and $x=0$ for $t>(1/4) ~\text s$ . Then:

(a)	The force at $t=(1/8) ~\text s$ on the particle is $-16 \pi^2\text{Am}$
(b)	The particle is acted upon by an impulse of magnitude $4 \pi^2 \text{Am}$ at $t=0 ~\text s$ and $t=(1/4) ~\text s$
(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

Which of the following statement/s is/are true?

1.	(a, c, d, e)	2.	(a, c)
3.	(b, c, d)	4.	(a, b, d)

(a, b, d) Hint: Using the equation of position of the particle, we can find the acceleration the particle.

Given,

Step 1: Find the force acting on the particle.

$\begin{array}{l} x = 0 for t < 0 s \\ x (t) = Asin 4 πt; for 0 < t < \frac{1}{4} s \\ x = 0; for t > \frac{1}{4} s \\ For, 0 < t < \frac{1}{4} s v (t) = \frac{dx}{dt} = 4 πAcos 4 πt \end{array}$

$\begin{array}{l} a (t) = acceleration \\ = \frac{dv (t)}{dt} = - 16 π^{2} Asin 4 πt \\ At t = \frac{1}{8} s, a (t) = - 16 π^{2} Asin 4 π \times \frac{1}{8} = - 16 π^{2} A \\ F = ma (t) = - 16 π^{2} A \times m = - 16 π^{2} mA \\ = Change in linear momentum \\ = F \times t = (- 16 π^{2} Am) \times \frac{1}{4} \\ = - 4 π^{2} Am \end{array}$

Step 2: Find the impulse of the particle.

The impulse (Change in linear momentum)

$at t = 0 is same as, t = \frac{1}{4} s$
Clearly, toræ A which is not constant. Hence. force is also not constant.

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