Ncert Exercises & Solutions

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

(a) The force at t = (1/8)s on the particle is $- 16 π^{2} A m$
(b) The particle is acted upon by an impulse of magnitude 4 $π$ Am 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

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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