Question

A \[1 \times {10^{ - 20}}kg\] particle is vibrating with simple harmonic motion with a period of $1 \times {10^{ - 5}}\sec $ and a maximum speed of $1 \times {10^3}m{s^{ - 1}}$.The maximum displacement of the particle from mean position is
A. 1.59 mm
B. 1.00 m
C. 10 m
D. 3.18 mm

Answer 1

Hint:In this, first we have to write an expression for displacement in case of simple harmonic motion(SHM) following which we have to take its derivative. Take into account the magnitude for maximum displacement. Then substitute the values accordingly as per given in question.

Complete step by step answer:
We know that displacement $y = A\sin wt$.
Taking its derivative, we get $y' = Aw\cos wt$.
Here y’ represents velocity
${V_{\max }} = Aw$
So the maximum displacement will be
$A = \dfrac{{{V_{\max }}}}{w}$-------- (1)
Now to find value of w, we know
$w = \dfrac{{2\pi }}{T}$
Substitute the given value of T in above equation i.e. $T = 1 \times {10^{ - 5}}s$
and solve to get $w = 6.28 \times {10^{^5}}$--------(2)
Given $V = {10^3}m{s^{ - 1}}$------- (3)
Put value of equation (2) and (3) in (1) to get
$A = \dfrac{V_{\max}}{w} \\
\Rightarrow A= \dfrac{{{{10}^3}}}{{6.28 \times {{10}^5}}} \\
\Rightarrow A= 0.159 \times {10^{ - 2}}m \\
\therefore A= 1.59\,mm$
The maximum displacement of the particle from mean position is 1.59 mm.

So option A is correct.

Note:Here we have taken the displacement of the particle in terms of sine angle, it does not mean that we cannot take it in terms of cos angle. If the displacement is taken in terms of sine then it means the particle displacement is zero at initial time.