Question

Time period of a particle executing SHM is $8\,\sec .$At $t = 0$ it is at the mean position. The ration of the distance covered by the particle in the 1st second to the 2nd second is:
$A.\dfrac{1}{{\sqrt 2 }} + 1 \\\ B.\sqrt 2 \\\ C.\dfrac{1}{{\sqrt 2 }} \\\ D.\sqrt 2 + 1 \\\$

Answer 1

Concept of simple harmonic motion and the displacements from mean positions at different times like at t= 1 sec and t = 2 sec from equation of motion of SHM.
$x = A\,\sin \,\omega t$
And $\omega = \dfrac{{2\pi }}{T}$

Complete step by step answer:
Simple Harmonic Motion: A particle is in SHM if it moves to and fro about the mean position under the action of restoring force proportional to its displacement from mean position and is always directed towards the mean position.
i.e. $F \times x \Rightarrow F = - kx$
Equation of SHM
In SHM, $F = - kx$……..(i)
The negative sign shows force and $x$ are in opposite direction.
By Newton’s second law,
$F = ma = \dfrac{{m{d^2}x}}{{dt}}$…. (ii)
From (i) and (ii)
$\dfrac{{m{d^2}x}}{{d{t^2}}} = - kx \\\ \dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{{ - k}}{m}x \\\$
Let $\dfrac{k}{m} = {\omega ^2},{\text{ }}\,\omega \to$angular frequency
$So, \\\ \dfrac{{{d^2}x}}{{d{t^2}}} = - {\omega ^2}x \\\$
$\dfrac{{{d^2}x}}{{d{t^2}}} + {\omega ^2}x = 0$
Which is the differential of SHM
The solution of such equation is given by
$x = A\sin \omega t$ …. (iii)
Where x is the displacement from mean position at time t
A is amplitude
$\omega$ is angular frequency
And $\omega = \dfrac{{2\pi }}{T}$where T is time period of SHM
So, equation (iii) becomes
$x = A\,\sin \left( {\dfrac{{2\pi }}{T}} \right)t$
According to question, $T = 8\,\sec$
At
$t = 1\,\sec ,x(t = 1\,\sec ) = A\,\sin \,\left( {\dfrac{{2\pi }}{8}} \right)(1) \\\ x(t = 1\,\sec ) = A\,\sin \left( {\dfrac{\pi }{4}} \right) \\\ x(t = 1\,\sec ) = A \times \dfrac{1}{{\sqrt 2 }} \\\ at\,t = 2\,\sec ,\,x(t = 2\,\sec ) = A\,\sin \,\left( {\dfrac{{2\pi }}{8}} \right)(2) \\\ x(t - 2\,\sec ) = A\,\sin \,\left( {\dfrac{\pi }{4}} \right) \times 2 \\\ x(t - 2\,\sec ) = A\, \\\$
The required ratio is $= \dfrac{{{\text{Distance vocered in }}{{\text{1}}^{st}}\,{\text{second}}}}{{Dis\tan ce{\text{ covered in }}{{\text{2}}^{nd}}{\text{ second}}}} = \dfrac{{x(t = 1\,\sec )}}{{x(t = 2) - x(t = 1)}}$

$$= \dfrac{{A \times \dfrac{1}{{\sqrt 2 }}}}{{A - \dfrac{A}{{\sqrt 2 }}}} \\\ = \dfrac{{\dfrac{A}{{\sqrt 2 }}}}{{\dfrac{{A\sqrt 2 - A}}{{\sqrt 2 }}}} = \dfrac{A}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{A(\sqrt 2 - 1)}} \\\$$

Required ratio:
$= \dfrac{1}{{\sqrt 2 - 1}} \\\ = \dfrac{1}{{\sqrt 2 - 1}} \times \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 + 1}} \\\ = \dfrac{{\sqrt 2 + 1}}{{{{(\sqrt 2 )}^2} - {1^2}}} \\\ = \dfrac{{\sqrt 2 + 1}}{{2 - 1}} \\\$
Required ratio $= \sqrt {2 + 1}$

So, the correct answer is “Option D”.

Note:
Distance covered in 1 second $=$ distance covered in first two sec $-$ distance covered in first second.
$\Rightarrow x(t = 2\sec ) - x(t = 1\sec )$