Question

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

Answer 1

Hint: This problem can be solved by writing the equation for the displacement in SHM in terms of the amplitude, the angular frequency and the time. Then by plugging in the respective values given in the question, we can find out the displacement in the first second and the second and compare to get the required ratio.

Formula Used:
$x=A\sin \omega t$
$\omega =\dfrac{2\pi }{T}$

Complete step-by-step answer:
Let us write the displacement equation for a body performing simple harmonic motion (SHM).
The displacement $x$ from the main position of a body in time $t$, performing SHM with amplitude $A$ and angular frequency $\omega $ is given by
$x=A\sin \omega t$ --(1)
Now, the time period $T$ and the angular frequency $\omega $ of a body in SHM are related by
$\omega =\dfrac{2\pi }{T}$ --(2)
Hence, now let us analyze the question.
Let the amplitude of motion of the body be $A$ and the angular frequency be $\omega $.
Let the time period of motion of the body be $T=8s$.
Therefore, using (2) we get
$\omega =\dfrac{2\pi }{8}=\dfrac{\pi }{4}$ --(3)
Let the displacement of the body from the mean position in time $t$ be $x$.
Since, at $t=0$, the body is at the mean position, its displacement $x$ from the mean position can be written as
$x=A\sin \omega t$ --(4)
Using (3) in (4), we get
$x=A\sin \left( \dfrac{\pi }{4}t \right)$ --(5)
Now, let the displacement of the body at $t=1s$ be ${{x}_{1}}$.
Using (5), we get
${{x}_{1}}=A\sin \left( \dfrac{\pi }{4}1 \right)=A\sin \left( \dfrac{\pi }{4} \right)=A\left( \dfrac{1}{\sqrt{2}} \right)=\dfrac{A}{\sqrt{2}}$ --(6) $\left( \because \sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} \right)$
It is given that the displacement of the body at $t=0$ is ${{x}_{0}}=0$.
Therefore, the displacement of the body in the first second will be ${{x}_{1st}}={{x}_{1}}-{{x}_{0}}$
$\therefore {{x}_{1st}}=\dfrac{A}{\sqrt{2}}-0=\dfrac{A}{\sqrt{2}}$ --(7) [Using (6)]
Now, let the displacement of the body at $t=2s$ be ${{x}_{2}}$.
Using (5), we get
${{x}_{2}}=A\sin \left( \dfrac{\pi }{4}2 \right)=A\sin \left( \dfrac{\pi }{2} \right)=A\left( 1 \right)=A$ --(8) $\left( \because \sin \dfrac{\pi }{2}=1 \right)$
From (6), the displacement of the body at $t=1s$ is ${{x}_{1}}=\dfrac{A}{\sqrt{2}}$
Therefore, the displacement of the body in the second second will be ${{x}_{2nd}}={{x}_{2}}-{{x}_{1}}$
$\therefore {{x}_{2nd}}=A-\dfrac{A}{\sqrt{2}}=\dfrac{A\left( \sqrt{2}-1 \right)}{\sqrt{2}}$ --(9) [Using (8)]
Now, we get the required ratio by dividing (7) by (9)
$\dfrac{{{x}_{1st}}}{{{x}_{2nd}}}=\dfrac{\dfrac{A}{\sqrt{2}}}{\dfrac{A\left( \sqrt{2}-1 \right)}{\sqrt{2}}}=\dfrac{1}{\sqrt{2}-1}$ --(10)
Rationalizing (10), we get
$\dfrac{{{x}_{1st}}}{{{x}_{2nd}}}=\dfrac{\dfrac{A}{\sqrt{2}}}{\dfrac{A\left( \sqrt{2}-1 \right)}{\sqrt{2}}}=\dfrac{1}{\sqrt{2}-1}\times \dfrac{\sqrt{2}+1}{\sqrt{2}+1}=\dfrac{\sqrt{2}+1}{\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)}$
$\therefore \dfrac{{{x}_{1st}}}{{{x}_{2nd}}}=\dfrac{\sqrt{2}+1}{\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)}=\dfrac{\sqrt{2}+1}{{{\left( \sqrt{2} \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\dfrac{\sqrt{2}+1}{2-1}=\dfrac{\sqrt{2}+1}{1}=\sqrt{2}+1$ $\left( \because \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} \right)$
$\therefore \dfrac{{{x}_{1st}}}{{{x}_{2nd}}}=\sqrt{2}+1$
Hence, we have got the required ratio.
Therefore, the correct option is $D)\text{ }\sqrt{2}+1$.

Note: Students must note that we wrote the equation for the displacement in SHM in sine terms and not cosine terms, that is, as $x=A\sin \omega t$ and not $x=A\cos \omega t$ since according to the question, at time $t=0$, the body is at the mean position, that is, it has zero displacement. This condition is satisfied only if we write $x=A\sin \omega t$. If we had written $x=A\cos \omega t$, it would have implied that at time $t=0$, the body is at the amplitude or extreme position and then obviously, its displacement with respect to the mean position would not have been zero.