Question

The time period of a simple harmonic motion is $ 8s $ . At $ t=0 $ , it is at its equilibrium position. The ratio of distance transversed by it in the first and second seconds is:
 $ \begin{align}
  & \text{A}\text{. }\dfrac{1}{\sqrt{2}} \\
 & \text{B}\text{. }\dfrac{1}{\sqrt{2}-1} \\
 & \text{C}\text{. }\dfrac{1}{\sqrt{3}} \\
 & \text{D}\text{. }\dfrac{1}{2} \\
\end{align} $

Answer 1

Hint: In simple harmonic motion, the time period of oscillation is the time taken by particle to complete full oscillation. We will apply the general equation of a simple harmonic motion to calculate the displacement of particles in first and second seconds. For finding displacement of particle in $ {{2}^{nd}} $ second, we have to subtract the displacement of particle in first second from the total displacement of particle in two seconds.

Complete step-by-step answer:
Simple harmonic motion is a type of periodic motion in which restoring force on the moving particle is directly proportional to the particle’s displacement magnitude and acts towards the particle’s equilibrium position.

Equation of simple harmonic motion (SHM):
 $ y=A\sin \omega t $
Where,
 $ y $ is the displacement of the particle
 $ A $ is the amplitude of the particle
 $ \omega $ is the angular frequency of oscillation
 $ t $ is the time
Equilibrium position is the position at which no net force acts on the particle. Displacement is the distance of the particle from its equilibrium position. Amplitude is the maximum value of displacement of a particle on either side of equilibrium position.
Expression for finding displacement of particle from mean position in SHM motion is given as:
 $ Y=A\sin \omega t $
Also, time period of SHM is given as $ 8s $ ,
 $ \begin{align}
  & T=8s \\
 & \dfrac{2\pi }{\omega }=8s \\
\end{align} $
Therefore,
 $ \omega =\dfrac{2\pi }{8} $
Now, displacement of particle when time $ t=1\sec $ will be,
 $ {{Y}_{1}}=A\sin \omega {{t}_{1}} $
Put $ {{t}_{1}}=1\sec $ , we get,
 $ \begin{align}
  & {{Y}_{1}}=A\sin \omega \left( 1 \right) \\
 & {{Y}_{1}}=A\sin \omega \\
\end{align} $
Put $ \omega =\dfrac{2\pi }{8} $
We get,
 $ {{Y}_{1}}=A\sin \left( \dfrac{2\pi }{8} \right)=A\sin \left( \dfrac{\pi }{4} \right) $
We have,
 $ \sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}} $
Therefore,
 $ {{Y}_{1}}=\dfrac{A}{\sqrt{2}} $
Now, displacement of particle when time $ t=2\sec $ will be,
 $ {{Y}_{2}}=A\sin \omega {{t}_{2}} $
Put $ {{t}_{2}}=2\sec $ , we get,
 $ \begin{align}
  & {{Y}_{2}}=A\sin \omega \left( 2 \right) \\
 & {{Y}_{2}}=A\sin 2\omega \\
\end{align} $
Put $ \omega =\dfrac{2\pi }{8} $
We get,
 $ {{Y}_{2}}=A\sin \left[ 2\times \left( \dfrac{2\pi }{8} \right) \right]=A\sin \left( \dfrac{\pi }{2} \right) $
We have,
 $ \sin \left( \dfrac{\pi }{2} \right)=1 $
Therefore,
 $ {{Y}_{2}}=A $
 $ {{Y}_{2}} $ is the displacement of the particle in two seconds, but we have to calculated the displacement of particle in $ {{2}^{nd}} $ second only; therefore,
Displacement of particle in $ {{2}^{nd}} $ second is given by: Displacement of particle in two seconds minus Displacement of particle in one second,

 $ \begin{align}
  & {{Y}_{2'}}={{Y}_{2}}-{{Y}_{1}} \\
 & {{Y}_{2'}}=A-\dfrac{A}{\sqrt{2}}=A\left( 1-\dfrac{1}{\sqrt{2}} \right) \\
\end{align} $
Ratio of $ {{Y}_{1}} $ and $ {{Y}_{2'}} $ ,
 $ \dfrac{{{Y}_{1}}}{{{Y}_{2'}}}=\dfrac{\dfrac{A}{\sqrt{2}}}{A\left( 1-\dfrac{1}{\sqrt{2}} \right)}=\dfrac{1}{\sqrt{2}-1} $
The ratio of distance transversed by the particle in the first and second seconds is $ \dfrac{1}{\sqrt{2}-1} $
Hence, the correct option is B.

Note: Students should not get confused between the displacement of particles in $ n $ seconds, and the displacement of particles in $ {{n}^{th}} $ second. For calculating the displacement in $ {{n}^{th}} $ second, we have to subtract the displacement of the particle in $ n-1 $ seconds from the total displacement in $ n $ seconds.