Time period of a particle executing SHM is $8\sec $. At $t = 0$, it is at the mean position. The ratio of distance covered by the particle in the $1^{st}$ second to the $2^{nd}$ second is:
A) $\dfrac{1}{{\sqrt 2 - 1}}$
B) $\sqrt 2 $
C) $\dfrac{1}{{\sqrt 2 }}$
D) $\sqrt 2 + 1$
Answer
249.9k+ views
Hint: Recall that the simple harmonic motion is defined as the type of motion in which a restoring force acts. This restoring force is directly proportional to the displacement that a body experiences from its mean position. The direction of the restoring force is towards the mean position. In the simple harmonic motion, the path of the particle will always be in a straight line.
Complete step by step solution:
The equation of the simple harmonic motion for a particle is written as
$x = A\sin \omega t$---(i)
Where $\omega $ is the angular frequency
And t is the time taken by the particle
Also it is known that $\omega = \dfrac{{2\pi }}{T}$
At $t = 0\sec $the particle is at its mean position.
Given $T = 8\sec $
When $t = 1\sec $
${x_1} = A\sin (\dfrac{{2\pi }}{T}t)$
$\Rightarrow {x_1} = A\sin \dfrac{{2\pi }}{8}$
$\Rightarrow {x_1} = A\sin \dfrac{\pi }{4}$
Substitute the value of $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
The position of the particle at the 1st second is
$ \Rightarrow {x_1} = \dfrac{A}{{\sqrt 2 }}$
Similarly when $t = 2\sec $
${x_2} = A\sin (\dfrac{{2\pi }}{T}t)$
$\Rightarrow {x_2} = A\sin (\dfrac{{2\pi }}{8} \times 2)$
$\Rightarrow {x_2} = A\sin \dfrac{\pi }{2}$
Substitute the value of $\sin \dfrac{\pi }{2} = 1$
The position of the particle at the 2nd second is
$ \Rightarrow {x_2} = A$
The ratio of the distance travelled by the particle is given by
$\dfrac{{{x_1}}}{{{x_2} - {x_1}}} = \dfrac{{\dfrac{A}{{\sqrt 2 }}}}{{A - \dfrac{A}{{\sqrt 2 }}}}$
$ \Rightarrow \dfrac{A}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 A - A}}$
$ \Rightarrow \dfrac{A}{{A(\sqrt 2 - 1)}}$
$ \Rightarrow \dfrac{1}{{\sqrt 2 - 1}}$
Option A is the right answer.
Note: It is important to note that the simple harmonic motion is a type of oscillatory motion in which the acceleration of the particle is directly proportional to the displacement. Also all simple harmonic motions are oscillatory but all oscillatory motions are not simple harmonic. This is because oscillatory motion repeats itself after a fixed period of time. The mean position from which the particle started moving is in a stable equilibrium. This means that at the mean position the net force acting on the system will be zero.
Complete step by step solution:
The equation of the simple harmonic motion for a particle is written as
$x = A\sin \omega t$---(i)
Where $\omega $ is the angular frequency
And t is the time taken by the particle
Also it is known that $\omega = \dfrac{{2\pi }}{T}$
At $t = 0\sec $the particle is at its mean position.
Given $T = 8\sec $
When $t = 1\sec $
${x_1} = A\sin (\dfrac{{2\pi }}{T}t)$
$\Rightarrow {x_1} = A\sin \dfrac{{2\pi }}{8}$
$\Rightarrow {x_1} = A\sin \dfrac{\pi }{4}$
Substitute the value of $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
The position of the particle at the 1st second is
$ \Rightarrow {x_1} = \dfrac{A}{{\sqrt 2 }}$
Similarly when $t = 2\sec $
${x_2} = A\sin (\dfrac{{2\pi }}{T}t)$
$\Rightarrow {x_2} = A\sin (\dfrac{{2\pi }}{8} \times 2)$
$\Rightarrow {x_2} = A\sin \dfrac{\pi }{2}$
Substitute the value of $\sin \dfrac{\pi }{2} = 1$
The position of the particle at the 2nd second is
$ \Rightarrow {x_2} = A$
The ratio of the distance travelled by the particle is given by
$\dfrac{{{x_1}}}{{{x_2} - {x_1}}} = \dfrac{{\dfrac{A}{{\sqrt 2 }}}}{{A - \dfrac{A}{{\sqrt 2 }}}}$
$ \Rightarrow \dfrac{A}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 A - A}}$
$ \Rightarrow \dfrac{A}{{A(\sqrt 2 - 1)}}$
$ \Rightarrow \dfrac{1}{{\sqrt 2 - 1}}$
Option A is the right answer.
Note: It is important to note that the simple harmonic motion is a type of oscillatory motion in which the acceleration of the particle is directly proportional to the displacement. Also all simple harmonic motions are oscillatory but all oscillatory motions are not simple harmonic. This is because oscillatory motion repeats itself after a fixed period of time. The mean position from which the particle started moving is in a stable equilibrium. This means that at the mean position the net force acting on the system will be zero.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

