The displacement of a particle in simple harmonic motion in one time period is:
A. A
B. 2A
C. 4A
D. Zero
Answer
302.1k+ views
Hint:In this question we are given a particle which is in simple harmonic motion and we are asked to find the total displacement of the same particle from the given four options, we will use the formula for the initial position of the particle and then will find the position of the same particle after one time period and then will find the net displacement.
Formula used:
The formula for initial position of the particle in a simple harmonic motion is given as,
\[y = A\sin \left( {\omega t + \phi } \right)\]
Here in the equation,
\[\left( {\omega t + \phi } \right)\] is the phase of the motion, \[\phi \] is the initial phase of the motion of particles and \[A\] is the amplitude.
Complete step by step solution:
We can write this equation simple harmonic motion as
\[y = A\sin \left( {\dfrac{{2\pi }}{T}t + \phi } \right)\]
Since, \[\omega = \dfrac{{2\pi }}{T}\]..... (the time period of a particle is inversely proportional to frequency)
Now, let us consider the position of the same particle which is in simple harmonic motion after one time period, we have
\[{y_T} = A\sin \left( {\dfrac{{2\pi }}{T}\left( {t + T} \right) + \phi } \right)\]
Further solving this equation, we have
\[{y_T} = A\sin \left( {\dfrac{{2\pi }}{T}t + 2\pi + \phi } \right)\]
Now, the net displacement of the particle in simple harmonic motion in one time period will be,
\[d = {y_t} - y\\ \Rightarrow d = A\sin \left( {\dfrac{{2\pi }}{T}t + \phi } \right)\, - A\sin \left( {\dfrac{{2\pi }}{T}t + \phi } \right)\,\\ \therefore d = 0\]
Hence we can say that the net displacement of the particle is zero.
Therefore, option D is correct.
Note:In one time period the particle in harmonic motion comes to the same point from where it started, in this case, the initial position of the particle becomes the same as the final position so the displacement of the particle remains zero. The displacement of a particle or wave is measured from the equilibrium position of the particle.
Formula used:
The formula for initial position of the particle in a simple harmonic motion is given as,
\[y = A\sin \left( {\omega t + \phi } \right)\]
Here in the equation,
\[\left( {\omega t + \phi } \right)\] is the phase of the motion, \[\phi \] is the initial phase of the motion of particles and \[A\] is the amplitude.
Complete step by step solution:
We can write this equation simple harmonic motion as
\[y = A\sin \left( {\dfrac{{2\pi }}{T}t + \phi } \right)\]
Since, \[\omega = \dfrac{{2\pi }}{T}\]..... (the time period of a particle is inversely proportional to frequency)
Now, let us consider the position of the same particle which is in simple harmonic motion after one time period, we have
\[{y_T} = A\sin \left( {\dfrac{{2\pi }}{T}\left( {t + T} \right) + \phi } \right)\]
Further solving this equation, we have
\[{y_T} = A\sin \left( {\dfrac{{2\pi }}{T}t + 2\pi + \phi } \right)\]
Now, the net displacement of the particle in simple harmonic motion in one time period will be,
\[d = {y_t} - y\\ \Rightarrow d = A\sin \left( {\dfrac{{2\pi }}{T}t + \phi } \right)\, - A\sin \left( {\dfrac{{2\pi }}{T}t + \phi } \right)\,\\ \therefore d = 0\]
Hence we can say that the net displacement of the particle is zero.
Therefore, option D is correct.
Note:In one time period the particle in harmonic motion comes to the same point from where it started, in this case, the initial position of the particle becomes the same as the final position so the displacement of the particle remains zero. The displacement of a particle or wave is measured from the equilibrium position of the particle.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Understanding Uniform Acceleration in Physics

What Are Current and Potential Difference in Electricity?

Understanding Collisions: Types and Examples for Students

Understanding Average and RMS Value in Electrical Circuits

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding Inertial and Non-Inertial Frames of Reference

CBSE Notes Class 11 Physics Chapter 9 - Mechanical Properties of Fluids - 2026-27 Free PDF Download (Sign-in Required)

CBSE Notes Class 11 Physics Chapter 14 - Waves - 2026-27 Free PDF Download (Sign-in Required)

