
The displacement of a particle in simple harmonic motion in one time period is:
A. A
B. 2A
C. 4A
D. Zero
Answer
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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.
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