
A particle is executing S.H.M. The phase difference between velocity and displacement is then
(A) $0$
(B) $\dfrac{\pi }{2}$
(C) $\pi $
(D) $2\pi $
Answer
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Hint The motion of a particle is said to be periodic if it completes an equal number of oscillations in equal intervals of time. One of the simplest forms of such motion is called the simple harmonic motion. Here it is said that the particle is executing simple harmonic motion. We have to find the phase difference between the velocity and displacement of the particle.
Complete Step by step solution
For a particle executing simple harmonic motion, the displacement of the particle can be written as,
$x = A\cos \omega t$
Where $x$ stands for the displacement of the particle, $A$ represents the amplitude of the particle, $\omega t$ represents the phase.
We know that the velocity of a particle is the rate of change of displacement with respect to time.
Hence we can write,
$v = \dfrac{{dx}}{{dt}}$
$ \Rightarrow v = - A\omega \sin \omega t$
This can be written as, $ - A\omega \sin \omega t = A\omega \cos \left[ {\omega t + \dfrac{\pi }{2}} \right]$
The phase of displacement is, ${\phi _1} = \omega t$
The phase of velocity can be written as, ${\phi _2} = \omega t + \dfrac{\pi }{2}$
The phase difference can be written as,
${\phi _2} - {\phi _1} = \omega t + \dfrac{\pi }{2} - \omega t = \dfrac{\pi }{2}$
Therefore, the phase difference between displacement and velocity is $\dfrac{\pi }{2}$.
The answer is: Option (B): $\dfrac{\pi }{2}$
Additional Information
The oscillation of a mass suspended on a string, simple pendulum, etc. are examples of simple harmonic motions. The total energy of simple harmonic oscillators will always be constant. At the mean position, the total energy will be equal to the kinetic energy of the particle, and at the extreme position, the total energy will be equal to the potential energy of the particle.
Note
The maximum value of displacement will be $A$when the phase $\omega t = {0^ \circ }$. The minimum value of displacement will be zero when then the phase $\omega t = {90^ \circ }$. The displacement of a particle executing simple harmonic motion can be expressed as a sine function or a cosine function depending on the situation.
Complete Step by step solution
For a particle executing simple harmonic motion, the displacement of the particle can be written as,
$x = A\cos \omega t$
Where $x$ stands for the displacement of the particle, $A$ represents the amplitude of the particle, $\omega t$ represents the phase.
We know that the velocity of a particle is the rate of change of displacement with respect to time.
Hence we can write,
$v = \dfrac{{dx}}{{dt}}$
$ \Rightarrow v = - A\omega \sin \omega t$
This can be written as, $ - A\omega \sin \omega t = A\omega \cos \left[ {\omega t + \dfrac{\pi }{2}} \right]$
The phase of displacement is, ${\phi _1} = \omega t$
The phase of velocity can be written as, ${\phi _2} = \omega t + \dfrac{\pi }{2}$
The phase difference can be written as,
${\phi _2} - {\phi _1} = \omega t + \dfrac{\pi }{2} - \omega t = \dfrac{\pi }{2}$
Therefore, the phase difference between displacement and velocity is $\dfrac{\pi }{2}$.
The answer is: Option (B): $\dfrac{\pi }{2}$
Additional Information
The oscillation of a mass suspended on a string, simple pendulum, etc. are examples of simple harmonic motions. The total energy of simple harmonic oscillators will always be constant. At the mean position, the total energy will be equal to the kinetic energy of the particle, and at the extreme position, the total energy will be equal to the potential energy of the particle.
Note
The maximum value of displacement will be $A$when the phase $\omega t = {0^ \circ }$. The minimum value of displacement will be zero when then the phase $\omega t = {90^ \circ }$. The displacement of a particle executing simple harmonic motion can be expressed as a sine function or a cosine function depending on the situation.
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