Question

Displacement-time equation of a particle executing SHM is: \[X = A\sin \left( {\omega t + \dfrac{\pi }{6}} \right)\]. Time taken by the particle to go directly from \[X = - \dfrac{A}{2}\] to \[X = + \dfrac{A}{2}\] is
A. \[\dfrac{\pi }{{3\omega }}\]
B. \[\dfrac{\pi }{{2\omega }}\]
C. \[\dfrac{{2\pi }}{\omega }\]
D. \[\dfrac{\pi }{\omega }\]

Answer 1

Hint: Calculate the time in the displacement equation of SHM for the given values of X. The time taken by the particle between these two points is the difference in the time values you get for both values of X. Convert the angle into radians for the correct answer.

Complete step by step answer:
We have given the displacement of the wave executing SHM,
\[X = A\sin \left( {\omega t + \dfrac{\pi }{6}} \right)\] …… (1)
Here, A is the amplitude of the wave, \[\omega \] is the angular frequency and t is the time.
We know that the term \[\dfrac{\pi }{6}\] represents the phase difference.
We have to determine the time for which the particle is at distance\[X = - \dfrac{A}{2}\]. Therefore, we substitute \[X = - \dfrac{A}{2}\] in the above equation.
\[ - \dfrac{A}{2} = A\sin \left( {\omega {t_1} + \dfrac{\pi }{6}} \right)\]
\[ \Rightarrow {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \omega {t_1} + \dfrac{\pi }{6}\]
\[ \Rightarrow - \dfrac{\pi }{6} = \omega {t_1} + \dfrac{\pi }{6}\]
\[ \Rightarrow {t_1} = - \dfrac{{2\pi }}{{6\omega }}\]
\[ \Rightarrow {t_1} = - \dfrac{\pi }{{3\omega }}\]
Now, we have to determine the time at distance\[X = + \dfrac{A}{2}\]. Therefore, we substitute \[X = + \dfrac{A}{2}\] in equation (1).
\[\dfrac{A}{2} = A\sin \left( {\omega {t_2} + \dfrac{\pi }{6}} \right)\]
\[ \Rightarrow {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \omega {t_2} + \dfrac{\pi }{6}\]
\[ \Rightarrow \dfrac{\pi }{6} = \omega {t_2} + \dfrac{\pi }{6}\]
\[ \Rightarrow {t_2} = 0\]
Therefore, we can calculate the time taken by the particle as follows,
\[t = {t_2} - {t_1}\]
We have to substitute \[{t_1} = \dfrac{\pi }{{3\omega }}\] and \[{t_2} = 0\] in the above equation.
\[t = 0 - \left( { - \dfrac{\pi }{{3\omega }}} \right)\]
\[ \Rightarrow t = \dfrac{\pi }{{3\omega }}\]

Therefore, the correct option is (A).

Additional information:
We should know the important terms used in the SHM wave equation.
Amplitude: it is the maximum displacement of the particle that is performing simple harmonic motion from its mean position.
Period: It is the time taken by the particle to complete one oscillation. It is given as,\[T = \dfrac{{2\pi }}{\omega }\], where, \[\omega \] is angular frequency of the wave.
Frequency: Frequency of the wave is the total number of oscillations performed by the particle in one second.

Note:Students should note that sine inverse of any negative value gives a negative angle. It should not be taken positively even for calculation purposes. To convert the angle from degrees into radians, multiply the angle by \[\dfrac{\pi }{{180}}\].