Question

A particle executing SHM while moving from the one extremity is found at distance${x_1}$, ${x_2}$ and ${x_3}$ from the center at the end of three successive seconds. The time period of oscillation is:
(A) $\dfrac{{2\pi }}{\theta }$
(B) $\dfrac{\pi }{\theta }$
(C) $\theta$
(D) $\dfrac{\pi }{{2\theta }}$
Where $\theta = {\cos ^{ - 1}}(\dfrac{{{x_1} + {x_3}}}{{2{x_2}}})$

Answer 1

Hint It is given that a particle is executing a Simple harmonic motion. Then its displacement is presented as $y = A\cos \omega t$. Now using this, find the displacement equations of all displacements and find $\theta$ value. Using it, find the time period of oscillation.

Complete Step By Step Solution
It is mentioned that a particle undergoes SHM from moving from one extremity to another. Now it is said to travel in three distances in three different time periods namely ${x_1}$in 1 seconds and distance ${x_2}$in time of 2 seconds and finally distance ${x_3}$ in time 3 seconds.
Now, displacement of a simple harmonic motion is given as function
$x = A\cos \omega t$, Where x is the displacement , A is amplitude of the harmonics and t is the time period of oscillation. Applying these in the required equation, we get
$\Rightarrow {x_1} = A\cos \omega$
$\Rightarrow {x_2} = A\cos 2\omega$
$\Rightarrow {x_3} = A\cos 3\omega$
Now, equating as per the given $\theta$ value, we get
$\Rightarrow \dfrac{{{x_1} + {x_3}}}{{2{x_2}}} = \dfrac{{A\cos \omega + A\cos 3\omega }}{{2A\cos 2\omega }}$
Cancelling out the A term we get ,
$\Rightarrow \dfrac{{{x_1} + {x_3}}}{{2{x_2}}} = \dfrac{{\cos \omega + \cos 3\omega }}{{2\cos 2\omega }}$---------(1)
Now, the numerator is in the form of $\operatorname{Cos} A + \operatorname{Cos} B$, which can be trigonometrically expanded as,
$\operatorname{Cos} A + \operatorname{Cos} B = 2\cos (\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2})$, Applying this formula on the above equation we get,
$\Rightarrow 2\cos (\dfrac{{\omega + 3\omega }}{2})\cos (\dfrac{{\omega - 3\omega }}{2})$
On further simplifying the equation, we get
$\Rightarrow 2\cos (2\omega )\cos (\omega )$, Substituting this in the equation number (1) we get,
$\Rightarrow \dfrac{{{x_1} + {x_3}}}{{2{x_2}}} = \dfrac{{2\cos (2\omega )\cos (\omega )}}{{2\cos 2\omega }}$
Cancelling out the common terms, we get
$\Rightarrow \dfrac{{{x_1} + {x_3}}}{{2{x_2}}} = \cos \omega$
Now using this, we can find $\theta$ value and hence find the value for time period for oscillation. Now , on taking inverse of cos to the other side we get,
$\Rightarrow {\cos ^{ - 1}}(\dfrac{{{x_1} + {x_3}}}{{2{x_2}}}) = \omega = \theta$
We know that the time period of oscillation is mathematically given as a ratio of 2 times pi and the angular velocity value.
Thus , substituting the values we get time period as ,
$T = \dfrac{{2\pi }}{\omega } = \dfrac{{2\pi }}{\theta }$

Hence, Option (a) is the right answer for the given question.

Note A body or a particle is said to be undergoing simple harmonic motion, when its motion is periodic and repetitive. The restoring force of the particle or body is said to be directly proportional to that of its displacement , when the body is in simple harmonic motion.