Question

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is:
A. $\dfrac{7}{3}\pi$
B. $\dfrac{3}{8}\pi$
C. $\dfrac{4}{3}\pi$
D. $\dfrac{8}{3}\pi$

Answer 1

Hint: A body undergoing simple harmonic motion is bounded by certain laws that can be written in the form of mathematical equations. To undergo SHM the acceleration a, on the body must be opposite in direction to the displacement, x. Using these equations we can find the missing quantities and thus the answer.

Formula Used:
Necessary condition for SHM, $a = - {\omega ^2}x$
Relation between velocity, amplitude, displacement and angular velocity:
\eqalign{ & v = \omega \sqrt {{A^2} - {x^2}} \cr & {\text{where }}v{\text{ is the velocity of SHM,}} \cr & \omega {\text{ is the angular velocity of SHM}}, \cr & A{\text{ is the amplitude of SHM,}} \cr & {\text{and }}x{\text{ is the displacement from mean position}}{\text{.}} \cr}

Complete step-by-step answer:
Simple harmonic motion (SHM) is a special type of oscillation i.e. to and fro motion in which the particle oscillates on a straight line, the acceleration of the particle is always directed towards a fixed point on the line and its magnitude is proportional to the displacement of the particle from this point. This fixed point is called the centre of oscillation.
Taking this point as origin and the line of motion as X-axis we can write the defining equation of SHM as:
$a = - {\omega ^2}x \cdots \cdots \cdots \cdots \cdots \left( 1 \right)$
Also the relation between the velocity, angular velocity, displacement and amplitude of SHM is given as:
\eqalign{ & v = \omega \sqrt {{A^2} - {x^2}} \cdots \cdots \cdots \cdots \left( 2 \right) \cr & {\text{where }}v{\text{ is the velocity of SHM,}} \cr & \omega {\text{ is the angular velocity of SHM}}, \cr & A{\text{ is the amplitude of SHM,}} \cr & {\text{and }}x{\text{ is the displacement from mean position}}{\text{.}} \cr}
Given that:
Amplitude, $A = 5{\text{ }}cm$
Displacement, $x = 4cm$
Substituting these values in equation (1) we get:
\eqalign{ & a = - {\omega ^2}x \cr & \Rightarrow a = - {\omega ^2}4 \cr & \therefore a = - 4{\omega ^2} \cdots \cdots \cdots \left( 3 \right) \cr}
Substituting these values in equation (2) we get:
\eqalign{ & v = \omega \sqrt {{A^2} - {x^2}} \cr & \Rightarrow v = \omega \sqrt {{5^2} - {4^2}} \cr & \Rightarrow v = \omega \sqrt {25 - 16} \cr & \Rightarrow v = \omega \sqrt 9 \cr & \therefore v = 3\omega \cdots \cdots \cdots \cdots \left( 4 \right) \cr}
According to the question at x=4cm, the velocity and acceleration of the particle are equal in magnitude.
So taking modulus of equation (3) and (4) and equating them we get:
\eqalign{ & \left| a \right| = \left| v \right| \cr & \Rightarrow \left| { - 4{\omega ^2}} \right| = \left| {3\omega } \right| \cr & \Rightarrow 4{\omega ^2} = 3\omega \cr & \therefore \omega = \dfrac{3}{4} \cr}
We know that angular velocity is related to Time period by the relation:
\eqalign{ & \omega = \dfrac{{2\pi }}{T} \cr & \Rightarrow T = \dfrac{{2\pi }}{\omega } \cr}
Substituting value of angular momentum in the above equation we get:
\eqalign{ & T = \dfrac{{2\pi }}{\omega } \cr & \Rightarrow T = \dfrac{{2\pi }}{{\dfrac{3}{4}}} \cr & \Rightarrow T = 2\pi \times \dfrac{4}{3} \cr & \therefore T = \dfrac{{8\pi }}{3} \cr}
Therefore, the correct option is D. i.e., the time period of SHM in seconds is $\dfrac{8}{3}\pi$

Note: The force responsible for Simple harmonic motion is always directed towards the center i.e., the equilibrium position of the system. Thus the negative sign in the equation of the necessary condition for SHM only shows the direction motion of the body, i..e., it is opposite to the direction of the acceleration on the system.