Question

A particle is describing simple harmonic motion. If its velocities are ${v_1}$​ and ${v_2}$when the displacements from the mean position are ${y_1}$and ${y_2}$respectively, then its time period is
a. $2\pi \sqrt {\dfrac{{y_1^2 + y_2^2}}{{v_1^2 + v_2^2}}}$
b. $2\pi \sqrt {\dfrac{{v_2^2 - v_1^2}}{{y_1^2 - y_2^2}}}$
c. $2\pi \sqrt {\dfrac{{v_2^2 + v_1^2}}{{y_1^2 + y_2^2}}}$
d. $2\pi \sqrt {\dfrac{{y_1^2 - y_2^2}}{{v_2^2 - v_1^2}}}$

Answer 1

We can write Simple harmonic motion as SHM. SHM is known as a special type of periodic motion. In question the time period of a particle is calculated. The period is defined by the time it takes for one oscillation. In question, we are using the phenomenon of SHM to calculate the time period of moving particles.

Formula used:
Simple harmonic motion is also known as a periodic oscillation. we can measure its period. To calculate time period, first we find from, velocity formula $v = \omega \sqrt {{A^2} - {y^2}}$.${v_1}$ and ${v_2}$can be calculated by putting and on above formula. Then using angular frequency formula$\omega = \dfrac{{2\pi }}{T}$, we can calculate time period of particle as $T = \dfrac{{2\pi }}{\omega }$.

Complete step by step answer:
In simple harmonic motion, the velocity of moving particle is given by$v = \omega \sqrt {{A^2} - {y^2}}$
Using above formula, Velocity of particle with displacement is given as ${v_1} = \omega \sqrt {{A^2} - y_1^2}$ ….(i)
And Velocity${v_2}$ of particle with displacement ${y_2}$is given as ${v_2} = \omega \sqrt {{A^2} - y_2^2}$ ….(ii)
On squaring equations (i) and (ii), we get
$v_1^2 = {\omega ^2}\left( {{A^2} - y_1^2} \right)$ ….(iii)
$v_2^2 = {\omega ^2}\left( {{A^2} - y_2^2} \right)$ ….(iv)
Subtracting (iii) from (iv), we get
$v_2^2 - v_1^2 = {\omega ^2}\left( {{A^2} - y_2^2} \right) - {\omega ^2}\left( {{A^2} - y_1^2} \right)$
$\Rightarrow v_2^2 - v_1^2 = {\omega ^2}{A^2} - {\omega ^2}y_2^2 - {\omega ^2}{A^2} + {\omega ^2}y_1^2$
$\Rightarrow v_2^2 - v_1^2 = {\omega ^2}\left[ {y_1^2 - y_2^2} \right]$
$\Rightarrow {\omega ^2} = \dfrac{{v_2^2 - v_1^2}}{{y_1^2 - y_2^2}}$
$\therefore \omega = \sqrt {\dfrac{{v_2^2 - v_1^2}}{{y_1^2 - y_2^2}}}$
Here is the angular frequency of particles in Simple Harmonic Motion. It is given by $\omega = \dfrac{{2\pi }}{T}$
Hence, time period is given by, $T = \dfrac{{2\pi }}{\omega }$
$T = \dfrac{{2\pi }}{{\sqrt {\dfrac{{v_2^2 - v_1^2}}{{y_1^2 - y_2^2}}} }}$
$\Rightarrow T = 2\pi \sqrt {\dfrac{{y_1^2 - y_2^2}}{{v_2^2 - v_1^2}}}$
$\therefore$, Time period is $T = 2\pi \sqrt {\dfrac{{y_1^2 - y_2^2}}{{v_2^2 - v_1^2}}}$

Hence, the correct answer is option (D).

Additional information:
In SHM, on the moving object, the restoring force occurs. This force is directly proportional to the displacement. The restoring force always acts towards the object's equilibrium position. It results in an oscillation. Simple harmonic motion is used to define a mathematical model for different types of motion. Most terminologies used for SHM are time period (or simply we write period), angular frequency, amplitude, displacement.

Note: A particle is moving in simple harmonic motion with 2 different velocities, So students should be clear that the angular frequency and amplitude will be the same. Students can see all the options. We can see that amplitude A is not included. So when they solve equations of velocities, they should remove amplitude.