Question

Particle executes simple harmonic motion between $x=-A$ and $x=+A$. The time taken by it to go from O to $\dfrac{A}{2}$ is $T_{1}$ and to go from $\dfrac{A}{2}$ to A is $T_{2}$. Then
a) $T_{1} \lt T_{2}$
b) $T_{1} \gt T_{2}$
c) $T_{1} = T_{2}$
d) $T_{1} =2 T_{2}$

Answer 1

Hint: A particle executes simple harmonic motion, so first we write harmonic wave equation, then we write equation according to the distance travelled. As motion starts from the mean position, then time and distance coordinate are zero at the mean position. At maximum point, velocity becomes zero.

Complete step-by-step solution:
The equation of motion when it starts from mean position :
$x=A sinwt$
At mean position $x = 0, t_{1} = 0$
$\dfrac{A}{2} =A sinwt_{2}$
$\implies \dfrac{1}{2} = sinwt_{2} = sin \dfrac{\pi}{6}$
$\implies wt_{2} = \dfrac{\pi}{6} or t_{2} = \dfrac{\pi}{6w} $
At $x=A$, $t_{3}$ is the time for travel from 0 to A.
$A=A sinwt_{3}$
$\implies 1 = sinwt_{3} = sin \dfrac{\pi}{2}$
$\implies wt_{3} = \dfrac{\pi}{2} or t_{3} = \dfrac{\pi}{2w} $
The time taken by it to go from O to is $T_{3}$.
$T_{1} = t_{2} – t{1} = \dfrac{\pi}{6w} $
$\implies T_{2} = t_{3} – t{2} = \dfrac{\pi}{2w} - \dfrac{\pi}{6w} = \dfrac{\pi}{3w}$
$\implies \dfrac{T_{2}}{T_{1}} = \dfrac{\dfrac{\pi}{3w}}{\dfrac{\pi}{6w}} = \dfrac{2}{1}$
$\implies T_{1} \lt T_{2}$
Option (a) is correct.

Additional Information Relation among the instant displacement of a particle dealing with SHM is called an equation of progressive wave. Angular frequency, frequency and time period are the properties of the source generating the wave and are not dependent on the nature of the medium in which the wave travels.

Note: Qualitatively, the particle's velocity is maximum at its mean position, and it decreases as it moves towards extreme vibration. Hence, the particle will take less time to travel the first half of the displacement compared to the second-half displacement. So, time travel from O to $\dfrac{A}{2}$ is less than to go from $\dfrac{A}{2}$ to A.