Q. A particle is undergoing a one dimensional simple harmonic oscillation of amplitude $X_{M}$ about the origin on $X$-axis with time period $T$ and is at –$X_{M}$ at. What is the position of the particle after a time interval $t = 3.15\ T$?
A) Between $–X_{M}$ and $O$
B) Between $O$ and $+ X_{M}$
C) At the origin
D) At $+ X_{M}$
Explanation
(b) Amplitude of SHM of a particle =$X_{M}$
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At $t = 0$ position of particle=$A$(given)
At any instant $t$ displacement of particle=$y$ (say)
Angular velocity of particle=$\omega$
Then, $y = X_{M}\sin{\omega t}$
$y = X_{M}\sin\frac{2\pi t}{T}$
At $t = 3.15\ T$
$y = X_{M}\sin\frac{2\pi}{T}(3.15T)$
$\ \ \ \ = X_{M}\sin{(2\pi \times 3.15)}$
$y = X_{M}\sin{(6\pi \times 54{^\circ})}$
$= X_{M}\sin{54{^\circ}}$
$y = X_{M}\ \ \frac{(\sqrt{5} - 1)}{4}$
Since, the measurement starts from position $A$, then after 3.15 $T$ particle will be between $O\ and\ X_{m}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.