A particle executes simple harmonic motion be | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For a particle executing simple harmonic motion starting from the mean position $(x=0)$,the displacement is given by $x(t) = A \sin(\omega t)$.For

Vedclass · Accepted Answer

$T_1 = 2T_2$

Vedclass · Answer

(D) For a particle executing simple harmonic motion starting from the mean position $(x=0)$,the displacement is given by $x(t) = A \sin(\omega t)$.For $x = 0$,$\omega t_0 = 0 \implies t_0 = 0$.For $x = \frac{A}{2}$,$\frac{A}{2} = A \sin(\omega t_1) \implies \sin(\omega t_1) = \frac{1}{2} \implies \omega t_1 = \frac{\pi}{6} \implies t_1 = \frac{\pi}{6\omega}$.For $x = \frac{A}{\sqrt{2}}$,$\frac{A}{\sqrt{2}} = A \sin(\omega t_2) \implies \sin(\omega t_2) = \frac{1}{\sqrt{2}} \implies \omega t_2 = \frac{\pi}{4} \implies t_2 = \frac{\pi}{4\omega}$.The time taken to move from $x=0$ to $x=\frac{A}{2}$ is $T_1 = t_1 - t_0 = \frac{\pi}{6\omega}$.The time taken to move from $x=\frac{A}{2}$ to $x=\frac{A}{\sqrt{2}}$ is $T_2 = t_2 - t_1 = \frac{\pi}{4\omega} - \frac{\pi}{6\omega} = \frac{3\pi - 2\pi}{12\omega} = \frac{\pi}{12\omega}$.Comparing $T_1$ and $T_2$,we see that $T_1 = \frac{\pi}{6\omega} = \frac{2\pi}{12\omega} = 2 	imes \frac{\pi}{12\omega} = 2T_2$.Thus,$T_1 > T_2$,which implies $T_1$ is greater than $T_2$. Looking at the options,the relationship $T_1 > T_2$ is not explicitly listed as an option,but checking the provided options,$T_1 = 2T_2$ is the correct mathematical result.

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. It starts from $x = 0$ and moves in the $+x$ direction. The time taken for it to move from $x = 0$ to $x = \frac{A}{2}$ is $T_1$,and the time taken to move from $x = \frac{A}{2}$ to $x = \frac{A}{\sqrt{2}}$ is $T_2$. Then:

Similar Questions

Explain with plots the position of a particle executing simple harmonic motion at different times.

Two bodies performing $S.H.M.$ have the same amplitude and frequency. Their positions and directions of motion at a certain instant are as shown in the figure. The phase difference between them is

Time period of a particle executing $SHM$ is $8 \, s$. At $t = 0$ it is at the mean position. The ratio of the distance covered by the particle in the $1^{st}$ second to the $2^{nd}$ second is:

$A$ body is in simple harmonic motion with a time period of $0.5 \ s$ and an amplitude of $1 \ cm$. Find the average velocity in the interval in which it moves from the equilibrium position to half of its amplitude (in $cm/s$).

$A$ particle executes $SHM$ with a period of $1.2 \, s$ and an amplitude of $8 \, cm$. Find the time it takes to travel $3 \, cm$ from the positive extremity of its oscillation.

Vedclass Test Series

Exam Paper Generator

Online Exam Module