A particle executes S.H.M. between x = -A and | vedclass

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

Question

(A) The displacement equation for $S.H.M.$ starting from the mean position is $x = A \sin(\omega t)$.For the interval $0$ to $A/2$,we have $A/2 = A \s

Vedclass · Accepted Answer

$T_1 < T_2$

Vedclass · Answer

(A) The displacement equation for $S.H.M.$ starting from the mean position is $x = A \sin(\omega t)$.For the interval $0$ to $A/2$,we have $A/2 = A \sin(\omega T_1)$,which gives $\sin(\omega T_1) = 1/2 = \sin(\pi/6)$.Thus,$\omega T_1 = \pi/6$,so $T_1 = \pi/(6\omega)$.For the interval $0$ to $A$,the time taken is $T_1 + T_2$. At $x = A$,we have $A = A \sin(\omega(T_1 + T_2))$,which gives $\sin(\omega(T_1 + T_2)) = 1 = \sin(\pi/2)$.Thus,$\omega(T_1 + T_2) = \pi/2$,so $T_1 + T_2 = \pi/(2\omega)$.Substituting $T_1 = \pi/(6\omega)$,we get $T_2 = \pi/(2\omega) - \pi/(6\omega) = (3\pi - \pi)/(6\omega) = 2\pi/(6\omega) = \pi/(3\omega)$.Comparing the two,$T_1 = \pi/(6\omega)$ and $T_2 = \pi/(3\omega)$,we find that $T_1 < T_2$.

$A$ particle executes $S.H.M.$ between $x = -A$ and $x = +A$. The time taken for it to go from $x = 0$ to $x = A/2$ is $T_1$,and from $x = A/2$ to $x = A$ is $T_2$. Then:

Similar Questions

$A$ particle executes $S.H.M.$ of amplitude $A$ along the $x$-axis. At $t = 0$,the position of the particle is $x = \frac{A}{2}$ and it moves along the positive $x$-axis. If the displacement of the particle in time $t$ is $x = A \sin (\omega t + \delta)$,then the value of $\delta$ will be:

The period of $S.H.M.$ of a particle is $16 \ s$. The phase difference between the positions at $t = 2 \ s$ and $t = 4 \ s$ will be

$A$ simple harmonic oscillator of frequency $1 \ Hz$ has a phase of $1$ radian. By how much should the origin be shifted in time so as to make the phase of the oscillator vanish? ($t$ in seconds).

$A$ particle executes $SHM$ of amplitude $25\, cm$ and time period $3\, s$. What is the minimum time required for the particle to move between two points $12.5\, cm$ on either side of the mean position?

$A$ particle executing $S.H.M.$ with an amplitude of $4 \, cm$ and a time period $T = 4 \, s$. The time taken by it to move from the positive extreme position to half the amplitude is ..... $s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module