A particle starts from a point P at a distanc | vedclass

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

Question

(B) The general equation for $SHM$ is $x = A \sin(\omega t + \phi)$,where $\omega = \frac{2\pi}{T}$.At $t = 0$,the particle is at $x = +A/2$ and movin

Vedclass · Accepted Answer

$x = A \sin \left( \frac{2\pi}{T}t + \frac{5\pi}{6} \right)$

Vedclass · Answer

(B) The general equation for $SHM$ is $x = A \sin(\omega t + \phi)$,where $\omega = \frac{2\pi}{T}$.At $t = 0$,the particle is at $x = +A/2$ and moving towards the left (negative direction).Substituting $t = 0$ into the equation: $A/2 = A \sin(\phi)$,which gives $\sin(\phi) = 1/2$.This implies $\phi = \pi/6$ or $\phi = 5\pi/6$.Since the particle is moving towards the left (towards the mean position from the positive side),its velocity $v = dx/dt = A\omega \cos(\omega t + \phi)$ must be negative at $t = 0$.For $\phi = \pi/6$,$v = A\omega \cos(\pi/6) > 0$.For $\phi = 5\pi/6$,$v = A\omega \cos(5\pi/6) Thus,the correct phase constant is $\phi = 5\pi/6$.The equation of motion is $x = A \sin \left( \frac{2\pi}{T}t + \frac{5\pi}{6} \right)$.

$A$ particle starts from a point $P$ at a distance of $A/2$ from the mean position $O$ and travels towards the left as shown in the figure. If the time period of $SHM$ executed about $O$ is $T$ and amplitude is $A$,then the equation of motion of the particle is:

Similar Questions

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. If the time taken by the particle to go from $x = 0$ to $x = A/2$ is $2 \, s$,then the time taken by the particle in going from $x = A/2$ to $x = A$ is $......... \, s$.

The phase difference between two $SHM$ equations $y_1 = 10 \sin(10\pi t + \frac{\pi}{3})$ and $y_2 = 12 \sin(8\pi t + \frac{\pi}{4})$ at $t = 0.5 \ s$ is:

$A$ particle starts executing simple harmonic motion from one extreme position. If $a, b$ and $c$ are the displacements of the particle from the mean position at the ends of three successive seconds,the frequency of simple harmonic motion is

The time taken by a particle executing simple harmonic motion of period $T$ to move from the mean position to half the maximum displacement is

$A$ 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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module