What is the phase difference between the velo | vedclass

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

Question

(B) Let the displacement of a particle executing $SHM$ be $x = A \sin(\omega t + \phi)$.The velocity $v$ is given by $v = \frac{dx}{dt} = A \omega \co

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) Let the displacement of a particle executing $SHM$ be $x = A \sin(\omega t + \phi)$.The velocity $v$ is given by $v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) = A \omega \sin(\omega t + \phi + \frac{\pi}{2})$.The acceleration $a$ is given by $a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \phi) = A \omega^2 \sin(\omega t + \phi + \pi)$.The phase of velocity is $(\omega t + \phi + \frac{\pi}{2})$ and the phase of acceleration is $(\omega t + \phi + \pi)$.The phase difference is $(\omega t + \phi + \pi) - (\omega t + \phi + \frac{\pi}{2}) = \frac{\pi}{2}$ radians or $90^{\circ}$.

$A \ (mm \ s^{-2})$	$16$	$8$	$0$	$-8$	$-16$
$x \ (mm)$	$-4$	$-2$	$0$	$2$	$4$

What is the phase difference between the velocity and acceleration of a particle executing $SHM$ (in $^{\circ}$)?

Similar Questions

Values of the acceleration $A$ of a particle moving in simple harmonic motion as a function of its displacement $x$ are given in the table below:
$A \ (mm \ s^{-2})$ $16$ $8$ $0$ $-8$ $-16$
$x \ (mm)$ $-4$ $-2$ $0$ $2$ $4$

The period of the motion is:

$A$ particle moving along the $X$-axis executes simple harmonic motion. The force acting on it is given by: (Where $A$ and $K$ are positive constants)

For a particle performing $S.H.M.$ given by the equation $y = 2 \sin \left( \frac{\pi t}{2} + \phi \right) \, (cm)$,what is the maximum acceleration of the particle?

For a particle executing simple harmonic motion given by $y(cm) = \sin \frac{\pi}{2} \left( \frac{t}{2} + \frac{1}{3} \right)$,what is the maximum acceleration in $cm/sec^2$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module