A point mass oscillates along the X-axis acco | vedclass

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

Question

(A) The displacement of the particle is given by $x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right)$.Velocity $v$ is the first derivative of displacement

Vedclass · Accepted Answer

$A=x_0 \omega^2, \delta=-\frac{3 \pi}{4}$

Vedclass · Answer

(A) The displacement of the particle is given by $x=x_0 \cos \left(\omega t-\frac{\pi}{4}ight)$.Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = -x_0 \omega \sin \left(\omega t-\frac{\pi}{4}ight)$.Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = -x_0 \omega^2 \cos \left(\omega t-\frac{\pi}{4}ight)$.Using the trigonometric identity $\cos(	heta + \pi) = -\cos(	heta)$, we can rewrite the acceleration as:$a = x_0 \omega^2 \cos \left(\omega t - \frac{\pi}{4} + \piight) = x_0 \omega^2 \cos \left(\omega t + \frac{3\pi}{4}ight)$.To match the form $a = A \cos(\omega t - \delta)$, we write the phase as $\omega t - (-\frac{3\pi}{4})$.Thus, $A = x_0 \omega^2$ and $\delta = -\frac{3\pi}{4}$.

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

$A$ point mass oscillates along the $X$-axis according to the law $x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right)$. If the acceleration of the particle is written as $a=A \cos (\omega t-\delta)$, then

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 of mass $1 \ kg$ is moving in $SHM$ with a path length of $0.01 \ m$ and frequency of $50 \ Hz$. The maximum force in newton,acting on the particle is (in $\pi^2$)

$A$ coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency $\omega$. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

Two simple harmonic motions of angular frequency $100 \, rad \, s^{-1}$ and $1000 \, rad \, s^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration is

$Assertion :$ In $SHM$,acceleration is always directed towards the mean position.
$Reason :$ In $SHM$,the body has to stop momentarily at the extreme position and move back to the mean position.

Vedclass Test Series

Exam Paper Generator

Online Exam Module