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) Given the displacement equation: $x = x_0 \cos(\omega t - \frac{\pi}{4})$.Velocity $v = \frac{dx}{dt} = -x_0 \omega \sin(\omega t - \frac{\pi}{4})

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the displacement equation: $x = x_0 \cos(\omega t - \frac{\pi}{4})$.Velocity $v = \frac{dx}{dt} = -x_0 \omega \sin(\omega t - \frac{\pi}{4})$.Acceleration $a = \frac{dv}{dt} = -x_0 \omega^2 \cos(\omega t - \frac{\pi}{4})$.Using the trigonometric identity $\cos(	heta + \pi) = -\cos(	heta)$,we can rewrite the acceleration as:$a = x_0 \omega^2 \cos(\omega t - \frac{\pi}{4} + \pi) = x_0 \omega^2 \cos(\omega t + \frac{3\pi}{4})$.Comparing this with the given form $a = A \cos(\omega t + \delta)$,we find:$A = x_0 \omega^2$ and $\delta = \frac{3\pi}{4}$.

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

Similar Questions

What is the maximum acceleration of a particle performing $SHM$ given by the equation $y = 2\sin \left( {\frac{{\pi t}}{2} + \phi } \right)$,where $y$ is in $cm$?

$A$ particle is performing $S.H.M.$ starting from the extreme position. The graphical representation shows that between displacement and acceleration,there is a phase difference of:

Obtain the force law for $SHM$ from the displacement of $SHM$ particle.

$A$ piston is performing $S.H.M.$ in the vertical direction with a frequency of $0.5 \,Hz$. $A$ block of $10 \,kg$ is placed on the piston. The maximum amplitude of the system such that the block remains in contact with the piston is: (in $\,m$)

The weight suspended from a spring oscillates up and down. The acceleration of the weight will be zero at

Vedclass Test Series

Exam Paper Generator

Online Exam Module