The motion of a particle varies with time acc | vedclass

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

Question

(C) Given the equation of motion: $y = a(\sin \omega \,t + \cos \omega \,t)$.We can rewrite this by multiplying and dividing by $\sqrt{2}$:$y = a\sqrt

Vedclass · Accepted Answer

The motion is $S.H.M.$ with amplitude $a\sqrt{2}$

Vedclass · Answer

(C) Given the equation of motion: $y = a(\sin \omega \,t + \cos \omega \,t)$.We can rewrite this by multiplying and dividing by $\sqrt{2}$:$y = a\sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \omega \,t + \frac{1}{\sqrt{2}} \cos \omega \,t \right)$.Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,where $\sin 45^\circ = \frac{1}{\sqrt{2}}$ and $\cos 45^\circ = \frac{1}{\sqrt{2}}$:$y = a\sqrt{2} (\sin \omega \,t \cos 45^\circ + \cos \omega \,t \sin 45^\circ)$.$y = a\sqrt{2} \sin(\omega \,t + 45^\circ)$.This is the standard equation of Simple Harmonic Motion $(S.H.M.)$,which is $y = A \sin(\omega \,t + \phi)$,where the amplitude $A = a\sqrt{2}$.

The motion of a particle varies with time according to the relation $y = a(\sin \omega \,t + \cos \omega \,t)$,then

Similar Questions

Show that the motion of a particle represented by $y = \sin \omega t - \cos \omega t$ is simple harmonic with a period of $\frac{2\pi}{\omega}$.

Explain that simple harmonic motion is the projection of uniform circular motion on a diameter of the reference circle and obtain the velocity and acceleration.

$A$ particle executing $SHM$ of amplitude $a$ has a displacement of $-\frac{a}{2}$ at $t = \frac{T}{4}$ and a positive velocity. Find the initial phase of the particle.

Who decides the characteristics of $SHM$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module