If the displacement equation of a particle is | vedclass

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

Question

(C) Given equation: $y = A\sin PT + B\cos PT$Let $A = r\cos 	heta$ and $B = r\sin 	heta$,where $r = \sqrt{A^2 + B^2}$ and $	an 	heta = B/A$.Substi

Vedclass · Accepted Answer

$A$ $S$.$H$.$M$.

Vedclass · Answer

(C) Given equation: $y = A\sin PT + B\cos PT$Let $A = r\cos 	heta$ and $B = r\sin 	heta$,where $r = \sqrt{A^2 + B^2}$ and $	an 	heta = B/A$.Substituting these into the equation,we get:$y = r\cos 	heta \sin PT + r\sin 	heta \cos PT$Using the trigonometric identity $\sin(x + y) = \sin x \cos y + \cos x \sin y$,we have:$y = r\sin(PT + 	heta)$This is the standard form of the equation for Simple Harmonic Motion $(SHM)$,where $r$ is the amplitude,$P$ is the angular frequency,and $	heta$ is the phase constant.

If the displacement equation of a particle is represented by $y = A\sin PT + B\cos PT$,the particle executes

Similar Questions

The motion of a particle represented by $y = \sin \omega t - \cos \omega t$ is

What provides the restoring force in the following cases?
$(1)$ Compressed spring.
$(2)$ Displacement of water in a $U$-tube.
$(3)$ Displacement of a pendulum bob from its mean position.

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

$A$ particle of mass $0.4 \text{ kg}$ executes simple harmonic motion of amplitude $0.4 \text{ m}$. When it passes through the mean position,its kinetic energy is $256 \times 10^{-3} \text{ J}$. If the initial phase of the oscillation is $\pi / 4$,then the equation of its motion is

The displacement of a particle is given at time $t$,by: $x = A \sin (-2 \omega t) + B \sin^2 \omega t$. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module