The displacement of a particle varies with ti | vedclass

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

Question

(D) Given the displacement equation: $x = a \sin \omega t + b \cos \omega t$.To find the amplitude,we rewrite the equation in the form $x = A \sin(\om

Vedclass · Accepted Answer

The motion is $SHM$ with amplitude $\sqrt{a^2 + b^2}$.

Vedclass · Answer

(D) Given the displacement equation: $x = a \sin \omega t + b \cos \omega t$.To find the amplitude,we rewrite the equation in the form $x = A \sin(\omega t + \phi)$.Let $a = A \cos \phi$ and $b = A \sin \phi$.Then $x = A \cos \phi \sin \omega t + A \sin \phi \cos \omega t = A \sin(\omega t + \phi)$.Squaring and adding the expressions for $a$ and $b$:$a^2 + b^2 = A^2 \cos^2 \phi + A^2 \sin^2 \phi = A^2(\cos^2 \phi + \sin^2 \phi) = A^2$.Thus,the amplitude $A = \sqrt{a^2 + b^2}$.Since the equation can be expressed as a single sine function,the motion is $SHM$ with amplitude $\sqrt{a^2 + b^2}$.

The displacement of a particle varies with time according to the relation $x = a \sin \omega t + b \cos \omega t$.

Similar Questions

The function $x = A \sin^2 \omega t + B \cos^2 \omega t + C \sin \omega t \cos \omega t$ does not represent $SHM$ for which of the following conditions?

What is periodic motion? Give its examples.

The displacement $x$ (in metres) of a particle performing simple harmonic motion is related to time $t$ (in seconds) as $x = 0.05 \cos \left( 4 \pi t + \frac{\pi}{4} \right)$. The frequency of the motion will be ..... $Hz$.

What is a linear harmonic oscillator? And what is a non-linear oscillator?

The equation of motion of a particle is $x = a \cos(\alpha t)$. The motion is

Vedclass Test Series

Exam Paper Generator

Online Exam Module