The amplitude of a wave represented by the di | vedclass

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

Question

(D) The given equation is $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$.We can rewrite $\cos \omega t$ as $\sin(\omega t

Vedclass · Accepted Answer

$\sqrt{\frac{a + b}{ab}}$

Vedclass · Answer

(D) The given equation is $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$.We can rewrite $\cos \omega t$ as $\sin(\omega t + \frac{\pi}{2})$.Thus,$y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \sin(\omega t + \frac{\pi}{2})$.This represents the superposition of two simple harmonic motions with amplitudes $A_1 = \frac{1}{\sqrt{a}}$ and $A_2 = \frac{1}{\sqrt{b}}$,and a phase difference of $\phi = \frac{\pi}{2}$.The resultant amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.Since $\cos(\frac{\pi}{2}) = 0$,the formula simplifies to $A = \sqrt{A_1^2 + A_2^2}$.Substituting the values,$A = \sqrt{(\frac{1}{\sqrt{a}})^2 + (\frac{1}{\sqrt{b}})^2} = \sqrt{\frac{1}{a} + \frac{1}{b}}$.Simplifying the expression,$A = \sqrt{\frac{a + b}{ab}}$.

The amplitude of a wave represented by the displacement equation $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$ will be

Similar Questions

$ICBN$ stands for

$100 \ g$ of water is heated from $30^\circ C$ to $50^\circ C$. Ignoring the slight expansion of the water,the change in its internal energy is ...... $kJ$ (specific heat of water is $4184 \ J/kg/K$).

The value of $\lambda$ with $|\lambda| < 16$ such that $2 x^2 - 10 x y + 12 y^2 + 5 x + \lambda y - 3 = 0$ represents a pair of straight lines,is

How many geometrical isomers are possible for the compound shown below?

Vedclass Test Series

Exam Paper Generator

Online Exam Module