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 know that $\cos \omega t = \sin(\omega t + \fra

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 know that $\cos \omega t = \sin(\omega t + \frac{\pi}{2})$.Substituting this,the equation becomes $y = \frac{1}{\sqrt{a}}\sin \omega t \pm \frac{1}{\sqrt{b}}\sin(\omega t + \frac{\pi}{2})$.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 resultant amplitude is $A = \sqrt{A_1^2 + A_2^2}$.Substituting the values: $A = \sqrt{(\frac{1}{\sqrt{a}})^2 + (\frac{1}{\sqrt{b}})^2} = \sqrt{\frac{1}{a} + \frac{1}{b}} = \sqrt{\frac{a + b}{ab}}$.

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

$A$ solution of concentration $C \ g \ \text{equiv}/L$ has a specific resistance $R$. The equivalent conductance of the solution is

During which phase of meiosis does crossing over occur?

Which of the following is a level of biological organization?

An $\alpha$-particle of energy $E$ is liberated during the decay of a nucleus of mass number $236$. The total energy released in this process is

Vedclass Test Series

Exam Paper Generator

Online Exam Module