The amplitude of a wave represented by the di | vedclass

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(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

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

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(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: $A = \sqrt{(\frac{1}{\sqrt{a}})^2 + (\frac{1}{\sqrt{b}})^2} = \sqrt{\frac{1}{a} + \frac{1}{b}} = \sqrt{\frac{a + b}{ab}}$.

$A$. $Pa \cdot s$	$(i)$ $[L^2 \ T^{-2} \ K^{-1}]$
$B$. $N \cdot m \cdot K^{-1}$	$(ii)$ $[M \ L \ T^{-3} \ K^{-1}]$
$C$. $J \cdot kg^{-1} \cdot K^{-1}$	$(iii)$ $[M \ L^{-1} \ T^{-1}]$
$D$. $W \cdot m^{-1} \cdot K^{-1}$	$(iv)$ $[M \ L^2 \ T^{-2} \ K^{-1}]$

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

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