Question

The amplitude of a wave is represented by displacement equation $y = \dfrac{1}{{\sqrt a }}\sin \omega t \pm \dfrac{1}{{\sqrt b }}\cos \omega t$ will be
A. $\dfrac{{a - b}}{{ab}}$
B. $\dfrac{{\sqrt a + \sqrt b }}{{ab}}$
C. $\dfrac{{\sqrt a \pm \sqrt b }}{{ab}}$
D. $\sqrt {\dfrac{{a + b}}{{ab}}}$

Answer 1

In the question, we are provided with a displacement equation which is the form of a sine wave. On comparing with the general sine wave which is represented as,
$y = A\sin (\omega t + \theta )$
Hence, squaring and adding the two respective amplitudes gave us the required resultant amplitude.

Complete step by step answer:
According to the question, the wave is,
$y = \dfrac{1}{{\sqrt a }}\sin \omega t \pm \dfrac{1}{{\sqrt b }}\cos \omega t$
Rewriting the wave equation in the form of $\sin$. So, converting the $\cos$ to $\sin$ by simple trigonometric property.
$y = \dfrac{1}{{\sqrt a }}\sin \omega t \pm \dfrac{1}{{\sqrt b }}\sin \left( {\omega t + \dfrac{\pi }{2}} \right)$
On looking at the second term, we come to know that the phase difference is $\dfrac{\pi }{2}$.General equation of the wave is,
$y = A\sin \left( {\omega t + \theta } \right)$ where A represents the amplitude.
Comparing with the above equation,
Now, we want to find the amplitude so writing amplitudes of both the terms in the form of ${A_{1,}}{A_2}$ respectively.
${A_1} = \dfrac{1}{{\sqrt a }}$ ${A_2} = \dfrac{1}{{\sqrt b }}$
Resultant amplitude be given by
$A = \sqrt {{A_1}^2 + {A_2}^2}$
Substituting the value
A = \sqrt {{{\left( {\dfrac{1}{{\sqrt a }}} \right)}^2} + {{\left( {\dfrac{1}{{\sqrt b }}} \right)}^2}} \\\ \Rightarrow A = \sqrt {\dfrac{1}{a} + \dfrac{1}{b}} \\\ \therefore A = \sqrt {\dfrac{{a + b}}{{ab}}} \\\
This is our required answer.

Hence,the correct option is D.

Note: Alternative method: on comparing our given equation, we got to know that $\dfrac{1}{{\sqrt a }} = A\cos \theta$ and $\dfrac{1}{{\sqrt b }} = A\sin \theta$ $\ldots \left( r \right)$
Squaring and adding the two equation $\left( r \right)$ gave
$A = \sqrt {\dfrac{{a + b}}{{ab}}}$ $\left( {\because {{\cos }^2}\theta + {{\sin }^2} = 1} \right)$
From simplifying equation $\left( r \right)$
$y = A\sin \left( {\omega t + \theta } \right)$ which is the general equation of a sine wave.