Question

if y1=5cos​Wt and y2= 5(sq root 3 cos Wt + sinWt)find ratio of amplitudes of two waves

Answer 1

Waves

We are given a simple question on waves. Let's write the question again for clarity.

Question:

If $y_1=5\cos\omega t$ and $y_2=5(\sqrt{3}\cos\omega t+\sin \omega t)$, find the ratio of amplitudes of the two waves.

Answer:

A wave can be written in general as:

$y = A \cos (\omega t+\phi)$

where

A = Amplitude

$\omega$ = Angular Frequency

$\phi$ = Phase constant

In order to be able to take ratio of amplitudes, we first need to bring both waves in the form given.

$y_1$ is already in standard form, but $y_2$ needs to be slightly manipulated to bring into the standard wave form.

We can use the following technique:

$\displaystyle\textsf{If }y=A\sin x+B\cos x\\\\\\\textsf{Divide and multiply with }\sqrt{A^2+B^2}\\\\\\\implies y=\sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}}\sin x+\frac{B}{\sqrt{A^2+B^2}}\cos x\right)\\\\\\\textsf{Now we can put }\sin\phi=\frac{A}{\sqrt{A^2+B^2}}\textsf{ and }\cos\phi=\frac{B}{\sqrt{A^2+B^2}}\\\\\\\implies y =\sqrt{A^2+B^2}(\sin x\sin\phi+\cos x\cos\phi)\\\\\\\implies y=\sqrt{A^2+B^2}\cos (x-\phi) \\\\\\ \textsf{where } \phi = \tan^{-1} \left(\frac{A}{B}\right)$

Hence, here we can do:

$\sf \displaystyle y_2 = 5(\sqrt{3}\cos\omega t+\sin \omega t) \\\\\\ \textsf{Multiple and divide with }\sqrt{(\sqrt{3})^2+(1)^2} = \sqrt{4} = 2 \\\\\\ \implies y_2 = 5\times 2\left(\frac{\sqrt{3}}{2}\cos\omega t + \frac{1}{2}\sin\omega t\right) \\\\\\ \implies y_2 = 10\left(\cos\omega t \cos \frac{\pi}{6} + \sin\omega t \sin \frac{\pi}{6}\right) \\\\\\ \implies y_2 = 10\cos\left(\omega t - \frac{\pi}{6}\right)$

$\rule{300}{1}$

So, we have our Data:

$\begin{aligned} &y_1 = 5\cos \omega t &\implies A_1 = 5 \\\\ &y_2 = 10 \cos\left(\omega t-\frac{\pi}{6}\right)&\implies A_2 = 10\end{aligned}$

And so we have our ratio:

$\sf Ratio = \dfrac{A_1}{A_2} = \dfrac{5}{10} \\\\\\ \implies \Large \boxed{\sf Ratio = \frac{A_1}{A_2}=\frac{1}{2}}$