Question

Find the ratio of y1 and y2​

Answer 1

Answer:

• The ratio of Y₁ : Y₂ is 1 : 2

Given:

1. Y₁ = 4 sin [ 3 π t + π/3 ]
2. Y₂ = 4 [sin (3 π t) + √3 cos (3 π t)]

Explanation:

$\rule{300}{1.5}$

The given first equation represents the standard from of equation, But second one is not representing the Standard equation.

First Equation:

$\displaystyle\Rightarrow\sf Y_{1}=4\;\sin\Bigg\lgroup 3\;\pi\;t+\Bigg(\dfrac{\pi}{3}\Bigg)\Bigg\rgroup\\\\\\\Rightarrow\sf Y_{1}=4\;\sin\Bigg\lgroup 3\;\pi\;t+\Bigg(\dfrac{\pi}{3}\Bigg)\Bigg\rgroup\quad\dfrac{\quad}{}\;[1]$

Second Equation:

$\displaystyle\Rightarrow\sf Y_{2}=4\;\bigg\lgroup \sin (3\;\pi\;t)+\sqrt{3}\;\cos(3\;\pi\;t)\bigg\rgroup$

Solving it,

$\displaystyle\Rightarrow\sf Y_{2}=\bigg\lgroup 4\; \sin (3\;\pi\;t)+4\;\sqrt{3}\;\cos(3\;\pi\;t)\bigg\rgroup$

• Multiply and divide it by 2

$\displaystyle\Rightarrow\sf Y_{2}=\bigg\lgroup 4\; \sin (3\;\pi\;t)+4\;\sqrt{3}\;\cos(3\;\pi\;t)\bigg\rgroup\times \dfrac{2}{2}\\\\\\\Rightarrow\sf Y_{2}=\Bigg\lgroup \dfrac{(4\times 2)\; \sin (3\;\pi\;t)}{2}+\dfrac{(4\times 2) \;\sqrt{3}\;\cos(3\;\pi\;t)}{2}\Bigg\rgroup\\\\\\\Rightarrow\sf Y_{2}=\Bigg\lgroup \dfrac{8\; \sin (3\;\pi\;t)}{2}+\dfrac{8\;\sqrt{3}\;\cos(3\;\pi\;t)}{2}\Bigg\rgroup$

$\displaystyle\Rightarrow\sf Y_{2}=8\Bigg\lgroup \dfrac{\sin (3\;\pi\;t)}{2}+\dfrac{\sqrt{3}\;\cos(3\;\pi\;t)}{2}\Bigg\rgroup\\\\\\\Rightarrow\sf Y_{2}=8\Bigg\lgroup \dfrac{1}{2}\times \sin (3\;\pi\;t)+\dfrac{\sqrt{3}}{2}\times\cos(3\;\pi\;t)\Bigg\rgroup$

Since we know,

• 1/2 = cos 60° = cos (π/3)
• √(3)/2 = sin 60° = sin (π/3)

Substituting,

$\displaystyle\Rightarrow\sf Y_{2}=8\Bigg\lgroup\cos\Bigg(\dfrac{\pi}{3}\Bigg)\times \sin (3\;\pi\;t)+\sin\Bigg(\dfrac{\pi}{3}\Bigg) \times\cos(3\;\pi\;t)\Bigg\rgroup$

From the Identity we know,

• $\displaystyle\sf (\cos\;A\;.\;\sin\;B) + (\sin\;A\;.\;\cos\;B)=\sin(a+b)$

Comparing with the above equation we get,

$\displaystyle\Rightarrow\sf Y_{2}=8\;\sin\Bigg\lgroup 3\;\pi\;t+\Bigg(\dfrac{\pi}{3}\Bigg)\Bigg\rgroup\\\\\\\Rightarrow\sf Y_{2}=8\;\sin\Bigg\lgroup 3\;\pi\;t+\Bigg(\dfrac{\pi}{3}\Bigg)\Bigg\rgroup\quad\dfrac{\quad}{}\;[2]$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, Ratio

$\displaystyle\Rightarrow\sf Equation\;[1]\;\colon Equation\;[2]\\\\\\\Rightarrow\sf Y_{1}\;\colon Y_{2}=4\;\sin\Bigg\lgroup 3\;\pi\;t+\Bigg(\dfrac{\pi}{3}\Bigg)\Bigg\rgroup\;\colon 8\;\sin\Bigg\lgroup 3\;\pi\;t+\Bigg(\dfrac{\pi}{3}\Bigg)\Bigg\rgroup\\\\\\\Rightarrow\sf Y_{1}\;\colon Y_{2}=4\;\cancel{\sin\Bigg\lgroup 3\;\pi\;t+\Bigg(\dfrac{\pi}{3}\Bigg)\Bigg\rgroup}\;\colon 8\;\cancel{\sin\Bigg\lgroup 3\;\pi\;t+\Bigg(\dfrac{\pi}{3}\Bigg)\Bigg\rgroup}$

$\displaystyle\Rightarrow\sf Y_{1}\;\colon Y_{2}=4\;\colon 8\\\\\\\Rightarrow\sf Y_{1}\;\colon Y_{2}=1\;\colon 2\\\\\\\Rightarrow \large{\underline{\boxed{\red{\sf Y_{1}\;\colon Y_{2}=1\;\colon 2}}}}$

∴ The ratio of Y₁ : Y₂ is 1 : 2.

$\rule{300}{1.5}$

Answer 2

$\huge\underline\mathtt\red{Answer:-}$

The ratio of Y₁ : Y₂ is 1 : 2