A four bar mechanism is shown in the figure. The link numbers are mentioned near the links. Input link 2 is rotating anti-clockwise with a constant angular speed ω₂. Length of different links are :

O₂O₄ = O₂A = L,

AB = O₄B = √(2)L.

The magnitude of the angular speed of the output link 4 is ω₄ at the instant when link 2 makes an angle of 90° with O₂O₄ as shown. The ratio ω₄/ω₂ is __________ (round off to two decimal places).

To find the ratio of the angular speeds ${\omega }_4/{\omega }_2$, we can use the concept of instantaneous centers of rotation (I-centers).

Step 1: Identify the links and their configurations from the given image
Let us define the links as follows:
• Link 1: Ground link $O_2{O}_4=L$ (horizontal)
• Link 2: Input link $O_{2}A=L$ (vertical, making an angle of $90°$ with $O_2{O}_4$)
• Link 3: Coupler link $AB=\sqrt{2}L$
• Link 4: Output link $O_{4}B=\sqrt{2}L$

Step 2: Establish the Coordinate System
Let the pivot $O_2$ be at the origin $(0,0)$.
Since $O_2{O}_4$ is along the horizontal axis, the coordinates of $O_4$ are $(L,0)$.
Since the input link $O_{2}A$ is vertical, the coordinates of point $A$ are $(0,L)$.

Step 3: Analyze the Geometry of Triangle $A{O}_{4}B$
The distance between point $A$ and point $O_4$ is:
$A{O}_4=\sqrt{{(L-0)}^2+{(0-L)}^2}=\sqrt{2}L$
Since $AB=O_{4}B=A{O}_4=\sqrt{2}L$, the triangle $A{O}_{4}B$ is an equilateral triangle.
Therefore, all interior angles of $∆A{O}_{4}B$ are $60°$, so:
$\angle O_{4}AB=60°$

In the right-angled isosceles triangle $∆O_2{O}_{4}A$:
$\angle O_{2}A{O}_4=45°$
Therefore, the total angle that the line $AB$ makes with the vertical link $O_{2}A$ is:
$\angle O_{2}AB=\angle O_{2}A{O}_4+\angle O_{4}AB=45°+60°=105°$

Step 4: Locate the Instantaneous Center $I_{24}$
By Kennedy's theorem, the instantaneous center $I_{24}$ lies at the intersection of the line passing through $I_{12}$ and $I_{14}$ (which is the ground line $O_2{O}_4$) and the line passing through $I_{23}$ and $I_{34}$ (which is the coupler line $AB$).
Let $I_{24}$ be the intersection of the line $AB$ with the horizontal line $O_2{O}_4$.
In the right-angled triangle $∆I_{24}{O}_{2}A$, the angle at $A$ is:
$\angle I_{24}A{O}_2=180°-\angle O_{2}AB=180°-105°=75°$

Using trigonometry in the right-angled triangle $∆I_{24}{O}_{2}A$:
$I_{24}{I}_{12}=O_2{I}_{24}=O_{2}A·tan(75°)=L·tan(75°)$

The distance from $I_{24}$ to $I_{14}$ (which is $O_4$) is:
$I_{24}{I}_{14}=O_4{I}_{24}=O_2{I}_{24}+O_2{O}_4=L·tan(75°)+L=L(1+tan(75°\left)\right)$

Step 5: Calculate the Angular Velocity Ratio
According to the angular velocity ratio theorem:
${\omega }_2\left(I_{24}{I}_{12}\right)={\omega }_4\left(I_{24}{I}_{14}\right)$

Substituting the values:
$\frac{{\omega }_4}{{\omega }_2}=\frac{I_{24}{I}_{12}}{I_{24}{I}_{14}}=\frac{Ltan(75°)}{L(1+tan(75°\left)\right)}=\frac{tan(75°)}{1+tan(75°)}$

Since $tan(75°)=2+\sqrt{3}\approx 3.732$:
$\frac{{\omega }_4}{{\omega }_2}=\frac{3.732}{1+3.732}=\frac{3.732}{4.732}\approx 0.7887$

Rounding off to two decimal places, we get:
$\frac{{\omega }_4}{{\omega }_2}\approx 0.79$