A planar four-bar linkage mechanism with 3 revolute kinematic pairs and 1 prismatic kinematic pair is shown in the figure, where AB ⊥ CE and FD ⊥ CE. The T-shaped link CDEF is constructed such that the slider B can cross the point D, and CE is sufficiently long. For the given lengths as shown, the mechanism is

The correct answer is: a non-Grashof chain with all oscillating links.

Step-by-step Explanation:

1. Identify the given parameters from the diagram:
Let us note the lengths and configuration of the links as shown in the figure:
• Ground link $\mathrm{FG}$ (distance between fixed pivots $G$ and $F$): $L_{\mathrm{FG}}=3 \mathrm{cm}$
• Link $\mathrm{AG}$ (connected to ground pivot $G$): $L_{\mathrm{AG}}=5 \mathrm{cm}$
• Link $\mathrm{AB}$ (connecting rod): $L_{\mathrm{AB}}=3 \mathrm{cm}$
• T-shaped link $\mathrm{CDEF}$ is pivoted at ground point $F$. The line $\mathrm{FD}$ is perpendicular to the slider guide line $\mathrm{CE}$ ($\mathrm{FD}\perp \mathrm{CE}$), with length: $L_{\mathrm{FD}}=1.5 \mathrm{cm}$

2. Analyze the range of motion and limits:
The slider $B$ is constrained to slide along the guide line $\mathrm{CE}$ of the rotating T-shaped link $\mathrm{CDEF}$. Since $\mathrm{FD}\perp \mathrm{CE}$ and the distance $\mathrm{FD}=1.5 \mathrm{cm}$, the straight line $\mathrm{CE}$ always remains tangent to a circle of radius $1.5 \mathrm{cm}$ centered at the fixed pivot $F$.

Let us define a coordinate system centered at the pivot $F(0,0)$ with the x-axis aligned along the line $\mathrm{FG}$, placing the fixed pivot $G$ at coordinates $(-3,0)$.

The equation of the line $\mathrm{CE}$ oriented at an angle $\theta$ relative to the x-axis is:
$x\cos \theta +y\sin \theta =1.5$

The distance from the fixed pivot $G(-3,0)$ to the line $\mathrm{CE}$ is given by:
$d(G,\mathrm{CE})=\left(-3\cos \theta -1.5\right)$

Since point $A$ rotates about $G$ with a radius of $5 \mathrm{cm}$, it traces a circle of radius $5 \mathrm{cm}$ centered at $G$. For a configuration to be physically possible, the link $\mathrm{AB}$ of length $3 \mathrm{cm}$ must be able to span the distance between the circle traced by $A$ and the slider line $\mathrm{CE}$.

This means the minimum distance from the circle centered at $G$ with radius $5 \mathrm{cm}$ to the line $\mathrm{CE}$ must not exceed the length of the link $\mathrm{AB}$ ($3 \mathrm{cm}$). Mathematically:
$5-d(G,\mathrm{CE})\le 3\Rightarrow d(G,\mathrm{CE})\ge 2 \mathrm{cm}$

Substituting the distance formula:
$\left|3\cos \theta +1.5\right|\ge 2$

However, if the angle $\theta$ is such that:
$-2<3\cos \theta +1.5<2\Rightarrow -1.17<\cos \theta <0.167$
the distance $d(G,\mathrm{CE})$ becomes less than $2 \mathrm{cm}$. In this range of angles, the slider line $\mathrm{CE}$ passes too close to the pivot $G$, making it impossible for the $3 \mathrm{cm}$ link $\mathrm{AB}$ to reach it from the $5 \mathrm{cm}$ circle. Consequently, the T-shaped link $\mathrm{CDEF}$ cannot rotate fully and is constrained to oscillate.

Similarly, due to the geometric limits at the extreme extensions of the links, the crank link $\mathrm{AG}$ also cannot complete a full rotation without causing the transmission angles to lock. Therefore, none of the links can undergo complete rotation, meaning they all oscillate.

Conclusion:
Since no link is capable of performing a full $360°$ rotation, the mechanism is classified as a non-Grashof chain with all oscillating links.