Interpret the setup

This is the standard arrangement of a mass $M$ attached between two springs of spring constants $k_1$ and $k_2$, with the junction point between the two springs labeled $P$.

Let:

• displacement of mass $M$ from equilibrium be $x$
• displacement of point $P$ from equilibrium be $y$

We are given that the mass oscillates with amplitude $A$, so the amplitude of $x$ is $A$.

Relation between spring forces

Point $P$ is massless, so at every instant the net force on it must be zero.

Hence the force in the left spring equals the force in the right spring.

• Extension/compression of spring $k_1$ is $y$
• Extension/compression of spring $k_2$ is $x-y$

Therefore, $k_1 y = k_2 (x-y)$

Solve for $y$ in terms of $x$

$k_1 y = k_2 x - k_2 y$ $k_1 y + k_2 y = k_2 x$ $(k_1+k_2)y = k_2 x$ $y = \frac{k_2}{k_1+k_2}x$

Find amplitude of point $P$

Since $y$ is proportional to $x$, the amplitude of $P$ is $\frac{k_2}{k_1+k_2}A$

Match with options

This corresponds to: $\boxed{\frac{k_2 A}{k_1+k_2}}$

So the correct option is D.