Amplitude of Point Between Two Springs in Simple Harmonic Motion

To find the amplitude of the point $\text{P}$, we analyze the system of two massless springs connected in series:

1. System Configuration:

   • Spring 1 has a spring constant $k_1$ and is connected between the rigid wall and the point $\text{P}$.
   • Spring 2 has a spring constant $k_2$ and is connected between the point $\text{P}$ and the mass $M$.
   • The mass $M$ undergoes simple harmonic motion with amplitude $A$.

2. Equation of Motion for Point $\text{P}$:
   Let the displacements from the respective equilibrium positions at any instant be:

   • $x$ for the mass $M$, where the maximum value of $x$ is the amplitude $A$.
   • $x_{\text{P}}$ for the point $\text{P}$.

   Since the springs are ideal and massless, the net force on the junction point $\text{P}$ must be zero at all times:
   $\Sigma F_{\text{P}} = 0$

   The restoring force exerted by spring 1 on point $\text{P}$ is:
   $F_1 = -k_1 x_{\text{P}}$

   The restoring force exerted by spring 2 on point $\text{P}$ is:
   $F_2 = k_2 (x - x_{\text{P}})$

   Equating these forces gives:
   $k_1 x_{\text{P}} = k_2 (x - x_{\text{P}})$

3. Solving for the Displacement and Amplitude of Point $\text{P}$:
   Rearranging the equation:
   $(k_1 + k_2) x_{\text{P}} = k_2 x$
   $x_{\text{P}} = \frac{k_2}{k_1 + k_2} x$

   Since the mass $M$ oscillates with an amplitude of $A$ (so $x_{\max} = A$), the amplitude of the point $\text{P}$ is:
   $A_{\text{P}} = (x_{\text{P}})_{\max} = \frac{k_2 A}{k_1 + k_2}$

Therefore, the correct option is (D).