A mass  M , attached to a horizontal spring, executes S.H.M. with amplitude  A1 . When the mass  M  passes through its mean position then a smaller mass  m  is placed over it and both of them move together with amplitude  A2 . The ratio of  (A1A2)  is.

The problem is about Simple Harmonic Motion (S.H.M.) and the conservation of momentum.

Step 1: Understand the problem The mass M is initially oscillating with amplitude A1. When it passes through the mean position, a smaller mass m is placed on it. Now, both the masses move together with a new amplitude A2. We need to find the ratio A1/A2.

Step 2: Apply the conservation of momentum At the mean position, the velocity of the mass M is maximum. Let's denote it as V. When the smaller mass m is placed on it, the system of two masses moves with a common velocity v. According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Therefore, MV = (M + m)v.

Step 3: Relate the amplitudes with the velocities In S.H.M., the maximum velocity V is related to the amplitude A1 and the angular frequency ω by the equation V = ωA1. Similarly, the maximum velocity v of the system of two masses is related to the amplitude A2 by the equation v = ωA2.

Step 4: Substitute the velocities in the momentum equation Substituting the expressions of V and v from step 3 in the momentum equation from step 2, we get ωA1M = ωA2(M + m).

Step 5: Solve for the ratio A1/A2 From the equation in step 4, we can solve for the ratio A1/A2 = (M + m) / M.

So, the ratio of the amplitudes A1 and A2 is equal to the total mass divided by the initial mass.