a) Correct. The new equilibrium position occurs at distance $x=mg/k$ below the initial position (where the collision occurred). Alternatively it is $X=(M+m)g/k$ below the top of the unloaded spring (with no cymbal attached).

b) The new angular frequency of oscillations can be written down without any calculation : $\omega=\sqrt{\frac{k}{m+M}}$. Correct.

However, **your calculation of amplitude is not correct**.

Peak-to-trough is **twice the amplitude** - ie $2A$. Amplitude is the distance of a peak or trough from the new equilibrium position. Peaks and troughs occur when kinetic energy is zero - ie "when it is not oscillating" as you put it. See **Energy Method** below. You had the right idea, but your calculation neglects the fact that the cymbal **is not released from rest** - it has some initial kinetic energy due to the collision. Also your solution does not deal with elastic energy correctly.

**CALCULATION OF AMPLITUDE OF OSCILLATIONS**

**1. Equation of Motion Method**

First work out the speed of the cymbal immediately after the inelastic collision. From conservation of momentum this is $v=\frac{m}{m+M}u$ where $u=\sqrt{2gh}$ is the speed of $m$ immediately before the collision.

Next, the oscillation about the new equilibrium position can be described by the equation of motion $$\xi=A\sin(\omega t+\phi), \dot \xi=\omega A\cos(\omega t+\phi)$$ where $\phi$ is some unknown phase angle. Suppose the collision occurs at $t=0, \xi=x$ with $\dot \xi=v$. (It does not matter what time it occurs, we will only get a different phase angle $\phi$.) Then $$x=A\sin\phi, v=\omega A\cos\phi$$ $$A^2=x^2+(\frac{v}{\omega})^2=(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)}$$

**2. Energy Method**

Let point U be the top of the spring when it has no load. I shall measure all potential and elastic energies from this point. Let P be the starting equilibrium position where the load is $M$, let O be the new equilibrium position when the load is $M+m$, and Q be the lowest point of the subsequent oscillations. The amplitude of oscillations is $OQ=A$. Distance $OP=x$ as already calculated. Other distances are $UP=UO-PO=X-x$ and $UQ=UO+OQ=X+A$.

At P there is kinetic energy of $\frac12 (m+M)v^2$. The spring is compressed by distance $UP=X-x$ so the elastic energy stored in the spring is $\frac12 k(X-x)^2$ where $X=(M+m)g/k$ as above. Gravitational PE at P relative to U is $-(m+M)g(X-x)=-kX(X-x)$.

At Q there is no KE, and gravitational PE is $-(M+m)g(X+A)=-kX(X+A)$. The elastic energy stored here is $\frac12 k(X+A)^2$.

By the conservation of energy, the total energy at P is the same as that at Q. Therefore $$\frac12 (m+M)v^2-kX(X-x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)-kX(X+A)$$ $$(m+M)v^2=k(A^2-x^2)$$ $$A^2=x^2+\frac{(m+M)v^2}{k}=x^2+(\frac{v}{\omega})^2$$ as found using Method 1.

(c) This part of the question is not clear. I assume that the block moves down with the cymbal as in (a) and (b). However, because it is not fixed to the cymbal it can separate from the cymbal when it rises from Q back above O. The block and cymbal do not 'stick together' because of the collision, they only 'move together' after it.

Separation occurs when the downward acceleration of the SHM becomes greater than $g$. Gravity is the only force holding the block in contact with the cymbal, so when gravity is no longer able to supply the required restoring force on the block, it leaves contact with the cymbal.

Suppose separation occurs at point R which has displacement $\xi$ above equilibrium position O. The downward acceleration at point R is $\omega^2 \xi=g$. So $$\xi=\frac{g}{\omega^2}=\frac{(m+M)g}{k}=X$$ This means that the lift-off point is **always** at the top of the spring when it has no load (R=U), whatever the values of $m, M, h, k$. This is a surprising result.

The explanation is that at this instant the only force acting on the cymbal and block is gravity, because the spring is no longer compressed or stretched so it exerts no force. Both cymbal and block are in 'free fall' so the force between them is zero. Just before this instant both cymbal and block are moving upwards but accelerating downwards at just less than $g$ because of a small upward push from the spring. After this instant the cymbal is being accelerated downwards at slightly more than $g$ because of a small pull from the spring. But the spring does not pull down on the block so the block is still accelerating downwards at $g$. There is relative acceleration, so the block and cymbal separate.

**Note** that if $A\lt X$ then the cymbal does not rebound to the relaxed position of the spring at U, then there is no lift-off. The condition for lift-off is that $A \gt X$. Substituting from the equations for $A^2$ and $X$ given above we get $$A^2 \gt X^2$$ $$\frac{m^2g^2}{k^2}+\frac{2ghm^2}{k(M+m)} \gt \frac{(m+M)^2g^2}{k^2}$$ $$1+\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}$$ $$\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}-1=\frac{m^2+2mM+M^2-m^2}{m^2}=\frac{M(2m+M)}{m^2}$$ $$h \gt \frac{M(M+m)(M+2m)g}{2km^2}$$

We can also find the maximum height $H$ above R=U reached by the mass $m$ after lift-off.

The total energy at P is the same as at U, so from above (**Energy Method**) we have $$\frac12 (m+M)v^2+\frac12 k(X-x)^2-kX(X-x)=\frac12 (m+M)V^2$$ $$\frac12 (m+M)(v^2-V^2)=k(X^2-xX)-\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2-x^2)$$ $$v^2-V^2=\frac{k}{(m+M)}(X-x)(X+x)$$ From earlier results we have $$v=\frac{m}{m+M}u=\mu u=\mu \sqrt{2gh}$$ $$v^2=2gh \mu^2$$ $$X-x=\frac{(m+M)g}{k}-\frac{mg}{k}=\frac{Mg}{k}$$ $$X+x=\frac{(m+M)g}{k}+\frac{mg}{k}=\frac{(m+M+m)g}{k}$$ The block rises to a height $H$ above U given by $2gH=V^2$. Making the above substitutions into the equation for $v^2-V^2$ we get $$2g(h\mu^2-H)=\frac{k}{m+M}\frac{Mg}{k} \frac{(m+M+m)g}{k}=\frac{Mg^2}{k}(1+\mu)$$ $$H=h\mu^2-\frac{Mg}{2k}(1+\mu)$$

Does it mean the collision is not completely inelastic? If so, then the block and symbol do not stick together, so they cannot later separate.

The best interpretation which I can make is that after the collision the block moves at the same velocity as the cymbal, so the collision is totally inelastic, but it does not 'stick' to the cymbal, then it can later separate from it.