Oscillation of a drumhead membrane

[A13235378]

Homework Statement

A horizontal membrane oscillates harmoniously along a vertical axis with a frequency equal to w. Determine the amplitude of the oscillations, if a grain of sand that is on the membrane, when jumping from it, reaches a maximum height of H in relation to the equilibrium position of the membrane.

Relevant Equations

Energy conservation.

w^2 = k/m.

My attempt,

Considering that it jumps in the maximum compression position:

$$\frac{kA^2}{2} = mg(H+A)$$

replacing k / m with w ^ 2 :

$$A^2 w^2-2gA-2gH=0$$

Solving the second degree equation:

$$A=\frac{2g+\sqrt{4g^2+8gHw^2}}{2w^2}$$

But the answer is:

$$A=\frac{g}{w}\sqrt{\frac{2H}{g}-\frac{1}{w^2}}$$

After that, I can't go on. I've tried to develop to arrive at the equality of both expressions, but I couldn't. Where am I going wrong?

 

[PeroK]

Homework Statement:: A horizontal membrane oscillates harmoniously along a vertical axis with a frequency equal to w. Determine the amplitude of the oscillations, if a grain of sand that is on the membrane, when jumping from it, reaches a maximum height of H in relation to the equilibrium position of the membrane.
Relevant Equations:: Energy conservation.

w^2 = k/m.

My attempt,

Considering that it jumps in the maximum compression position:

$$\frac{kA^2}{2} = mg(H+A)$$

That's a false assumption.

The first thing to try is to assume that the sand leaves the membrane at the equilibrium point. See what you get.

If that's not the right answer, then have another think about when the sand does actually leave the membrane.

 

[haruspex]

Not sure what your thinking is wrt the k and m you refer to.
From the RHS of your equation, it looks like m is the mass of the grain of sand. But that is unrelated to w. The oscillation of the membrane is not affected by the sand grain.

As @PeroK notes, you need to figure out the point in the cycle where the grain will lose contact with the surface. Think about forces and accelerations at that point.

 

[kuruman]

@A13235378:
To elaborate on what has already been said.

As @PeroK notes, you need to figure out the point in the cycle where the grain will lose contact with the surface. Think about forces and accelerations at that point.

A free body diagram of the grain at the time of separation will be extremely helpful when following the suggestions above. Remember that, if an object oscillates harmoniously, its acceleration $a$ as a function of displacement $x$ from the equilibrium position is $a(x)=-\omega^2 x$.