Oscillating bucket, variable mass

[amiras]

Homework Statement

A 2kg bucket containing 10kg of water is hanging from a vertical ideal spring of force constant 125N/m and oscillating up and down with amplitude 3cm. Suddenly the bucket springs a leak in the bottom such that water drops out of the bucket at a steady rate of 2g/s.

Homework Equations

When the bucket is half full find the period of oscillation.

The Attempt at a Solution

What confuses me about this problem is that book solutions manual simply find the mass of the half full bucket (7kg) and plugs to the equation T = 2pi*sqrt(m/k).

What I initially tried to do is write the general equation of 2nd Newtons law:

dp/dt = -kx => x''m + x'm' + kx = 0, where x' means a derivative with respect of time.

And this is basically the equation with exact form as equation for damped oscillations. It only has replaced damping constant with m'. And it oscillates with frequency
ω' = √[k/m-(1/2m*dm/dt)^2]

However. the book assumes that the angular frequency is ω=√[k/m]

The answers in both cases agree with 5 significant figures, this is because dm/dt is very small. But in the case of dm/dt is significantly big, is this is the right way to attack this kind of problem?

 

[Simon Bridge]

Pretty much what you did was the correct approach - the book used an approximate method which, as you saw, was quite good.

Suspect a misplaced minus sign though ... let's see:
m(t).g - k.x(t) = v(t).dm(t)/dt + m(t).a(t): m(t) = 10 - 2t

 

[takudo_1912]

I think that the period of oscillation didn't depend on amplitude or other effects.It only depend on m and k.So that T=2π.\sqrt{\frac{m}{2k}}

 

[ougoah]

Are you sure you can use the damping equation? The derivative of m(t) is a constant in this example, but m(t) is not. Isn't this required for the damping equation?

 

[Infinitum]

The answers in both cases agree with 5 significant figures, this is because dm/dt is very small. But in the case of dm/dt is significantly big, is this is the right way to attack this kind of problem?

If dm/dt is quite big, then the approximation used by the book will become inaccurate.

Though the rate of mass change, from a small hole, wouldn't normally exceed lot more than that in practical situations.

 

[ehild]

The Attempt at a Solution

What confuses me about this problem is that book solutions manual simply find the mass of the half full bucket (7kg) and plugs to the equation T = 2pi*sqrt(m/k).

What I initially tried to do is write the general equation of 2nd Newtons law:

dp/dt = -kx => x''m + x'm' + kx = 0, where x' means a derivative with respect of time.

The equation is not correct. See: http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Derivation

The momentum of the system bucket and water conserves during that very short time when the drop leaves the bucket. As the drop has the same velocity as the bucket, its falling out has no influence on the velocity of the bucket. The equation of motion is

mx"=F(x),

but you need to take into account that the mass is function of time.
The external force includes both the elastic force and gravity. The change of mass will change the "equilibrium" position of the bucket, so it will move upward and oscillate with changing frequency and amplitude.

 

[amiras]

The equation is not correct. See: http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Derivation

I don't really see why I could not do the same steps with bucket-water as it was done with the rocket-mass and get the same equation?

 

[ehild]

I don't really see why I could not do the same steps with bucket-water as it was done with the rocket-mass and get the same equation?

You can do it, and you get the same equation. But in that equation, the relative velocity of the drop (with respect to to bucket) would appear. And that relative velocity is zero.

ehild

 

[amiras]

Lets say a different case... Imagine a pendulum with a hollow sphere filled with water at the end of the spring. There are two holes in a sphere such that, the water leaks out in the tangential direction of the the sphere motion.

Will the equation of motion be the same as in the case where water cannot leak out?

 

[ehild]

You misunderstood something. The rocket equation is

mdV/dt+v_reldm/dt=ƩF_i,

where v_rel is the relative velocity of the exhaust with respect to the rocket. This is not the same equation as without the exhaust.

In your problem, "leaking" means that the water drops leave the rocket with zero relative velocity. But that is not quite true, as the speed of the water pouring out from a hole depends on the height of the water level in the bucket and the phase of the oscillation. That would be a different system and a different equation, too complicated for me.

ehild

 

[amiras]

I get it now, thanks for your effort. :)