Damped Oscillators: Homework Solution

[Buffu]

Homework Statement

Homework Equations

The Attempt at a Solution

After the release the block will move towards right and friction will be towards the left.

$M\ddot x = f - kx$

Solving for $x$,

$x = A\cos (\omega t) + B\sin(\omega t) + f/k$

Initial conditions are $x(0) = x_0, \dot x(0) = 0$

$\therefore x(t) = \left(x_0 - \dfrac fk\right)\cos (\omega t) + f/k$

But now how shall I show that amplitude decreases with each oscillation at a constant rate ?

 

[scottdave]

One error that I see is you have Sum of forces = f - kx, which means that you are assuming f is always pointing in the same direction. This is not the case. The friction force (f) will be in the opposite direction of motion.

 

[scottdave]

Here is a way to approach it. You can derive (or get formulas from the textbook) that energy of a spring is (1/2)*k*x², where x is the displacement from equilibrium. Could you make an equation that takes into account the work done by friction (f*d)? Sorry I forgot the squared after x, which has been corrected now.

 

[Buffu]

One error that I see is you have Sum of forces = f - kx, which means that you are assuming f is always pointing in the same direction. This is not the case. The friction force (f) will be in the opposite direction of motion.

I took the left side of the equilibrium point as +ve axis and other side as negative. Since the spring force is pulling towards right, it is -ve and since friction is towards left it is +ve, but I guess the signs will change with each oscillation.

Here is a way to approach it. You can derive (or get formulas from the textbook) that energy of a spring is (1/2)*k*x², where x is the displacement from equilibrium. Could you make an equation that takes into account the work done by friction (f*d)? Sorry I forgot the squared after x, which has been corrected now.

Ok I will try.

 

[Buffu]

One error that I see is you have Sum of forces = f - kx, which means that you are assuming f is always pointing in the same direction. This is not the case. The friction force (f) will be in the opposite direction of motion.

Here is what I did,

$\dfrac 12 kx_0^2 = -\int_0^{x_0 + A} f\cdot dr + \dfrac 12k A^2 \implies kx_0^2 = -2 f(x_0 + A) + k A^2$

Solving for $A$ I got, $A = - x_0$ or $A = \dfrac {2f}k + x_0$,

Second solution say amplitude increased. Where I am wrong ?

 

[haruspex]

Where I am wrong ?

The work done against friction is $+ \int_0^{x_0 + A} f\cdot dr$

 

[Buffu]

The work done against friction is $+ \int_0^{x_0 + A} f\cdot dr$

I felt that should be the case but think of it like this, if friction was not there then the enegry would be $1/2kA^2$, friction is taking some energy out of it and converting it into heat. So I thought it should be -ve.

 

[haruspex]

I felt that should be the case but think of it like this, if friction was not there then the enegry would be $1/2kA^2$, friction is taking some energy out of it and converting it into heat. So I thought it should be -ve.

As far as I can see, you have not defined A. x₀ is the initial amplitude. From your integral, I deduced A represents the amplitude after one half cycle, i.e. from max displacement one way to max the other way.
The initial energy is $\frac 12 kx_0^2$. After one half cycle, $\int _{-x_0}^Af.dr$ has gone to friction and the remainder is $\frac 12kA^2$.

 

[Buffu]

As far as I can see, you have not defined A. x₀ is the initial amplitude. From your integral, I deduced A represents the amplitude after one half cycle, i.e. from max displacement one way to max the other way.
The initial energy is $\frac 12 kx_0^2$. After one half cycle, $\int _{-x_0}^Af.dr$ has gone to friction and the remainder is $\frac 12kA^2$.

Ok so after one full oscillation the amplitude will be $A = x_0 - \dfrac{4f}k$.

 

[haruspex]

Ok so after one full oscillation the amplitude will be $A = x_0 - \dfrac{4f}k$.

Yes.