Block Hitting Spring: Time to Travel x=0 to x1

[Tanero]

Homework Statement

A small block of mass m=1 kg and v0=6.32m/s hits a massless relaxed spring with spring constant k= 3 N/m, which starts to be compressed as the block continues to move horizontally. There is friction between the block and the horizontal surface, and it is not uniform. As a function of distance, the friction coefficient varies like μ(x)=αx, with α= 0.7 m^−1. For simplicity that static and dynamic friction coefficients are the same.

What time t1 does it take for the block to travel between x=0 (relaxed spring) and x=x1 (block at first stop)? (in seconds)

https://courses.edx.org/static/content-mit-801x~2013_SOND/html/final_p06.png

Homework Equations

I chose x-axis positive to the right. x0=0 at point of contact, x1=first momentarily stop
F_friction=-μmg=-αxmg
F_spring=-kx

Newton's Second Law
∑F=m*d2x/dt2

The Attempt at a Solution

-kx-αxmg=m*d2x/dt2
-(k+αmg)x=m*d2x/dt2

new konstant K=k+αmg
-K/m=d2x/dt2

This is SHO with general solution x(t)=Acos(ωt+θ)
ω=√(K/m)=3.16

To find A, and θ our initial conditions are:
x0=x(t=0)=0
v0=v(t=0)=6.32 m/s

Thus, 0=x(t=0)=Acos(θ)
This is true when θ=pi/2 or -3/2pi . Question: which angle should I use ?

v0=v(t=0)=6.32/-ω=Asin(θ)

A=√(x(t=o)^2+(v(t=0)/ω)^2)=√(0+(6.32/3.16)^2/)=2

So, I know that my amplitude is 2, x1=2m.

(Trying with 3/2pi. pi/2 can't feet because I am getting negative time.)

2=x(t1)=2*cos(3.16t-3/2pi)

1=cos(3.16t-3/2pi)
This is true when (3.16t-3/2pi)=0
t1=1.49 sec.

Please, check my solution if it seemed to be right.
And my t1 should be also half a period T=pi/ω=pi/3.16=0.99s. Isn't it? I am confused.

 

[voko]

This is not SHO because there is friction between the block and the surface.

 

[Tanero]

So, what approach should I take instead?

 

[voko]

Do you have any ideas?

 

[Tanero]

I have found how far it travels using work - energy theorem. May be using equations of motion: x(t)=x0+v0t+1/2at^2, v(t)=v0+at. and for acceleration put acceleration from N2L -(k+αmg)/m*x=d2x/dt2. May be then I can find my t?

 

[voko]

I have re-read the problem and your original solution, and I must say that I misunderstood what you were doing.

While the system is not an SHO, it is not incorrect to treat it as an SHO until the block stops because the equation is the same.

You have found the angular frequency $\omega$ of this system. How does that correspond to the period? What part of the period is spent by an SHO between max kinetic energy and max potential energy?

 

[Tanero]

You have found the angular frequency $\omega$ of this system. How does that correspond to the period? What part of the period is spent by an SHO between max kinetic energy and max potential energy?

T=2pi/ω. Between those two conditions block on spring spends half period T/2=pi/ω ?

 

[voko]

Correct. So what is your answer?

 

[Tanero]

T=pi/ω=pi/3.16=0.99 s.

But I tried using equations of motion: v(t)=v0+at. for acceleration I put acceleration from N2L -(k+αmg)/m*x=d2x/dt2.

0=v(t1)=vo-(k+αmg)/m*x*t
x1=2
t1=.316 s.

Something wrong.

 

[voko]

Your equation of motion is $$ - {k + \alpha m g \over m} x = {d^2 x \over dt^2} .$$ This is a differential equation and its solution cannot be obtained by multiplying it by $t$. In fact, its solution is given by $x = A \cos (\omega t + \theta)$ as you wrote originally.

 

[Tanero]

But this equation v(t)=v0+at is also differential v(t)=v0+d2x/dt2*t . I have acceleration in both equations. I substitute one into another. Wanna understand why this is not right thing to do?

 

[voko]

$v(t) = v_0 + at$ is only correct when $a = \mathrm{const}$. Which is not the case here. In this case, we have $v(t) = -A \omega \sin (\omega t + \theta)$, and $a = -A \omega^2 \cos (\omega t + \theta) \ne \mathrm{const}$.

 

[Tanero]

ahhh! sure, a is constant! that's why you can't. Thanks!

 

[Geldof]

T=2pi/ω. Between those two conditions block on spring spends half period T/2=pi/ω ?

Could either of you please explain why half of the period should be used?

 

[dummyano]

I found the max displacement for the mass via the energy work theorem, basically
-> Potential Energy became Kinetic Energy at the end of the track
-> in the collision no energy is lost as it is stated by the problem, so
-> it is KE = Work F_spring + Work F_friction
-> they both change with x so i took the integral from x0 to x1 for both the Work done
that way I found max displacement.

1) is that correct?
2) is it possible to find the time without SHO equations?

 

[Tanero]

@ dummyano
1) Yes, the way you found x is correct.
2) It seems you can't find t without SHO

I am very much in doubt about finding t from T/2 . It doesn't agree with t found from solving SHO general solution equation. See above. I am stuck.

 

[Tanero]

Could either of you please explain why half of the period should be used?

because it when object slides to the right it makes half a period.

 

[Tanero]

You have found the angular frequency $\omega$ of this system. How does that correspond to the period? What part of the period is spent by an SHO between max kinetic energy and max potential energy?

I am very much in doubt about finding t from T/2 . It doesn't agree with t found from solving SHO general solution equation. See above.

Can you please comment?

 

[dummyano]

@ dummyano
1) Yes, the way you found x is correct.
2) It seems you can't find t without SHO

I am very much in doubt about finding t from T/2 . It doesn't agree with t found from solving SHO general solution equation. See above. I am stuck.

Thank you for answering.
I will try SHO for t and post the result then

 

[TSny]

What fraction of a period does it take for a mass in SHM to travel from the equilibrium position to the first time it comes to rest?

 

[TSny]

To find A, and θ our initial conditions are:
x0=x(t=0)=0
v0=v(t=0)=6.32 m/s

Thus, 0=x(t=0)=Acos(θ)
This is true when θ=pi/2 or -3/2pi .

pi/2 and -3pi/2 are effectively the same angle. (They differ by a full cycle.)

Try to find another choice for θ.

 

[voko]

I am very much in doubt about finding t from T/2 . It doesn't agree with t found from solving SHO general solution equation. See above.

Can you please comment?

Your doubts are right on. A simple harmonic oscillator goes like this:

(0) max kinetic energy, velocity directed to the right;

(1) max potential energy, no velocity;

(2) max kinetic energy, velocity directed to the left;

(3) max potential energy, no velocity;

(4) max kinetic energy, velocity directed to the right; this completes the period.

In this case, you are interested in the transition from (0) to (1). How much of the period is that?

 

[Tanero]

Your doubts are right on. A simple harmonic oscillator goes like this:

(0) max kinetic energy, velocity directed to the right;

(1) max potential energy, no velocity;

(2) max kinetic energy, velocity directed to the left;

(3) max potential energy, no velocity;

(4) max kinetic energy, velocity directed to the right; this completes the period.

In this case, you are interested in the transition from (0) to (1). How much of the period is that?

wow! voko! you are opening my eyes on SHO! I never thought of period like this. So, it's T/4? right?

 

[Tanero]

pi/2 and -3pi/2 are effectively the same angle. (They differ by a full cycle.)

Try to find another choice for θ.

Well, I was trying to find other choices, but only these two angles make x(t=0)=0 .
Moreover, problem is that when I put pi/2 to find my time I get negative time. That's why i chose 3/2pi to overcome it.

Can you suggest the angle? I don't see. May be I can find angle from v0=v(t=0) part of solution. I know A, v0, w, then try to find angle through arcsin...

 

[voko]

So, it's T/4? right?

Yes, T/4 should be right. Sorry for not spotting the mistake earlier.

 

[voko]

Can you suggest the angle? I don't see. May be I can find angle from v0=v(t=0) part of solution. I know A, v0, w, then try to find angle through arcsin...

Try thinking logically. You know that the block is at $x = 0$ at $t_0 = 0$. Thus you know that $\cos (\phi_0) = 0$, where $\phi_0 = \omega t_0 + \theta = \theta$ is the initial phase of the oscillation. The block then goes to the farthest point at $x = A$; this clearly must correspond to $\cos (\phi_1) = 1$, where $\phi_1 = \omega t_1 + \theta$ is the final phase of the oscillation. Look at the unit circle: where does the cosine function goes from zero to 1 as the angle increases? What is the angle difference? This difference then must be equal to the phase difference $\phi_1 - \phi_0 = \omega t_1$.

 

[dummyano]

Thank you for answering.
I will try SHO for t and post the result then

This is my result for SHO, thank both you and voko for the hint on period!

Data
V_0= sqrt(40)
K = 8
%mu = 0.8 x

Results
Max displacement = sqrt(5/2)

for SHO I did
m(d^2x/dt^2) = -kx - Ff
(d^2x/dt^2) = -16x
w = 4
T = pi / 8

 

[TSny]

Can you suggest the angle? I don't see. May be I can find angle from v0=v(t=0) part of solution. I know A, v0, w, then try to find angle through arcsin...

Voko has given a good suggestion. You could also graph the cosine function. You know that the cosine equals 0 when its argument is -3pi/2 or pi/2. Is there an value between those which will also make the cosine 0?