How Does the Period of Oscillation Change When a Block Detaches from the Wall?

[rude man]

d²x₁/dt² = (k/m)(x2-x1-L)

d²x₂/dt² = -(k/m)(x2-x1-L)

What next ?

Yay team! OK now, you need to subtract one equation from the other to form a single diff. eq. in (x2 - x1). Think carefully about your initial conditions. They're very simple for x1, more elaborate for x2 since x2 starts at x2=L and it also has an initial velocity when x2 = L.

In other words, you're forming a new variable x = x2 - x1. x is now the distance between the two masses.

PS you don't need to compute the initial velocity of the right-hand mass. Just call it x-dot-0 or whatever.

 

[Vibhor]

Yay team! OK now, you need to subtract one equation from the other to form a single diff. eq. in (x2 - x1). Think carefully about your initial conditions. They're very simple for x1, more elaborate for x2 since x2 starts at x2=L and it also has an initial velocity when x2 = L.

In other words, you're forming a new variable x = x2 - x1. x is now the distance between the two masses.

d²(x₂-x₁)/dt² = -(2k/m)(x2-x1-L)

If x=x2-x1 ,then

d²x/dt² = -(2k/m)(x-L)

or, d²x/dt² + (2k/m)x = 2kL/m

At t=0 (i.e when left block leaves the wall) , x = L , but how do I know what is dx/dt at t=0 ?

What next ?

 

[rude man]

d²(x₂-x₁)/dt² = -(2k/m)(x2-x1-L)

If x=x2-x1 ,then

d²x/dt² = -(2k/m)(x-L)

At t=0 (i.e when left block leaves the wall) , x = L , but how do I know what is dx/dt at t=0 ?

What next ?

You don't have to know the actual number of dx/dt(t=0). Just call it something (like x-dot-0).
Next is solving your diff. eq. Have you done diff eq's? If not you should look at the diff. eq. for simple harmonic motion and the frequency should pop right out for you. Your big hint is the term 2k/M. Remember what the frequency was if the diff. eq. is d²x/dt² + (k/M)x = constant? That would be a single mass dangling on a single spring with some initial deflection of the mass away from its rest position. That's all you really need to answer the problem.

 

[Vibhor]

Thank you rude man for the guidance.

You don't have to know the actual number of dx/dt(t=0). Just call it something (like x-dot-0).

We can find dx/dt(t=0) .The right block is compressed by D initially .Applying the conservation of energy ,

(1/2)mv₂² = (1/2)kD²

dx/dt at t=0 is equal to the speed of the right block when the left block leaves the wall.

dx/dt(at t=0) = D√(k/m)

So now we have the initial conditions at t=0 -> x=L ,dx/dt = D√(k/m) , d²x/dt² = 0

Do you think this is correct ?

 

[SammyS]

...

In this case, it's only in the frame of the CM, that there is truly periodic motion.

I should have said,

The only inertial frame for which the motion is truly periodic, is the Center of Mass frame.


A frame of reference fixed to either mass is a non-inertial frame. The problem can certainly be solved in such a reference frame, but beware of the possibility of fictitious forces.

 

[Vibhor]

Thank you Sammy .

Is post#34 correct ? Are the initial conditions correct ?

I have basic knowledge about differential equations so I have a doubt .

The solution of d²x/dt² + (2k/m)x =0 is of the form is x=Asin(ωt+θ) but what is the solution of d²x/dt² + (2k/m)x = (2kL/m) considering there is a constant term (2kL/m) on the right hand side ?

 

[ehild]

In case of such two-body problems one can introduce new coordinates. One is connected to the centre of mass X_CM=(m1x1+m2x2)/(m1+m2). The other is the difference of the position: here it is L=x2-x1.

Transforming the equations of motion you get that (m1+m2)a_CM=∑Fi=N where N is the normal force of the wall, till contact is maintained. After loosing contact, the CM moves with constant velocity. That correspond to zero angular frequency.

After loosing contact with the wall, m1a1=k(x2-x1-Lo), m2a2=-k(x2-x1-Lo) Using the notation u=x2-x1-Lo→d^2u/dt^2= -k(1/m1+1/m2). 1/m1+1/m2=1/μ, the reduced mass. u obeys the equation characteristic to SHM: μd²u/dt²=-ku. In our case μ=M/2, ω²=k/μ=2k/M.

ehild

 

[ehild]

Thank you Sammy .

Is post#34 correct ? Are the initial conditions correct ?

I have basic knowledge about differential equations so I have a doubt .

The solution of d²x/dt² + (2k/m)x =0 is of the form is x=Asin(ωt+θ) but what is the solution of d²x/dt² + (2k/m)x = (2kL/m) considering there is a constant term (2kL/m) on the right hand side ?

The problem does not ask the exact motion, you need to give the time period only.

The solution of an inhomogeneous linear differential equation is equal to x_h+X, where x_h is the general solution of the homogeneous equation, (with zero on the right-hand side). It is x_h=A(sinωt+θ) now.X is the particular solution of the original equation. In this case, it is a constant:X=L. So the whole solution is x=L+Asin(ωt+θ) .

ehild

 

[Vibhor]

Thank you ehild .This is probably the first time I am dealing with DE's .You have very nicely explained it .

Please have a look at post#34 and let me know if I have correctly found the initial conditions i.e at t=0 x=L ,dx/dt = D√(k/m) , d²x/dt² = 0 .

 

[rude man]

Thank you rude man for the guidance.



We can find dx/dt(t=0) .The right block is compressed by D initially .Applying the conservation of energy ,

(1/2)mv₂² = (1/2)kD²

dx/dt at t=0 is equal to the speed of the right block when the left block leaves the wall.

dx/dt(at t=0) = D√(k/m)

So now we have the initial conditions at t=0 -> x=L ,dx/dt = D√(k/m) , d²x/dt² = 0

Do you think this is correct ?

A second-degree diff. eq. only has 2 independent initial conditions, any other is redundant. But yes, what you wrote otherwise is correct. As I said before, you didn't have to actually compute dx/dt(0) but of course it doesn't hurt!

 

[rude man]

The solution of d²x/dt² + (2k/m)x =0 is of the form is x=Asin(ωt+θ) but what is the solution of d²x/dt² + (2k/m)x = (2kL/m) considering there is a constant term (2kL/m) on the right hand side ?

Hint: x can never be < 0. Yet a sinusoidal motion in x goes negative as well as positive. The presence of the 2kL/m term keeps x > 0 for all t. BTW all along we have assumed that L > D, obviously.

 

[ehild]

We can find dx/dt(t=0) .The right block is compressed by D initially .Applying the conservation of energy ,

(1/2)mv₂² = (1/2)kD²

dx/dt at t=0 is equal to the speed of the right block when the left block leaves the wall.

dx/dt(at t=0) = D√(k/m)

When the left block leaves the wall, also the compression of the spring is zero. So the total energy is that of the kinetic energy of the right block, your equation is correct.

ehild

 

[Vibhor]

When the left block leaves the wall, also the compression of the spring is zero. So the total energy is that of the kinetic energy of the right block, your equation is correct.

ehild

Can I write the solution of DE as x=(D/√2)sin(2k/m)t + L ? The angular frequency is (2k/m) and the amplitude as (D/√2) ?

Can I infer that the maximum distance between the blocks during the motion will be L+ (D/√2) and the minimum distance will be L - (D/√2) ?

 

[ehild]

Can I write the solution of DE as x=(D/√2)sin(2k/m)t + L ? The angular frequency is (2k/m) and the amplitude as (D/√2) ?

The angular frequency is $\omega=\sqrt{2k/m}$, so the solution is $x=(D/\sqrt2)sin\left(\sqrt{\left(2k/m\right)}\cdot t\right)+L$

Can I infer that the maximum distance between the blocks during the motion will be L+ (D/√2) and the minimum distance will be L - (D/√2) ?

Yes, it is true.

ehild

 

[rude man]

Can I write the solution of DE as x=(D/√2)sin(2k/m)t + L ? The angular frequency is (2k/m) and the amplitude as (D/√2) ?

Can I infer that the maximum distance between the blocks during the motion will be L+ (D/√2) and the minimum distance will be L - (D/√2) ?

Looks fine!

 

[Vibhor]

Sorry for the late response . I was busy with mid term tests .

Thanks everyone for helping me with the problem ,especially ehild and rude man :)

One thing that still bothers me is that we have calculated frequency of oscillations of the distance between the blocks , rather than displacement of a block from its equilibrium position .

Are the two frequencies same ?

 

[ehild]

One thing that still bothers me is that we have calculated frequency of oscillations of the distance between the blocks , rather than displacement of a block from its equilibrium position .

Are the two frequencies same ?

The two blocks are connected to the ends of the vibrating spring, so they must vibrate with the same frequency as the spring length change.

There is no equilibrium position for any of the blocks, as the CM of the system translates. The blocks vibrate with respect to the CM.


ehild

 

[rude man]

Sorry for the late response . I was busy with mid term tests .

Thanks everyone for helping me with the problem ,especially ehild and rude man :)

One thing that still bothers me is that we have calculated frequency of oscillations of the distance between the blocks , rather than displacement of a block from its equilibrium position .

Are the two frequencies same ?

Each block vibrates with the same frequency as that of the difference distance, but in addition each block moves with constant velocity in the direction of initial motion of the right-hand block.