Maximum mechanical energy in the spring-block system

[Vibhor]

Homework Statement

Homework Equations

The Attempt at a Solution

Suppose the free end of hte spring moves towards right .Consider a reference frame moving with speed 'v' along with the free end of the spring .

From this frame ,the block has an initial speed 'v' towards left . At maximum elongation of spring ,the speed of the block is zero .

Applying Conservation of energy $\frac{1}{2}mv^2 = \frac{1}{2}kx^2$ .

Total mechanical energy at any instant from this frame is $\frac{1}{2}mv^2$ . But this is incorrect .

Could someone help me with the problem .

Thanks .

 

[BvU]

At maximum elongation of spring ,the speed of the block is zero

If the speed is zero in your moving frame, then what is the kinetic energy of the block in the laboratory frame ?

 

[Vibhor]

If the speed is zero in your moving frame, then what is the kinetic energy of the block in the laboratory frame ?

The speed of the block in the lab frame would be 'v' and kinetic energy = $\frac{1}{2}mv^2$

 

[BvU]

So what is the maximum mechanical energy in the spring-block system ?

Apparently it depends on the reference frame...

Did your transformation to the spring-end system help you or did it ultimately just confuse you ?

 

[Vibhor]

So what is the maximum mechanical energy in the spring-block system ?

Does that mean maximum mechanical energy is $\frac{1}{2}mv^2+\frac{1}{2}kx^2 = \frac{1}{2}mv^2+\frac{1}{2}mv^2 = mv^2$ ?

 

[Doc Al]

Does that mean maximum mechanical energy is $\frac{1}{2}mv^2+\frac{1}{2}kx^2 = \frac{1}{2}mv^2+\frac{1}{2}mv^2 = mv^2$ ?

That's what I would say.

 

[Vibhor]

The answer given is $2mv^2$ .

 

[Doc Al]

Hmm... Let me think about this one. What book is this problem from?

 

[Vibhor]

The problem was given in one of the practice sheets . No idea about the book .

Please take your time.

 

[TSny]

Does that mean maximum mechanical energy is $\frac{1}{2}mv^2+\frac{1}{2}kx^2 = \frac{1}{2}mv^2+\frac{1}{2}mv^2 = mv^2$ ?

No. You need to determine at what point of the motion the mass has the most energy in the the "lab" frame of reference.

 

[Doc Al]

No. You need to determine at what point of the motion the mass has the most energy in the the "lab" frame of reference.

TSny is right. The mechanical energy is frame dependent and so is the point of maximum mechanical energy. (My prior statement was wrong. )

 

[Vibhor]

No. You need to determine at what point of the motion the mass has the most energy in the the "lab" frame of reference.

The block has maximum speed when during the motion , the spring becomes unstretched .

How should I approach this problem ?

 

[TSny]

There are different ways to approach it.

In the moving frame, can you write expressions for the kinetic energy and the potential energy as functions of time? (What type of motion does the block have in the moving frame?)

If so, you could try transforming each of these expressions individually to the lab frame and then construct the total energy as a function of time in the lab frame.

[EDIT: Or, instead of worrying with the expressions as functions of time, you might try the following. Express the total instantaneous energy in the lab frame in terms of the total energy in the moving frame and additional terms that depend on v and v', where v is the speed of the moving frame and v' is the instantaneous speed of the mass in the moving frame. You should then be able to deduce the maximum energy in the lab frame.]

 

[Vibhor]

There are different ways to approach it.

In the moving frame, can you write expressions for the kinetic energy and the potential energy as functions of time? (What type of motion does the block have in the moving frame?)

If so, you could try transforming each of these expressions individually to the lab frame and then construct the total energy as a function of time in the lab frame.

In the moving frame block would have SHM .

$E(t) =\frac{1}{2}mv'^2 + \frac{1}{2}kx^2$ ,where $x=Asinωt$ and $v'=Aωcosωt$

In the lab frame , this could be written as ,

$E(t) =\frac{1}{2}m(v'+v)^2 + \frac{1}{2}kx^2$

Does this make any sense ?

 

[TSny]

In the moving frame block would have SHM .

$E(t) =\frac{1}{2}mv'^2 + \frac{1}{2}kx^2$ ,where $x=Asinωt$ and $v'=Aωcosωt$

In the lab frame , this could be written as ,

$E(t) =\frac{1}{2}m(v'+v)^2 + \frac{1}{2}kx^2$

Does this make any sense ?

Yes, that looks good.

Expand out your last equation (without substituting for x and v' in terms of time) and see what you get.

 

[Vibhor]

$E(t) =\frac{1}{2}mv^2 +\frac{1}{2}kA^2+ mvv'$

On putting $E'(t)=0$ , I get t=0 i.e E(t) is maximum at times t=0 , π/ω, 2π/ω ...

At t=0 , $E_{max} =\frac{1}{2}mv^2 +\frac{1}{2}kA^2+ mvAω$

Now from the moving frame we can deduce that maximum potential energy $\frac{1}{2}kA^2$ should be equal to kinetic energy at the mean position (at t=0 in this case) which is equal to $\frac{1}{2}mv^2$ . So $\frac{1}{2}kA^2 =\frac{1}{2}mv^2$ .

Again , from the Moving frame (i.e from SHM equation) , we can deduce maximum speed $Aω=v$ .

Putting everything back in the expression of $E_{max}$ above , $E_{max} = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 + mv^2 = 2mv^2$

Does the reasoning look alright ?

 

[TSny]

I didn't catch it earlier, but your expressions for $x$ and $v'$ in terms of $t$ have a sign error. In the primed frame, is the mass moving toward positive x or negative x at t = 0?

I think you'll find that $E_{max}$ in the lab frame does not occur at $t = 0$ or at $t = 2 \pi/\omega$.

Nevertheless, your reasoning overall looks correct to me and I agree with your final answer for $E_{max}$.

Note that if you go back to $E = \frac{1}{2}mv^2 + \frac{1}{2}kA^2 + mvv'$, you can see by inspection that max E occurs when $v'$ has its maximum value. This should get you the answer without worrying about values of $t$.

 

[Vibhor]

I didn't catch it earlier, but your expressions for $x$ and $v'$ in terms of $t$ have a sign error. In the primed frame, is the mass moving toward positive x or negative x at t = 0?

In the moving frame ,the mass is moving toward negative x (towards left ) at t = 0 .

So , should it be $x=-Asinωt$ and $v'=-Aωcosωt$ ?

Another thing, looking from the moving frame , at t=0 , isn't the mass at the mean position and moving towards the negative extreme position ?

 

[TSny]

In the moving frame ,the mass is moving toward negative x (towards left ) at t = 0 .

So , should it be $x=-Asinωt$ and $v'=-Aωcosωt$ ?

Yes.

Another thing, looking from the moving frame , at t=0 , isn't the mass at the mean position and moving towards the negative extreme position ?

Yes. That agrees with your equations for x and v' above.

 

[Vibhor]

I think you'll find that $E_{max}$ in the lab frame does not occur at $t = 0$ or at $t = 2 \pi/\omega$.

But , E'(t) = 0 gives $\omega t = n\pi$ OR in other words the maximum speed in the SHM is at the mean position ?

 

[TSny]

The derivative of a function is zero at minima as well as maxima. Does t = 0 correspond to a maximum of E(t) or a minimum?

Yes, the maximum speed in SHM (in the moving frame) occurs when the mass passes through the point where the spring is neither stretched nor compressed (i.e., the equilibrium position of the spring). In the lab frame, does the mass have maximum speed every time it passes through the equilibrium position of the spring?

 

[Vibhor]

The derivative of a function is zero at minima as well as maxima.

[EDIT: Or, instead of worrying with the expressions as functions of time, you might try the following. Express the total instantaneous energy in the lab frame in terms of the total energy in the moving frame and additional terms that depend on v and v', where v is the speed of the moving frame and v' is the instantaneous speed of the mass in the moving frame. You should then be able to deduce the maximum energy in the lab frame.]

How do i go with this approach ?

The total energy in the moving frame =$\frac{1}{2}kA^2$

But How do I write total instantaneous energy in the lab frame ? What are the additional terms that depend on v and v'?

 

[TSny]

How do i go with this approach ?

The total energy in the moving frame =$\frac{1}{2}kA^2$

But How do I write total instantaneous energy in the lab frame ?

You did that in post #16. See your first equation there. You can substitute for $v'$ in terms of $t$ to get the explicit time dependence of E. It will simplify nicely. See post #18 for $v'(t)$.

 

[Vibhor]

But how do i interpret $\frac{1}{2}mv^2 + mvv'$ ? What energy these two terms represent ?

 

[TSny]

But how do i interpret $\frac{1}{2}mv^2 + mvv'$ ?

These two terms together represent the difference in total energy in the lab frame and the total energy in the moving frame.

 

[Vibhor]

These two terms together represent the difference in total energy in the lab frame and the total energy in the moving frame.

I understand it now by backward reasoning ,since I know the expression for total energy in lab frame and moving frame . But what if I had approached with this method first .

How do we write total energy in the lab frame ?

Total energy in lab frame = Total energy in moving frame + ?

 

[TSny]

From an earlier post you had $E(t) = \frac{1}{2}m(v+v')^2 + \frac{1}{2}kx^2$. In another post, you have expressions for $x$ and $v'$ as functions of time. So, you should be able to get $E(t)$ as an explicit function of time.

 

[Vibhor]

I was thinking that in post#13 you were giving an alternative way of approaching the problem . But now I realize that the edited part was an alternate/faster way to analyze the expression for the total energy .

Right ?

 

[Vibhor]

.

 

[TSny]

I was thinking that in post#13 you were giving an alternative way of approaching the problem . But now I realize that the edited part was an alternate/faster way to analyze the expression for the total energy .

Right ?

Yes. The first way was to get E(t) explicitly as a function of time. The second was to express E in terms of v and v' and get the answer by inspection.

 

[Vibhor]

One last question , am I right in concluding that during the course of motion the work done by external agent pulling the free end would be both positive and negative ? Alternate positive and negative work would be done by the agent .

??

 

[TSny]

One last question , am I right in concluding that during the course of motion the work done by external agent pulling the freed end would be both positive and negative ? Alternate positive and negative work would be done by the agent .

??

Yes, there are times when the pulling force is doing positive work and times when it is doing negative work. The total work done (starting at t = 0) varies between 0 and 2mv² over and over again. So, the total work done since the start is never negative.

 

[Vibhor]

Thank you very much Sir .