# Maximum mechanical energy in the spring-block system

1. Aug 10, 2015

### Vibhor

1. The problem statement, all variables and given/known data

2. Relevant equations

3. 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 .

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2. Aug 10, 2015

### BvU

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

3. Aug 10, 2015

### Vibhor

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

4. Aug 10, 2015

### 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 ?

5. Aug 10, 2015

### Vibhor

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

6. Aug 10, 2015

### Staff: Mentor

That's what I would say.

7. Aug 10, 2015

### Vibhor

The answer given is $2mv^2$ .

8. Aug 10, 2015

### Staff: Mentor

9. Aug 10, 2015

### Vibhor

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

10. Aug 10, 2015

### TSny

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

11. Aug 10, 2015

### Staff: Mentor

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

12. Aug 10, 2015

### Vibhor

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

How should I approach this problem ?

Last edited: Aug 10, 2015
13. Aug 10, 2015

### 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.]

Last edited: Aug 10, 2015
14. Aug 10, 2015

### Vibhor

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 ?

15. Aug 10, 2015

### TSny

Yes, that looks good.

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

16. Aug 10, 2015

### 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 ?

Last edited: Aug 10, 2015
17. Aug 10, 2015

### 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 $t$.

18. Aug 10, 2015

### Vibhor

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 ?

Last edited: Aug 10, 2015
19. Aug 10, 2015

### TSny

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

20. Aug 10, 2015

### Vibhor

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

Last edited: Aug 10, 2015