Finding the maximum velocity and PE of the block between 2 springs

[Jenny0000]

Homework Statement

Ihave tried to solve but cannot get the ans for b and c which is 0.16 and 0.0363

Relevant Equations

1/2 kx^2 ，F=ma KE=1/2mv^2

 

[haruspex]

Homework Statement:: Ihave tried to solve but cannot get the ans for b and c which is 0.16 and 0.0363
Relevant Equations:: 1/2 kx^2 ，F=ma KE=1/2mv^2

View attachment 276957

Please post your attempt, per forum rules.

Attempt shown in new thread at https://www.physicsforums.com/threads/finding-the-maximum-velocity-and-pe-of-the-block-between-2-springs.999060/

 

[Jenny0000]

C) P.E=1/2 (12.5)(0.02)^2+1/2 (12.5)(0.02)^2
=0.005 J
B)P.E=1/2 (0.01)^2+1/2(12.5)*(0.01)^2
=0.00125 J
what is the next step?
both answer is 0.16(b) and 0.0363 (c)

 

[haruspex]

Homework Statement:: A block on a smooth table is attached to two stretched identical springs which are fixed to the walls are shown.
The mass of the block is 1.5 kg. The initial extension of each spring is 5.0cm.The effective force constant for the combined spring is 25 Nm^-1.The block is then displaced 4.0 cm to the left and released to perform a simple harmonic motion.Calculate
b)the maximum velocity of the block,
c)the potential energy of the system when the block is 2.0 cm from the mid point.
Relevant Equations:: U=1/2 kx^2+1/2 kx^2

C) P.E=1/2 (12.5)(0.02)^2+1/2 (12.5)(0.02)^2
=0.005 J
B)P.E=1/2 (0.01)^2+1/2(12.5)*(0.01)^2
=0.00125 J
what is the next step?
both answer is 0.16(b) and 0.0363 (c)

I don’t understand your attempt at b, since b asks for velocity. The 'attempt' looks like a PE calculation. Now, that could be a useful step, but it would help if you were to explain your thinking.

For c, please explain how you get the "0.01" displacements. Remember that in the equation U=1/2 kx^2 x is the displacement from the relaxed spring length. The springs in this question do not start relaxed.

For anyone else reading this thread, the original, with diagram, is at https://www.physicsforums.com/threa...-block-between-2-springs.999024/#post-6449965.
@Jenny0000 , it would have been better if you had edited that rather than start a new thread. I'll try to get them merged.

 

[Jenny0000]

b)0.05-0.04=0.01
c)move at 0.02 right?

 

[haruspex]

b)0.05-0.04=0.01

For both springs? And how do you plan to get the max velocity from the initial PE?

c)move at 0.02 right?

For both springs?

 

[kuruman]

I need a clarification. The statement of the problem, after mentioning that each spring is extended from the relaxed position by 5.0 cm, goes on to say that "the effective spring constant for the combined springs is 25 N m^-1." I interpret this to mean that if the mass is displaced from the equilibrium position by $x$, the net force on it due to the combined springs will be $k_{\text{eff}}~x$ in the opposite direction. With the zero of elastic potential energy placed at the equilibrium position, it follows that $U=\frac{1}{2}k_{\text{eff}}~x^2,$ does it not?

 

[haruspex]

I need a clarification. The statement of the problem, after mentioning that each spring is extended from the relaxed position by 5.0 cm, goes on to say that "the effective spring constant for the combined springs is 25 N m^-1." I interpret this to mean that if the mass is displaced from the equilibrium position by $x$, the net force on it due to the combined springs will be $k_{\text{eff}}~x$ in the opposite direction. With the zero of elastic potential energy placed at the equilibrium position, it follows that $U=\frac{1}{2}k_{\text{eff}}~x^2,$ does it not?

Good point, but it leaves me uncertain. It says "effective spring constant for the combined springs", not "effective spring constant for the combined springs under that tension". So it is unclear exactly what is to be taken into account in assessing the effective value.
On balance, I think you are right.

 

[kuruman]

Good point, but it leaves me uncertain. It says "effective spring constant for the combined springs", not "effective spring constant for the combined springs under that tension". So it is unclear exactly what is to be taken into account in assessing the effective value.
On balance, I think you are right.

I shall attempt to remove your uncertainty. Let's look at the force balance in three different cases. Subscripts "L", "R" and "N" respectively stand for "Left", "Right" and "Net". Individual spring constants are the same $k$.
Case I.
Springs relaxed - mass in the middle
$F_L=0~;~~F_R=0~;~~F_N=0+0=0$.

Case II.
Spring anchoring points extended symmetrically to left and right by x_0 - mass in the same position
$F_L=-kx_0~;~~F_R=+kx_0~;~~F_N=-kx_0+kx_0=0$.

Case III.
Mass is displaced to the left by $x$ relative to its previous equilibrium position
$F_L=-k(x_0-x)~;~~F_R=+k(x_0+x)~;~~F_N=-k(x_0-x)+k(x_0+x)=2kx$.

It follows that we can write $F_N=-k_{\text{eff}}~x$, where $k_{\text{eff}}=2k$, and that the equation of motion is independent of whether the springs are initially extended, compressed or relaxed. If the spring constants were different, the equilibrium point will shift away from the mid point and $k_{\text{eff}}=k_L+k_R$.

 

[haruspex]

the equation of motion is independent of whether the springs are initially extended, compressed or relaxed.

Sure, but that doesn’t help with part c. The question remains, is the potential energy to include the residual EPE when at the central position? You are assuming not, but I'm less inclined to agree with you on that.

 

[kuruman]

Sure, but that doesn’t help with part c. The question remains, is the potential energy to include the residual EPE when at the central position? You are assuming not, but I'm less inclined to agree with you on that.

When a problem is asking for the "potential energy" without specifying the reference point, often the case in introductory physics problems, one has to look for the obvious reference point which sometimes may not be that "obvious". In electrostatics infinity is obvious unless otherwise specified. In oscillatory situations the equilibrium point might be obvious because the potential energy is minimum there; anything else is an additive constant which might as well be taken to be zero. A notable exception to this is the massive dipole oscillating in a uniform field where the zero of potential energy is not taken at the equilibrium position where the restoring torque is zero but at the point where the torque is maximum.

Having written the above, I can see how this particular case could be another such exception. We are asked to find the potential energy of the "system". The system consists of the mass and the two springs, therefore the "obvious" zero of potential energy is when the springs are relaxed. Work is done on the system when the springs are stretched, which increases the system's potential energy to $\frac{1}{2}k_{\text{eff}}~x_0^2$.

OK, I talked myself into adopting your point of view.