Springs and Splitting Mass Concept Confusion

[snowcrystal42]

Am I thinking about this problem correctly?

"A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 3 Hz. The amplitude of the motion is 5.08 x 10^-2 m. At the point where the block has its maximum speed, it suddenly splits into two parts, only one part remaining attached to the string. (a) What are the amplitude and frequency of the simple harmonic motion that exists after the block splits? (b) What about when the block splits while at one of the extreme ends?"

(a) First of all, the answer key I was given has an equation that looks like it relates to elastic potential energy (it says ½mv² = ½kA²). Isn't the point where the speed is at its max also where the box is at its equilibrium position (and in this case where the spring is unstretched) so the elastic potential energy is zero? How would elastic potential energy be related to this otherwise? Shouldn't the object/spring only have translational kinetic energy?

Since Vmax = Aω and ω = √(k/m)

A = Vmax√(m/k)

Since the mass is halved, A is divided by √2 so the answer is just the original amplitude divided by √2, right?

Then if Aω = Vmax, A(2πƒ) = Vmax so ƒ = Vmax / (2πA), so if A is divided by √2 then frequency is multiplied by √2, correct?

(b) This is the part that sort of confuses me. At the ends, V = 0 so there is no kinetic energy; all the energy is in the form of potential energy. If PE = ½(kx²), then mass isn't a part of determining the potential energy, so the potential energy stays the same? If the potential energy doesn't change, why does this mean that the amplitude doesn't change?

From a non-mathematical point of view, why does it matter where you break the block in half?

And if the amplitude doesn't change, why does the frequency change? And how do you know by how much? Is it because ω = 2πƒ = √(k/m) so if the mass is halved then the frequency is multiplied by √2 ?

Thanks!

 

[TSny]

(a) First of all, the answer key I was given has an equation that looks like it relates to elastic potential energy (it says ½mv² = ½kA²). Isn't the point where the speed is at its max also where the box is at its equilibrium position (and in this case where the spring is unstretched) so the elastic potential energy is zero? How would elastic potential energy be related to this otherwise? Shouldn't the object/spring only have translational kinetic energy?

The answer key is probably using conservation of energy and relating the energy at the point of maximum speed to the energy at maximum amplitude.

Since Vmax = Aω and ω = √(k/m)

A = Vmax√(m/k)

Since the mass is halved, A is divided by √2 so the answer is just the original amplitude divided by √2, right?

Yes. The half of the mass that is not attached to the spring has no effect on the motion of the half of the mass that is still attached to the spring.

Then if Aω = Vmax, A(2πƒ) = Vmax so ƒ = Vmax / (2πA), so if A is divided by √2 then frequency is multiplied by √2, correct?

Yes. You could also get this result by using the relation between f and √(k/m) as you note at the end of your post.

(b) This is the part that sort of confuses me. At the ends, V = 0 so there is no kinetic energy; all the energy is in the form of potential energy. If PE = ½(kx²), then mass isn't a part of determining the potential energy, so the potential energy stays the same?

Yes

If the potential energy doesn't change, why does this mean that the amplitude doesn't change?

Think about a mass on a spring and starting the motion by pulling the mass a certain distance D from equilibrium and letting it go from rest. How is the amplitude A related to the distance D? Does your answer depend on the amount of mass that is attached to the spring?

From a non-mathematical point of view, why does it matter where you break the block in half?

When you break the block at the point of maximum speed are you changing the energy of the mass-spring system? When you break the block at the point of maximum ampliltude, are you changing the energy of the mass-spring system?

And if the amplitude doesn't change, why does the frequency change? And how do you know by how much? Is it because ω = 2πƒ = √(k/m) so if the mass is halved then the frequency is multiplied by √2 ?

Yes to your final question.