# Ambuguous description of a problem

Hi guys!
I have a small problem about a... problem

My physics lecturer gave us a homework task (see below), which is the same as one problem in Paul Tipler's book of Physics (for scientists and engineers, fourth edition) [page 434, task 36]. In the book it is said: "An object of mass m is supported by a vertical spring...". However, my teacher changed just the word "object" with "weightlifter". Everything else is the same! Did he do that intentionally, or just to mislead us? Moreover, he provided a tip: "Don't neglect the gravitational force!"

I got a result, saying the mass is 1.5 kg!!! Can someone check if everything is correct, because I will vomit! :yuck: It took me three days to search for a different solution, but all the results I got were = 1.5 kg!

A "weightlifter" of mass m is supported by a vertical spring (k=1800 N/m). When pulled down 2.5 sm from equilibrium and released, it oscillates at 5.5 Hz. Find m!

OlderDan
Homework Helper
Ramses The Pharaoh said:
Hi guys!
I have a small problem about a... problem

My physics lecturer gave us a homework task (see below), which is the same as one problem in Paul Tipler's book of Physics (for scientists and engineers, fourth edition) [page 434, task 36]. In the book it is said: "An object of mass m is supported by a vertical spring...". However, my teacher changed just the word "object" with "weightlifter". Everything else is the same! Did he do that intentionally, or just to mislead us? Moreover, he provided a tip: "Don't neglect the gravitational force!"

I got a result, saying the mass is 1.5 kg!!! Can someone check if everything is correct, because I will vomit! :yuck: It took me three days to search for a different solution, but all the results I got were = 1.5 kg!

A "weightlifter" of mass m is supported by a vertical spring (k=1800 N/m). When pulled down 2.5 sm from equilibrium and released, it oscillates at 5.5 Hz. Find m!

No idea why the word was changed, but it does not matter what the object is, as long as the shape of the object is not changing during the motion. Perhaps the instructor started to change the problem to something else and got sidetracked. If your answer is correct for "object" it is correct for "weightlifter".

OlderDan, I agree with you up to "If your answer is correct for "object" it is correct for "weightlifter"." :shy: Because as I said I got 1.5 kg, which may be an object's mass, but not a mass of a weightlifter Exactly that is confusing.... following the semantics of coure.
Given the frequency and the spring constant, I use f=(1/2pi)*sqrt(k/m) to obtain the value for m. No gravity is included! Should I use a different formula to take his hint into consideration, or this one should be correct and I should argue with the teacher for better stated problems?

HallsofIvy
Homework Helper
I imagine that "weightlifter" just means some kind of platform that a weight is sitting on.
Unfortunately, the term gives me the image of a very small but powerful man sitting on the spring!

OlderDan
Homework Helper
Ramses The Pharaoh said:
OlderDan, I agree with you up to "If your answer is correct for "object" it is correct for "weightlifter"." :shy: Because as I said I got 1.5 kg, which may be an object's mass, but not a mass of a weightlifter Exactly that is confusing.... following the semantics of coure.
Given the frequency and the spring constant, I use f=(1/2pi)*sqrt(k/m) to obtain the value for m. No gravity is included! Should I use a different formula to take his hint into consideration, or this one should be correct and I should argue with the teacher for better stated problems?

Gravity is included in problems that involve a mass hanging on a spring, but it becomes invisible when you talk about oscillations relative to equlibrium. The potential energy of a spring is a quadratic function of the springs elongation. Gravitational potential energy is a linear function of height. When you combine the two, the potential energy of the combination of spring and gravity becomes a quadratic function of the displacement of the system from the new equlibrium position. The shape of the function is the same regardless of the strength of the gravitational force, but gravity determines the position of the vertex of the potential energy function, i.e. the equilibrium position, and the minimum potential energy.

It is a worthwhile exercise to plot a simple quadratic on your graphing calculator y = x^2 and see what happens when you add a linear term to form y = x^2 + bx (b can be negative). If you like you can also add a c term and adjust c to make the minimum y value always zero. This is effectively what is going on when we say the zero of potential energy of a spring mass system is the equilibrium position and increases with displacement in either direction.

HallsofIvy, I forgot to mention that my teacher even provided a picture with a man, who is standing on that spring and lifts weights ... for those of us, who can't imagine :tongue2: . That creates the imagination of an object with a very big mass!! But whatever...
I just want to ask if am allowed to use the formula in this situation (of an object on a vertical spring) -> f=(1/2pi)*sqrt(k/m) ? Also... is there a formula with the gravity constant included (speaking about "oscillations relative to equlibrium"), that will provide a different solution? :uhh:

Doc Al
Mentor
Ramses The Pharaoh said:
I just want to ask if am allowed to use the formula in this situation (of an object on a vertical spring) -> f=(1/2pi)*sqrt(k/m) ?
That's the one.
Also... is there a formula with the gravity constant included (speaking about "oscillations relative to equlibrium"), that will provide a different solution? :uhh:
No. That is the solution for the frequency of an oscillating spring/mass system. (Vertical or horizontal: same formula.)

Your teacher's tip "Don't neglect the gravitational force!" is a strange one. As OlderDan explains, the gravitational force determines where the equilibrium position will be, but you weren't asked to find that. (Perhaps he was having a bad day.)

OlderDan