Spring Problem Involving Variables and Constants Only

• Argonaut
Argonaut
Homework Statement
An experimental apparatus with mass ##m## is placed on a vertical spring of negligible mass and pushed down until the spring is compressed a distance ##x##. The apparatus is then released and reaches its maximum height at a distance ##h## above the point where it is released. The apparatus is not attached to the spring, and at its maximum height it is no longer in contact with the spring. The maximum magnitude of acceleration the apparatus can have without being damaged is ##a##, where ##a > g##. (a) What should the force constant of the spring be? (b) What distance ##x## must the spring be compressed initially?
Relevant Equations
$$F=ma$$
$$U_{\text{grav}}=mgh$$
$$U_{\text{el}}=\frac{1}{2}kx^2$$
Here is my attempt at the solution:

a) The apparatus may only experience acceleration ##a > g## while in contact with the spring. Since the spring exerts the greatest force when it is the most compressed, the apparatus will undergo the greatest acceleration at that point. So Newton's second law gives
$$\sum F = ma$$
$$kx-mg = ma$$
Therefore, the force constant of the spring should be $$k = \frac{m(a+g)}{x}$$.

b) There are only conservative forces in the system, so energy is conserved. Let point 1 (with ##y=0##) be the point where the apparatus is released and let point 2 be the point where it reaches height ##h##. Then
$$U_1=U_2$$
$$\frac{1}{2}kx^2 = mgh$$
Expressing ##x##
$$x=\sqrt{\frac{2mgh}{k}}$$

However, the official solution at the back of the book is
a)
$$k = \frac{m(g+a)^2}{2gh}$$
b)
$$x = \frac{2gh}{g+a}$$

I could 'reverse-engineer' both solutions. However, I don't understand how I should have known to express ##k## in terms of ##m##, ##a##, ##g## and ##h##, and not ##x##. Is it because of part b? Because essentially, both ##k## and ##x## are target variables and only the rest are known?

Argonaut said:
$$k = \frac{m(a+g)}{x}$$ $$x=\sqrt{\frac{2mgh}{k}}$$
These look good. Can you combine them so that ##k## is expressed in terms of ##m,g, a## and ##h## instead of ##m, g, a## and ##x##?

Argonaut
Argonaut said:
how I should have known to express k in terms of m, a, g and h, and not x.
The question ought to have stated, in part a, that the answer should be in terms of m, g, a and h.
I suppose you might have noticed that your answers expressed x in terms of k, then k in terms of x, in such a way that each could be expressed without the other; and since x usually refers to an unknown to be found, and you know k is to be found…

Argonaut
TSny said:
These look good. Can you combine them so that ##k## is expressed in terms of ##m,g, a## and ##h## instead of ##m, g, a## and ##x##?
Yes and they give the book solution.
haruspex said:
The question ought to have stated, in part a, that the answer should be in terms of m, g, a and h.
I suppose you might have noticed that your answers expressed x in terms of k, then k in terms of x, in such a way that each could be expressed without the other; and since x usually refers to an unknown to be found, and you know k is to be found…
Got it.

Thanks, both. It makes more sense now that I typed it up and pondered some more.

TSny

What is Hooke's Law and how is it related to spring problems?

Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, it is expressed as F = -kx, where F is the force, k is the spring constant, and x is the displacement. This law is fundamental in solving spring problems as it relates the variables of force and displacement through the constant k.

How do you determine the spring constant (k) in a spring problem?

The spring constant (k) can be determined experimentally by measuring the force applied to the spring and the resulting displacement. Using Hooke's Law (F = -kx), rearrange the equation to solve for k: k = -F/x. By plotting force versus displacement, the slope of the resulting linear graph gives the spring constant.

What is the potential energy stored in a spring?

The potential energy (U) stored in a compressed or stretched spring is given by the formula U = (1/2)kx^2, where k is the spring constant and x is the displacement from the equilibrium position. This formula is derived from the work done to compress or stretch the spring.

How do you solve a spring problem involving oscillations?

For a mass-spring system undergoing simple harmonic motion, the displacement x(t) can be described by the equation x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency (ω = sqrt(k/m)), t is time, and φ is the phase constant. By applying initial conditions, you can determine the specific values of A and φ to solve the problem.

What are the units of the spring constant (k) and how do they affect calculations?

The spring constant (k) is typically measured in Newtons per meter (N/m). The units are crucial for ensuring that calculations involving force, displacement, and energy are consistent. Using the correct units helps maintain dimensional consistency in equations and ensures accurate results.

Replies
22
Views
820
Replies
14
Views
779
Replies
7
Views
314
Replies
12
Views
1K
Replies
24
Views
2K
Replies
3
Views
2K
Replies
7
Views
1K
Replies
12
Views
2K
Replies
17
Views
699
Replies
4
Views
2K