# Hooke's law

by brake4country
Tags: hooke
 P: 32 I need help with a spring problem: Two identical springs hung from a ceiling have different masses. The first spring is 1 kg with a total length of 6 cm. The second spring is 1.5 kg and is 7 cm in length. The displacement between the two springs is 1 cm. Find the spring constant k. (A) 2 N/cm (B) 5 N/cm (C) 10 N/cm (D) 20 N/cm I started solving each equation for k, but ended up with different values. Thanks in advance.
 P: 403 The critical piece of information we miss is the length of the springs (since they both identical) in their natural position. Suppose x is that length, then you can make a system of two equations with two unknowns which are the x and k. The first equation is from the first spring and is 1*g=k*(6-x) find the 2nd equation from the second identical spring and solve the system of equations. (g=10m/s^2)
 P: 32 I am confused. In Hooke's problems, how is the original length of the spring applicable? Aren't we just concerned with F, difference in x and k? Thanks.
 P: 403 Hooke's law Yes it is F=-k*x where x is the displacement from the original length. The problem says that the total length is 6cm for the first spring, and not the displacement from the original length. So its not F=-k*6 but F=-k(6-x) where x is the original length. Alternatively you can put as x the displacement from the original length . Then it would be 1*g=k*x for the first spring. For the second spring the displacement from the original length will be x+1 so you can make the equation for the second spring and solve again the system of two equations for x and k.
Mentor
P: 41,323
 Quote by brake4country I need help with a spring problem: Two identical springs hung from a ceiling have different masses. The first spring is 1 kg with a total length of 6 cm. The second spring is 1.5 kg and is 7 cm in length. The displacement between the two springs is 1 cm. Find the spring constant k. (A) 2 N/cm (B) 5 N/cm (C) 10 N/cm (D) 20 N/cm I started solving each equation for k, but ended up with different values. Thanks in advance.
Show what you did.

I assume these are identical springs, the only difference being the amount of mass hung from each (not the mass of the spring itself).
HW Helper
Thanks
P: 5,166
 Quote by brake4country Two identical springs are hung from a ceiling. To the first spring is added 1 kg to leave that spring with a total length of 6 cm. To the second spring is added 1.5kg so that spring now is 7 cm in length. Find the spring constant k. The displacement between the two springs is 1 cm. (A) 2 N/cm (B) 5 N/cm (C) 10 N/cm (D) 20 N/cm
There, problem statement fixed! ..... I think.
P: 503
 Quote by brake4country how is the original length of the spring applicable? Aren't we just concerned with F, difference in x and k?
Exactly: We are concerned with the difference in x ... which is equal to the final x minus the original x

In the problem, though, you are only given the final length, not the original length.

That means there are two unknowns (original x and k)
(Two unknowns makes the equation for each spring individually unsolvable; you must solve them together)
 P: 32 I have tried this problem using two unknowns but it just doesn't come out. Here is what I did: a) Take the difference of the masses (1.5 kg-1 kg) = 0.5 kg b) Then, I used mg = kx c) (0.5 kg)(10 m/s^2) = k (1 cm) d) Now my answer turns out to be 5 N This is faster and easier than working a quadratic. Did anyone try to solve a system of equations and end up with a quadratic? I couldn't get a positive answer.
P: 503
 Quote by brake4country Did anyone try to solve a system of equations and end up with a quadratic?

$k(7-x)=1.5g$
$k(6-x)=g$

...

$x=7-\frac{1.5g}{k}$
$x=6-\frac{g}{k}$

...

$7-\frac{1.5g}{k}=6-\frac{g}{k}$

...

$k=0.5g$