EK Physics 1001 Question 252

[sillyjoe]

Question reads:

"If two springs (k=400N/m) are placed parallel to each other, what is the spring constant of the two spring system?

A. 100 N/m
B. 200 N/m
C. 400 N/m
D. 800 N/m

D is correct. Imagine hanging mass m/2 from one of the springs. Now hang a second mass m/2 from the second spring right next to the first. If you were to glue the two masses together, the change in spring length would not be affected. Now you have a two spring system. Write Hooke's law for both equations, solve for delta(x), and substitute. mg/2 = k1*delta(x) and mg = k2*delta(x) => k2 =2k1"

Can someone please provide me with a more in depth explanation of this question. I am confused at the algebra that was done and also do not understand how he got his answer. Moreover, why does delta X not change but the constant does? Thanks in advance!

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[DrknoSDN]

An easier way to think of this would be attach a spring to a wall with k=400N/m and stretch it out to 1 meter by pulling. If you attached a second spring to the wall and pulled them both together, what force would be required to pull the 2 spring system to 1 meter? (800N) per meter.

Using that example you can use hooke's law and avoid any gravitational forces.
In the given example delta x does not change because you are doubling the mass. For a single spring you use mass m/2 and for 2 springs you use 2*(m/2) or just M.

 

[sillyjoe]

Wow, thanks. You literally just distilled that question into an easy way to understand. You should write the answer explanations!

 

[DrknoSDN]

Hah, thanks.