How Do Different Spring Constants Affect Oscillation Frequency?

[smb62]

suppose that the two springs have different spring constants k1=280 N/m and k2=260 N/m and the mass of the object is m=14 kg. Find the frequency of oscillation of the block in Hz.

http://www.physics.umd.edu/rgroups/ripe/perg/abp/think/oscil/mos.htm
the first picture on this site is what the problem looks like. Its a mass between two springs that are attached to walls.

i know that angular frequency equals the square root of k/m. but i don't know how to do it when there are two springs involved. Please help Asap. Thanks so much

 

[arildno]

Welcome to PF!
Let your origin be at where both springs has their rest length.
If you displace the box a distance "x", what are the forces (with direction) acting on it from either box?

 

[smb62]

Welcome to PF!
Let your origin be at where both springs has their rest length.
If you displace the box a distance "x", what are the forces (with direction) acting on it from either box?

So if i have F1= -kx = -280x and F2= 260x do i then add them ? if so i would get Ft= -20x ...then should i use -20 as my value for k? and substitute it into the equation that i said before...where frequency equals the square root of (k divided by m)?

THanks so much for your help by the way...this forum is a wonderful idea and i am definitely going to spread the word!

 

[arildno]

No that is incorrect!
Let the rest lengths be $$L_{1},L_{2}$$
For clarity, "x" be a positive number.
Then, the new length of spring 1 is $$L_{1}+x$$
Hence, $$F_{1}=-k(L_{1}+x-L_{1})=-kx$$
Let's look at spring 2:
If spring 2 had been lengthened by a positive amount "y" (dragged out to the left), then, the force from it would drag the block to the right (the positive direction).
Hence, for positive displacement of spring 2 "y", $$F_{2}=2ky$$
Now, setting y=-x (spring 2 is actually shortened), we get:
$$F_{2}=-2kx$$

Hence, total force F on block is:
$$F=F_{1}+F_{2}=-3kx$$