How Do Different Spring Constants Affect Oscillation Frequency?

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The discussion focuses on calculating the oscillation frequency of a mass attached to two springs with different spring constants, k1=280 N/m and k2=260 N/m, and a mass of m=14 kg. The correct approach involves determining the effective spring constant when both springs are involved, which requires analyzing the forces acting on the mass when displaced. The total force on the block is derived from the individual forces of the springs, leading to an effective spring constant of -3k, where k is the average of the two spring constants. The frequency of oscillation can then be calculated using the formula for angular frequency, which incorporates this effective spring constant. Understanding the interaction of the two springs is crucial for accurate frequency determination.
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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
 
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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?
 
arildno said:
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!
 
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
 
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