# Oscillating Mass between Two Springs

## Homework Statement

A block on a frictionless table is connected to two springs having spring constants k1 and k2. Show that the block's oscillation frequency is given by f = (f12 + f22).5 where f1 and f2 are the frequencies at which it would oscillate if attached to spring 1 or spring 2 alone.

f = 1/T
T = 2pi(m/k).5

## The Attempt at a Solution

I have substituted the period into the frequency and I am stuck. It looks like they found f1 and f2 alone and put them into the Pythagorean theorem, but I do not know how to relate them.

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## Homework Statement

A block on a frictionless table is connected to two springs having spring constants k1 and k2. Show that the block's oscillation frequency is given by f = (f12 + f22).5 where f1 and f2 are the frequencies at which it would oscillate if attached to spring 1 or spring 2 alone.

f = 1/T
T = 2pi(m/k).5

## The Attempt at a Solution

I have substituted the period into the frequency and I am stuck. It looks like they found f1 and f2 alone and put them into the Pythagorean theorem, but I do not know how to relate them.
I do not see any relation to Pythagorean theorem. Notice that if you were to latch two springs onto a mass, you could model the same motion with one spring with a constant equal to the sum of the two springs. Knowing this, calculate the frequency of all three configurations(spring 1, spring 2, and spring 1 & 2 combined). Next, substitute your answer for spring 1 and spring 2 into that equation and see if you get the same answer you derived for spring 1 & 2 combined.

When I find the frequencies of the springs, I do not understand how to relate them.

When I find the frequencies of the springs, I do not understand how to relate them.
I told you how.

Calculate frequency 1 with k = k1, frequency 2 with k = k2, and f with k = k1 + k2.

$$f_1 = \frac{1}{2\pi} \sqrt{\frac{k_1}{m}}$$
$$f_2 = \frac{1}{2\pi} \sqrt{\frac{k_2}{m}}$$
$$f = \frac{1}{2\pi} \sqrt{\frac{k_1 + k_2}{m}}$$ because $$k = k_1 + k_2$$ and nothing else changes.

Next, plug $$f_1$$ and $$f_2$$ into $$f=\sqrt{f_1^2+f_2^2}$$ to see if you get the same expression as above for $$f$$.
(you do)

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Oh, I was going a different route using a different equation making it more complicated. Thank you for your help.