# Harmonic motion - Find the Mass held between two Springs

• JoeyBob
If ##\omega=\sqrt {\frac k m}##the value of that term ##k## must be equal to the ##k## of a system of two springs working in series.Therefore, ##k=k_1+k_2=m\omega^2##f

#### JoeyBob

Homework Statement
See attached
Relevant Equations
E=1/2kA^2=1/2m(wA)^2
So first I find the energy using the eqn (1/2)kA^2. Since there are two springs with the same k I multiply it by two to get kA^2. Energy I get is 2.0475,

Now I use E=(1/2)m(wA)^2 to find mass. Again since there are two springs I use E=m(wA)^2.

m=E/(wA)^2. w=(2(pi))/T btw.

I get the answer of 3.375 when the correct answer is 6.750. Why is it that multiplying my answer by 2 gives me the rights answer? what am i missing here

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since there are two springs I use E=m(wA)^2.
Given the mass and velocity, how does the number of springs or their constants affect the KE?
Btw, you don't need to know the amplitude. The calculation can be simplified.

Lnewqban
I believe that your error is when you replace ##k_{total~of~the~system}## with ##m\omega^2## in the equation.

Given the mass and velocity, how does the number of springs or their constants affect the KE?
Btw, you don't need to know the amplitude. The calculation can be simplified.
But the question gives amplitude? Kinetic E of a spring would be (1/2)kx^2.

I believe that your error is when you replace ##k_{total~of~the~system}## with ##m\omega^2## in the equation.
Isnt E=kA^2=m(wA)^2 for two springs? For one spring it would be E=(1/2)kA^2=(1/2)m(wA)^2 right?

Isnt E=kA^2=m(wA)^2 for two springs? For one spring it would be E=(1/2)kA^2=(1/2)m(wA)^2 right?
##E=\frac 12kA^2=\frac 12m(\omega A)^2## is for one spring.
##EPE_{max}=kA^2## is for two springs of constant k because the effective spring constant is 2k.
##KE_{max}=\frac 12mv_{max}^2=\frac 12m(\omega A)^2## because there is still only a mass m, not 2m.

Last edited:
JoeyBob and Lnewqban
Isnt E=kA^2=m(wA)^2 for two springs? For one spring it would be E=(1/2)kA^2=(1/2)m(wA)^2 right?
If ##\omega=\sqrt {\frac k m}##
the value of that ##k## must be equal to the ##k## of a system of two springs working in series.
Therefore, ##k=k_1+k_2=m\omega^2##

The value of that term ##m\omega^2## is the one to be used in the equation of elastic potential energy.