Frequency of Oscillation: Two Springs Connected to a Mass m | Homework Help

[toesockshoe]

Homework Statement

Two springs are joined and connected to a mass m such that they are all in a straight line. The two springs are connected first and then the mass last so that all three are in a row. If the springs have a stiffness of k1 and then k2, find the frequency of oscillation of m.

Homework Equations

T = 2 \pi \sqrt{\frac{m}{k}}

The Attempt at a Solution

so i tried making an F=ma for the mass and spring 1 (which is said was the spring closer to the mass)...

F=ma system mass

F_{el} = ma
k_1 x_1 = ma
max acceration happens at aplitude:
k_1 x_1 = mA \omega ^2

F=ma system spring 1.

F_{el mass} - F _{el 2} = M_{s1}
i am assuming the spring is massless ( i think we can do that)
so F_{elmass} = F_{el 2} <br /> k_1 x_1 = k_2 x_2 <br /> <br /> i suppose x_1 + x_2 = A when both x's are at maximum. ...<br /> so \frac {k_1}{k_1} = k_2 (A-x_1)<br /> <br /> x_1 = \frac {A}{ \frac {k_1}{k_2} + 1 }<br /> <br /> go back the the last equation we got in f=ma system mass and subtitute in the x_1 we just found...<br /> <br /> the A's cancel out and after we simply we get:<br /> <br /> \omega = \sqrt{ \frac{k_1 k_2}{ (k_1 + k_1) m }} ////<br /> and to get F... just divide it by 2pi... right?<br /> is this even correct?

 

[AYPHY]

try it with energy method you would be albe to solve it. if not reply i will post the solution.

 

[SammyS]

try it with energy method you would be albe to solve it. if not reply i will post the solution.

Be careful about this in the homework section of PF !

 

[toesockshoe]

Be careful about this in the homework section of PF !

do you see a problem with my solution?

 

[SammyS]

Homework Statement

Two springs are joined and connected to a mass m such that they are all in a straight line. The two springs are connected first and then the mass last so that all three are in a row. If the springs have a stiffness of k1 and then k2, find the frequency of oscillation of m.

Homework Equations

T = 2 \pi \sqrt{\frac{m}{k}}

The Attempt at a Solution

so i tried making an F=ma for the mass and spring 1 (which is said was the spring closer to the mass)...
F=ma system mass
F_{el} = ma
k_1 x_1 = ma
max acceration happens at aplitude:
k_1 x_1 = mA \omega ^2

F=ma system spring 1.
F_{el mass} - F _{el 2} = M_{s1}
i am assuming the spring is massless ( i think we can do that)
so F_{elmass} = F_{el 2} <br /> k_1 x_1 = k_2 x_2<br /> <br /> i suppose x_1 + x_2 = A when both x's are at maximum. ...<br /> so \frac {k_1}{k_1} = k_2 (A-x_1)<br /> <br /> x_1 = \frac {A}{ \frac {k_1}{k_2} + 1 }<br /> <br /> go back the the last equation we got in f=ma system mass and subtitute in the x_1 we just found...<br /> <br /> the A's cancel out and after we simply we get:<br /> <br /> \omega = \sqrt{ \frac{k_1 k_2}{ (k_1 + k_1) m }} ////<br /> and to get F... just divide it by 2pi... right?<br /> is this even correct?

<br /> toesockshoe,<br /> <br /> I haven't examined your entire solution, but I'm pretty sure that your final answer IS correct !<br /> <br /> Two springs connected in that manner have an effective spring constant of $\displaystyle\ k_\text{eff}=\frac{k_1\,k_2}{k_1+k_2}\ .$