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Mechanics}Normal Modes

Note that \omega =\sqrt{\frac{k_{eff}}{m_{eff}}}:

For figure 1, the potential energy is U=2\times 1/2kx^2=kx^2. The kinetic energy is just T=1/2 m\dot{x}^2. Thus, k_{eff} = 2k and m_{eff}=m. Thus, \omega_1 = \sqrt{2k/m}.

For figure 2, the potential energy is U=2\times 1/2 k (x/2)^2, since each spring travels only half as far. The kinetic energy is the same as in figure 1. Thus, k_{eff}=k/2 and m_{eff}=m. So, one has, \omega_2 =\sqrt{k/2m}

Since \omega = 2\pi f = 2\pi/T, the period T_1/T_2 = \omega_2/\omega_1 = \sqrt{(1/2)/(2)}=1/2, as in choice (A).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
jax
2005-11-30 18:02:40
Just remember that capacitors are analogous to springs, so springs and capacitors add in the same way.

Springs in series add like capacitors in series, etc.
Alternate Solution - Unverified
sf001k
2005-11-10 21:09:52
springs in series add like resistors in parallel, springs in parallel add like resistors in series. find k effective for system 1 and 2, then get the period. Alternate Solution - Unverified
Comments
joy
2008-11-04 18:40:00
I'm sorry but I don't understand why springs act like capacitors since k is equivalent to 1/C
Shoudln't it be
In series : 1/C1 + 1/C2 = 1/C \Rightarrow k1 + k2 = k
In parallel C1 + C2 = C \Rightarrow 1/k1 + 1/k2 = 1/k
?
Thanks for this awesome site
NEC
star
2008-03-01 04:03:29
just remember that you need mor force for springs in parallel NEC
jax
2005-11-30 18:02:40
Just remember that capacitors are analogous to springs, so springs and capacitors add in the same way.

Springs in series add like capacitors in series, etc.
poljen88
2007-10-27 12:26:12
The spring constant k is equivalent to 1/C, so it is the reverse. Springs acting in series are equivalent to capacitors acting in parallel.
Richard
2007-10-29 15:26:20
My thought exactly...
and if you don't remember the expression for \omega Yosun quoted, all you have to recall is that
F=-kx for springs (Hook's law) from which you can use \ddot{x}=-\omega^2 x
to find the ratio of the periods (T = \frac{2\pi}{\omega}).
r10101
2007-11-02 17:27:29
jax is correct and poljen88 is not, that is, spring constants add in series/parallel in the same fashion as capacitance. That is, the opposite from the normal for resistors:

R_{eq} = R_1 + R_2 (series)
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} (parallel)

So for springs:

k_{eq} = k_1 + k_2 (parallel)
\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} (series)

E.g. http://www.physics.brown.edu/physics/demopages/Demo/waves/demo/3a2020.htm
edhopkins
2009-06-28 19:48:12
Is it accurate to stake the assumption that both the effective spring constant adds in parallel and the distance traveled is half? It seems like one ought to come from the other. If you ignore the distance traveled you can still arrive at the ratio of periods purely by \omega = \sqrt{k/m}.

Offhand it seems that by inspection one could assume the distance traveled by the two springs in series is twice that of those in parallel, and intuitively reach the same conclusion.
Alternate Solution - Unverified
sf001k
2005-11-10 21:09:52
springs in series add like resistors in parallel, springs in parallel add like resistors in series. find k effective for system 1 and 2, then get the period. Alternate Solution - Unverified
sf001k
2005-11-10 21:08:03
NEC

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