# I Coupled pendulums with different masses

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1. Nov 22, 2016

### devansh rathi

I would like to know how to solve the coupled pendulums problem when the masses of the pendulums are different. BUT the ratio of the masses are known and all other factors are kept constant. Need to find its affect on the beating frequency, but i cannot an equation for this with different masses.

SO, it would be very helpful if someone could explain how they are related (and preferably point me towards a source from which i can read up more about this) it would be really helpful!

2. Nov 22, 2016

### hilbert2

If you're making an assumption of small oscillations/harmonic motion, then this problem is just a pair of coupled linear differential equations (which you solve by finding the normal modes of the system).

3. Nov 22, 2016

### devansh rathi

Okay yeah. So, i do not anything about this (I am in the 11th grade doing it just for fun). BUT, the higher of the normal modes depends on the mass of the pendulums, which is a problem because in the problem i have the masses are different, only the ratios are known. How do you find the two normal modes then?

4. Nov 22, 2016

### hilbert2

This is something people learn in second-year university physics curriculum. Even the problem of a single pendulum's motion can't be solved with the usual simple mathematical functions (like square root, sine, cosine, etc...) unless you make the assumption of small oscillations. The method of finding the normal modes is basically as simple as solving a pair of linear equations, though.

5. Nov 24, 2016

### devansh rathi

Could you explain to me how to find the normal modes if the ratio is 1:2.2 (108.71g and 239.16g) and the beat period is 10.03 seconds. (I also have the spring constant and length of string of the pendulums (if you need it too)).

6. Nov 24, 2016

### hilbert2

Now, because this is a classical mechanical system, the instantaneous state of the system is given by telling the coordinates and velocities of the two pendulums. The system has some equilibrium point, which means that the pendulums will stay motionless indefinitely if they are put to rest at the equilibrium positions. Let's choose coordinates as $x_1$ and $x_2$, which are the horizontal coordinates of pendulum 1 and pendulum 2 relative to their equilibrium points (at equilibrium $x_1 = x_2 = 0)$.

Now, the equations of motion, which tell the acceleration of both pendulums as a function of their positions, are:

$m_1 \frac{d^{2}x_{1}(t)}{dt^2} = -k x_1 (t) + k_c (x_2 (t) - x_1 (t))$
$m_2 \frac{d^{2}x_{2}(t)}{dt^2} = -k x_2 (t) + k_c (x_1 (t) - x_2 (t))$

Here $m_1$ and $m_2$ are the masses of the pendulums, $k$ is a Hooke's law constant that depends on the strength of gravity and on the length of the strings. The parameter $k_c$ is another Hooke's law constant that tells the strength of the coupling between the pendulums. In these equations it is assumed that the oscillations of the system are small enough for the motion to be harmonic.

Solving the normal modes means that we find out numbers $a, b, c, d, \omega_1$ and $\omega_2$, for which

$\frac{d^{2}}{dt^2}(ax_1 (t) + bx_2 (t)) = -(\omega_{1})^2 (ax_1 (t) + bx_2 (t))$
$\frac{d^{2}}{dt^2}(cx_1 (t) + dx_2 (t)) = -(\omega_{2})^2 (cx_1 (t) + dx_2 (t))$

Here the $ax_1 + bx_2$ and $cx_1 + dx_2$ are the two normal modes of the system. Mathematically they are weighted averages of the coordinates $x_1$ and $x_2$, which behave as if they were independent uncoupled oscillators that oscillate with frequencies $\omega_1$ and $\omega_2$.

With linear algebra that involves matrices and determinants, it can be shown that the two possible frequencies $\omega$ can be solved from a quadratic equation for $\omega^2$:

$\left(\frac{-k - k_c}{m_1} + \omega^2 \right)\left(\frac{-k - k_c}{m_2} + \omega^2 \right) - \frac{k_c^2}{m_1 m_2}= 0$

You can solve this with Wolfram Alpha if you want to. A more detailed explanation is here: http://users.physics.harvard.edu/~schwartz/15cFiles/Lecture3-Coupled-Oscillators.pdf

Last edited: Nov 24, 2016
7. Nov 24, 2016

### devansh rathi

Okay that was perfect. Thanks a ton!