Question

Consider N identical masses connected through identical springs of force constant k.
The free ends of the coupled system are rigidly fixed at x = 0 and x = l. The masses
are made to execute longitudinal oscillations on a frictionless table.
i) Depict the equilibrium as well as instantaneous configurations.
ii) Write down their equations of motion, decouple them and obtain frequencies of
normal modes.

Answer 1

Answer on Question #61614 - Physics - Mechanics | Relativity

Question:

Consider N identical masses connected through identical springs of force constant k. The free ends of the coupled system are rigidly fixed at $x = 0$ and $x = l$ . The masses are made to execute longitudinal oscillations on a frictionless table.

1) Depict the equilibrium as well as instantaneous configurations.

2) Write down their equations of motion, decouple them and obtain frequencies of normal modes.

Answer:

1) Suppose the length of each spring in its natural unextended or uncompressed form is a. Then the total length of $N + 1$ springs is $(N + 1)a$ . In the equilibrium position, the springs are not oscillating.

Mass1 is at $x = a$ .

Mass 2 is at $x = 2a$ .

Mass N is at $x = Na$ .

On either side of each mass the force acting on it is: $k \Delta$

Each is extended by:

$$\Delta = \frac {L}{N + 1} - a$$

Mass1 is at $x = a + \Delta$ .

Mass 2 is at $x = 2a + 2\Delta$ .

Mass N is at $x = Na + N\Delta$ .

2) Equations of motion of the masses are:

$$m \frac {d ^ {2} x _ {1}}{d t ^ {2}} = - k (2 x _ {1} - x _ {2})$$ $$m \frac {d ^ {2} x _ {2}}{d t ^ {2}} = - k (2 x _ {2} - x _ {1} - x _ {3})$$ $$m \frac {d ^ {2} x _ {3}}{d t ^ {2}} = - k (2 x _ {3} - x _ {2} - x _ {4})$$

··

$$m \frac {d ^ {2} x _ {n - 1}}{d t ^ {2}} = - k (2 x _ {n - 1} - x _ {n - 2} - x _ {n})$$ $$m \frac {d ^ {2} x _ {n}}{d t ^ {2}} = - k (2 x _ {n} - x _ {n - 1})$$

Suppose that: $x_{n}(t) = A_{n}\cos (\omega t)\Rightarrow \frac{d^{2}x_{n}}{dt^{2}} = -\omega^{2}x_{n}(t)$

Let $2 - \frac{m\omega^2}{k} = 2 - \frac{\omega}{\omega_0} = C$

We get equations like:

$$C x _ {1} = x _ {2}$$ $$C x _ {2} = x _ {1} + x _ {3}$$ $$C x _ {3} = x _ {2} + x _ {4}$$ $$C x _ {n - 1} = x _ {n - 2} + x _ {n}$$ $$C x _ {n} = x _ {n - 1}$$

The system can be solved knowing the number of masses in it:

For 3 masses we have:

$$C = \pm \sqrt {2}, \frac {\omega}{\omega_ {0}} = 2 \pm \sqrt {2}$$

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