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Two equal masses are connected by a spring satisfying Hooke’s law. The spring is elongated a little and allowed to let go.. Let the angular frequency of oscillations be $$\omega$$. Now one mass stopped. What is the square of the new frequency?

Discussion:

$$\mu=\frac{m}{2}$$
$$\omega^2=k/\mu=2k/m$$
When one block is stopped $$\omega_1^2=k/m=\omega^2/2$$