Masses Connected by a Spring

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 $$ \omega_1^2=k/m=\omega^2/2$$

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