
By the end of this lecture, you should:
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0.0172 m; 17.2 m
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\[\frac{l}{g} \sf{\mbox{has units}} \frac{[m]}{[m/{{s}^{2}}]}=\frac{[1]}{[1/{{s}^{2}}]}=[{{s}^{2}}]\]
\[\frac{m}{k} \sf{\mbox{has units}} \frac{[kg]}{[N/m]}=\frac{[kg]}{[kg\cdot m/{{s}^{2}}/m]}=[{{s}^{2}}]\]
The only further operation is a square root which gives [s].
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\[\omega =\sqrt{\frac{g}{l}}\]
makes sense since we would think the higher restoring force from a higher “g” would make a pendulum go faster, while a longer pendulum is know to vibrate more slowly
\[\omega =\sqrt{\frac{k}{m}}\]
makes sense since a stiffer spring vibrates with higher frequency, while a larger mass being pushed by a spring moves ponderously.
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377 rad/s
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360º or equivalently, 2π radians