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Find the dimensions of K in the relation `T = 2pi sqrt((KI^2g)/(mG))` where T is time period, - YouTube
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The period of oscillation of a simple pendulum is `T = 2 pi sqrt((L)/(g)) .L` is about `10 cm` a... - YouTube
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Sensors | Free Full-Text | Modeling Full-Field Transient Flexural Waves on Damaged Plates with Arbitrary Excitations Using Temporal Vibration Characteristics
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![For a two body oscillator system, prove the relation, `T = 2pi sqrt((mu)/(k))` where, `mu = (m_(... - YouTube For a two body oscillator system, prove the relation, `T = 2pi sqrt((mu)/(k))` where, `mu = (m_(... - YouTube](https://i.ytimg.com/vi/OpVMBxsEe4g/maxresdefault.jpg)
For a two body oscillator system, prove the relation, `T = 2pi sqrt((mu)/(k))` where, `mu = (m_(... - YouTube
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