The frequency of swinging pendulum in radians/second is Sqrt[g/L] where g is gravity and L is...
The frequency of swinging pendulum in radians/second is Sqrt[g/L] where g is gravity and L is the pendulum length. Find L in meters so that a pendulum has the same frequency as Taipei 101, about 0.15 Hz. Note that you must convert the frequency. (2)
L=_____________meters
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The frequency of swinging pendulum in radians/second is
Sqrt[g/L] where g is gravity and L is...
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(radians) from the vertical. It can be shown that as a function of time satisfies the (1 point) Suppose a pendulum with length L (meters) has angle differential equation: d20 + & sin 0 = 0 dt 2 L where g = 9.8 m/sec/sec is the acceleration due to gravity. For small values of we can use the approximation sin() ~ 0, and with that substitution, the differential equation becomes linear. A. Determine the equation of motion of a pendulum...
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Calculus question please help
<3 . (ignore the working)
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(1 point) Suppose a pendulum of length L meters makes an angle of θ radians with the vertical, as n the figure t can be shown that as a function of time, θ satisfies the differential equation d20 + sin θ-0, 9.8 m/s2 is the acceleration due to gravity For θ near zero we can use the linear approximation sine where g to get a linear di erential equa on d20 9 0 dt2 L Use the linear differential equation...
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<3 . (ignore the working)
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