Question

T = 2\pi \sqrt{\frac{L}{g}} 2. The restorative force of a pendulum is the part of gravity that acts perpendicular to the pendulum arm: F = -mg \sin \theta. For small angles, this force is directly proportional to displacement because \sin \theta \approx \theta. To verify this, convert 20\textdegree to radians, then compute the sine of the angle. (5 pts) \theta = \sin \theta = \% Diff = By the small angle approximation, then, the pendulum behaves as a simple harmonic oscillator: F = -(mg)\theta. 20(\frac{\pi}{180}) = 0.35 rad

          T = 2\pi \sqrt{\frac{L}{g}}
2. The restorative force of a pendulum is the part of
gravity that acts perpendicular to the pendulum arm:
F = -mg \sin \theta. For small angles, this force is directly
proportional to displacement because \sin \theta \approx \theta. To
verify this, convert 20\textdegree to radians, then compute the
sine of the angle. (5 pts)
\theta = 
\sin \theta = 
\% Diff = 
By the small angle approximation, then, the pendulum
behaves as a simple harmonic oscillator:
F = -(mg)\theta.
20(\frac{\pi}{180}) = 0.35 rad

T = 2π√((L)/(g))
2. The restorative force of a pendulum is the part of
gravity that acts perpendicular to the pendulum arm:
F = -mg sinθ. For small angles, this force is directly
proportional to displacement because sinθ≈θ. To
verify this, convert 20°to radians, then compute the
sine of the angle. (5 pts)
θ=
sinθ=
% Diff =
By the small angle approximation, then, the pendulum
behaves as a simple harmonic oscillator:
F = -(mg)θ.
20((π)/(180)) = 0.35 rad

Added by Cheryl M.

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