The figure below shows a pendulum of length L with a bob of mass M. The bob is a

Question

The figure below shows a pendulum of length L with a bob of mass M. The bob is attached to a spring that has a force constant k. When the bob is directly below the pendulum support, the spring is unstressed. Derive an expression for the period of this oscillating system for small-amplitude vibrations (assume there is no displacement from the horizontal). Suppose that M = 1.10 kg and L is such that in the absence of the spring the period is 2.98 s. What is the force constant k if the period of the oscillating system is 1.49 s?

Explanation / Answer

if only pendulum k = 0 w = 2Pi/T = 2Pi/2.98 = 2.108 w = sqrt(g/l) so g/l = 4.445 Now to the pendulum plus sring w = 2Pi/T = 2Pi/1.49 = 4.216 w = sqrt(k/M+g/L) 4.216 = sqrt(k/M+4.445) 17.782= k/M+4.445 k/M = 13.337 k = 14.670 N/m

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