Question

A certain simple pendulum consists of a small 750.0 ? bob that swings on the end of a 25.0 ?? string. The small amplitude of the oscillations of this pendulum decays to half its original value after 45.0 oscillations. The angular position of the pendulum as a function of time, ?(?), can be expressed as follows.

?(?) = ??₀ ? ^{− ??/2m} cos(? ′ ? + ?)

??₀ is the original angular amplitude. ? is the time, and ? is the mass of the bob. ? is a phase angle which helps set the initial angular position of the pendulum. ?′ is the angular frequency of the pendulum. This pendulum can be considered to be “lightly damped”, meaning that the damping does not appreciably affect the angular frequency, so ? ′ ≈ ????????? ????????.

a) Find the damping constant ? for the pendulum. Don’t forget units!

b) Find the time at which the energy of the pendulum is equal to one half its original value. Note: Similar to with a spring oscillator, the energy is proportional to angular amplitude squared.

Answer 1

Given mass energy = -bt 2 0.25 m= 750g = 0.75 kg 6 Initial 1 0² where kis constant length d= 25cm = 0.25m How energy at time to from given condition. T= 1.0035 sec time for 45 oscillation dx & Koo? t= us T = 45.16 sec. T= 20 S 9.8 Å KO? Now "Cos(wit+o) Oo Oo the In (3) = -bt t = m In (2) b bxus 16 In() 2x0.75 b = 2x0.75 x 0.693 us.16 b=0.093 kg/s = 233/s 0.023 kg/s 123 g/s to 0.750.693 0.023 t- 22.579 sec OY 22.58 Sec

Thanks for asking.