Question

12 8) (10 pts) Inspired by donuts, I decided to make a clock. The I0 9 2 3 pendulum for the clock is made of a donut (R-5.0 cm, M- 0.5 kg. eom-7.5x10 kg m2) attached to the end of a stick (L- 20 cm, m 0kg). Will my clock run on-time, fast, or slow? Support your answer with calculations. (note: assume the 20 cm length of stick is from the center af the donut to 765 9) (8 pts) An object oscillates in the manner described by "(2.0m)e쁘-Gs)) If the amplitude of the oscillation is 10m at time t#20s, find the damping constant of the oscillator. a) b) In the space below sketch a graph of the oscillations for 6.0s.

Answer 1

(8) My clock will be running fast.

According to given picture, it shows that the pendulum for my clock is made of donut which attached to the end of a stick. In this picture, i think that pendulum looks running fast.

Using a formula, we have

T = 2$\pi$_$\sqrt{}$L / g

where, L = length of a stick = 0.2 m

g = acceleration due to gravity = 9.8 m/s²

then, we get

T = (6.28 rad) _$\sqrt{}$[(0.2 m) / (9.8 m/s²)]

T = (6.28 rad) _$\sqrt{}$0.0204081 s²

T = [(6.28 rad) (0.1428 sec)]

T = 0.896 sec

(9) An expression for damped simple harmonic motion which will be given below as -

x (t) = A e^-bt/2m cos ($\omega$t + _$\phi$)

An object oscillates in the manner which is described by -

x = (2 m) e^-bt/(6kg) cos { [(2$\pi$/3) s^-1] t }

a. If the amplitude of an oscillation is 1 m at time t = 2 sec, then the damping constant of an oscillator will be given by -

(1 m) = (2 m) e^-bt/(6kg) cos { [(2$\pi$/3) s^-1] (2 s) }

[(1 m) / (2 m)] = e^-bt/(6kg) cos (4.186 rad)

ln (0.5) = - [b t / (6 kg)] (0.9973)

- [(0.6931) (6 kg)] = - b [(2 s) (0.9973)]

b = [(4.1586 kg) / (1.99465 s)]

b = 2.08 kg/s