Question

17 points) One simple pendulum and the physical pendulums (disk and rod) are suspended on the crossbar, as shown in figure. (a) Calculate the natural linear frequency of the simple pendulum, if the length of the simple pendulum is -1.6 m (b) Calculate the natural angular frequency of the disk. The radius of the disk is R-0.5 m; moment of inertia about an axis through the center of mass is ICM = mR2 - (c) Calculate the natural period of oscillations of the rod. The length of the rod is 1.5 m; moment of inertia about an axis through the center of mass is Icm = ml? 16 points) Two cagles fly directly in sequence one after another, the first eagle (11) at 25.0 m/s and the second m/s. Eagle #1 screechs, emitting a frequency of 3500 Hz. What frequency do the eagle #2 receives, if the speed Id is 330 m/s?

Answer 1

a)

l = length of simple pendulum = 1.6 m

natural frequency of simple pendulum is given as

f = (1/2$\pi$) sqrt(g/l)

f = (1/2(3.14)) sqrt(9.8/1.6)

f = 0.394 Hz

b)

For the disk, we have

f = (1/2$\pi$) sqrt(mgR/I_cm)

f = (1/2$\pi$) sqrt(mgR/((0.5) mR²))

f = (1/2$\pi$) sqrt(2g/R)

f = (1/(2(3.14)) sqrt(2(9.8)/(0.5))

f = 0.997 Hz

c)

For the disk, we have

f = (1/2$\pi$) sqrt(mg(L/2)/I_cm)

f = (1/2$\pi$) sqrt(mg(L/2)/((1/12)mL²))

f = (1/2$\pi$) sqrt(6g/L)

f = (1/(2(3.14)) sqrt(6 (9.8)/((1.5)))

f = 0.997 Hz