Physics (Part 1 - Part 3)

13.5 | Motion of a Pendulum 451
The angular frequency v is the factor in front of t in
Equations (1) and (2). Equate these factors:
Divide v by 2p to get the frequency f :
v52pf 5
f 5 v 5 0.062 5 Hz
2p
p rad/s 5 0.393 rad/s
8.00
The period T is the reciprocal of the frequency:
T 5 1 f 5 16.0 s
(b) Find the maximum magnitudes of the velocity and the
acceleration.
Calculate the maximum speed from the factor in front of
the sine function in Equation 13.14b:
Calculate the maximum acceleration from the factor in
front of the cosine function in Equation 13.14c:
v max
5 Av 5 (0.250 m)(0.393 rad/s) 5 0.098 3 m/s
a max
5 Av 2 5 (0.250 m)(0.393 rad/s) 2 5 0.038 6 m/s 2
(c) Find the position, velocity, and acceleration of the
object after 1.00 s.
Substitute t 5 1.00 s in the given equation:
Substitute values into the velocity equation:
Substitute values into the acceleration equation:
x 5 (0.250 m) cos (0.393 rad) 5 0.231 m
v 5 2Av sin (vt)
5 2(0.250 m)(0.393 rad/s) sin (0.393 rad/s ? 1.00 s)
v 5 20.037 6 m/s
a 5 2Av 2 cos (vt)
5 2(0.250 m)(0.393 rad/s 2 ) 2 cos (0.393 rad/s ? 1.00 s)
a 5 20.035 7 m/s 2
REMARKS In evaluating the sine or cosine function, the angle is in radians, so you should either set your calculator to
evaluate trigonometric functions based on radian measure or convert from radians to degrees.
QUESTION 13.6 If the mass is doubled, is the magnitude of the acceleration of the system at any position (a) doubled,
(b) halved, or (c) unchanged?
EXERCISE 13.6 If the object–spring system is described by x 5 (0.330 m) cos (1.50t), find (a) the amplitude, the angular
frequency, the frequency, and the period, (b) the maximum magnitudes of the velocity and acceleration, and (c) the position,
velocity, and acceleration when t 5 0.250 s.
ANSWERS (a) A 5 0.330 m, v 5 1.50 rad/s, f 5 0.239 Hz, T 5 4.18 s (b) v max
5 0.495 m/s, a max
5 0.743 m/s 2 (c) x 5
0.307 m, v 5 20.181 m/s, a 5 20.691 m/s 2
13.5 Motion of a Pendulum
A simple pendulum is another mechanical system that exhibits periodic motion.
It consists of a small bob of mass m suspended by a light string of length L fixed
at its upper end, as in Active Figure 13.15 (page 452). (By a light string, we mean
that the string’s mass is assumed to be very small compared with the mass of the
bob and hence can be ignored.) When released, the bob swings to and fro over the
same path, but is its motion simple harmonic?
Answering this question requires examining the restoring force—the force
of gravity—that acts on the pendulum. The pendulum bob moves along a circular
arc, rather than back and forth in a straight line. When the oscillations are
small, however, the motion of the bob is nearly straight, so Hooke’s law may apply
approximately.
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