QUICK QUIZ 15.1
(end of section 15.2)
You hang an object onto a vertically hanging spring and measure the stretch
length of the spring to be 1 meter. You then pull down on the object and release
it so that it oscillates in simple harmonic motion. The period of this oscillation
will be a) about half a second, b) about 1 second, c) about 2 seconds, or d)
impossible to determine without knowing the mass or spring constant.
QUICK QUIZ 15.1 ANSWER
(c). This problem illustrates an easy method for determining the
properties of a spring-object system. When you hang the object,
the spring force, kx, will be equal to the weight, mg, so that kx =
mg or x/g = m/k. From Equation 15.13,
T  2
m
k
 2
x
g
 2
1m
9 .8 m /s
2
~
6
3
s ~ 2s
QUICK QUIZ 15.1
(end of section 15.3)
You hang an object onto a vertically hanging spring.
You slowly let the object down and it stretches the
spring a distance d1. You then remove the object,
return the spring to its equilibrium position and
reattach the object. This time, instead of slowly
letting the object down, you release it all at once and
measure the distance the spring stretches, d2, before
the object springs back up again. The ratio d2/d1 will
be a) 1, b) 2, c) 4, or d) impossible to determine
without knowing the mass and spring constant.
QUICK QUIZ 15.2 ANSWER
(b). When you slowly let the object down, the object is in static equilibrium
and therefore the spring force, kx, is equal to the weight, mg, so that x = d1 =
mg/k. When you quickly release the mass, the mass will oscillate between its
release point and its lowest point of travel. The initial gravitational potential
energy of the mass is converted to spring potential energy when the mass is at
its lowest point. Therefore, mgx = kx2/2, so that x = d2 = 2mg/k.
QUICK QUIZ 15.3
(end of section 15.5)
You release a pendulum bob from an angle which is 5° from the
vertical and measure its period of oscillation. You then repeat the
experiment but from an angle which is 90° from the vertical. The
period of oscillation for the pendulum in the second case will be a)
greater than the first case, b) less than the first case, c) the same as the
first case, d) impossible to determine.
QUICK QUIZ 15.3 ANSWER
(a) and (d). Equation 15.24 for the simple pendulum is
valid for small angles when sin q ~ q. Only then is the
period independent of the amplitude. For q = 5° or 0.0873
radians, sin q is 0.0872 and Equation 15.24 is clearly valid.
For q = 90° or 1.57 radians, sin q is 1 and Equation 15.24 is
not valid. Otherwise, the equation,
d q
2
dt
2

g
sin q ,
L
valid for any angle must be used. This differential equation
can not be easily solved to determine the period of
oscillation. However, a simple experiment will show that
the period of oscillation for an angle of 90° is only about
15% greater than for an angle of 5°.
QUICK QUIZ 15.4
(end of section 15.6)
A damped oscillator undergoes
exponentially decaying oscillatory
motion, as in the figure shown here.
After 5 seconds of such motion, the
amplitude (the quantity Ae-bt/2m) is a
factor of 2 smaller than its initial value at
t = 0. After 15 seconds of motion, the
amplitude will be a) a factor of 4 smaller
than its initial value at t = 0, b) a factor of
6 smaller than its initial value at t = 0, c)
a factor of 8 smaller than its initial value
at t = 0, or d) impossible to determine
without knowing the constants b and m.
QUICK QUIZ 15.4 ANSWER
(c). The negative exponential function has the property that it will
decrease by the same factor over the same period of time. Relating the
amplitude after 5 seconds (t1) to the initial amplitude we have,

1
Ae

2
Ae
bt 1
2m
 e
0

bt 1
2m
.
After another 5 seconds we have,

b 2 t1
2m
Ae
Ae
e
0

bt 1
2m

e
bt 1
2m

1

2
1

2
1
.
4
So after 15 seconds we get,

Ae
Ae
b 3 t1
2m
0
e

bt 1
2m

e
bt 1
2m

e
bt 1
2m

1
2

1
2

1
2

1
8
.
QUICK QUIZ 15.5
(end of section 15.7)
Vibration isolation tables are designed to isolate sensitive instruments, for example lasers and
microscopes, from vibrations in the floor. The tables behave like a mass and a spring with
small damping and act like a forced oscillator with the floor as the driving force. Typical
floors vibrate with a frequency of about 20 Hz and this driving frequency is applied to the
table. It is desirable to keep the amplitude of vibration of the isolation table to a minimum
and the natural frequency of the table is adjusted to be significantly smaller than the driving
frequency of the floor. Isolation tables should therefore be designed to have a) a large mass
and a small spring constant, b) a small mass and a large spring constant, c) a large mass and a
large spring constant, or d) a small mass and a small spring constant.
QUICK QUIZ 15.5 ANSWER
(a). The equation for the natural frequency, wo= (k/m)1/2,
indicates that the spring constant should be small and the
mass large to produce a small natural frequency. From
examination of Equation 15.36,
F0 / m
A
(w  w 0 )  (
2
2
2
bw

F0
m (w  w 0 )  ( b w )
2
)
2
2
2
2
,
2
m
a large difference in w and wo, and in addition a large mass
m, will reduce the vibration amplitude of the table by a
substantial amount.