Chapter 15 Questions
1. (Question 15.3) Which of the following describe φ for the SHM of Fig. 15-20a:
x
(a) −π < φ < −π/2.
6
(b) π < φ < 3π/2.
- t
(c) −3π/2 < φ < −π.
Figure 15-20a (figure above) shows the displacement of a particle undergoing simple harmonic motion as
function of time, x = xm cos(ωt + φ). At t = 0 the particle’s position is below equilibrium position (negative)
so at that moment cos φ < 0. That means that φ should be between π/2 and 3π/2.
The particle’s velocity, v = −xm ω sin(ωt + φ), is positive at t = 0 (by inspecting the graph), which gives that
sin φ < 0. That means that φ should be between π and 2π.
The overlap of these two conditions is π < φ < 3π/2. That angle can be, also, represented as negative angle
by subtracting 2π, which gives another way to answer the same question, −π < φ < −π/2. So, the correct
answers are both (a) and (b).
The other way to answer the question is to draw x = xm cos ωt in the same plot as x = xm cos(ωt + φ) and
read the phase difference between, for example, the closest crests of the two functions. The phase difference,
φ, is between −π/2 and −π for the closest crest to the right of x = xm cos ωt or between π and 3π/2 for the
closest crest to the left of x = xm cos ωt. See the plot below.
x
φ>0
6φ < 0
-¾
- t
2. (a) What is the phase constant for the harmonic oscillation with the acceleration function a(t) given in the
figure below, if the position function x(t) has the form x = xm cos(ωt+φ)? (b) What is the angular frequency?
(c) What is xm ?
¾
T =12.0 s
2
Using: a(t) = −am cos(ωt + φ) and
a (cm/s )
da
6
a = −am cos ωt
slope(t) =
= am ω sin(ωt + φ) at t = 0
8 pppppp
dt
6
and the initial conditions given:
4
2
2
2
-t (s)
a(0) = +2 cm/s , slope(0) > 0, with am = 8 cm/s
-3.4
2.6
8.6
-2
-4
we have:
µ
¶
-6
p
p
p
p
p
p
a(0)
-8
¾
φ = cos−1 −
a
m
φ = +1.82 rad > 0
! (
Ã
2
φ
=
−(2π
−
1.82)
= −4.46 rad < 0
2 m/s
+1.82 rad = −4.46 rad
=
=cos−1 −
2π − 1.82 = +4.46 rad
8 m/s2
To choose between φ = +1.82 rad and +4.46 rad we need to check for which φ is the slope at t=0
positive: sin(1.82 rad) >0 while sin(+4.46 rad) <0, so the phase constant is φ = +1.82 rad = −4.46 rad.
To relate the phase costant φ to the time along the t axis one can use: t =
(b) ω =
φ
1.82 rad
T =
(12.0 s) = 3.47 s.
2π
2π rad
2π
2π rad
=
= 0.524 rad/s.
T
12.0 s
(c) Using am = xm ω 2 we have: xm =
am
8 cm/s2
=
= 29.1 cm = .291 m.
ω2
(0.524 rad/s)2
3. A 1.2 kg square block, with edge length of 20.0 cm is supported by a massless rod, 56.0 cm long, which is
pivoted at at its end.
Find the period of small angle oscillations.
s
T = 2π
IP
mgh
t
where h = L + d/2
1
m(d2 + d2 ) + m(L + d/2)2 =
12
d2
(0.2 m)2
m[ + (L + d/2)2 ] = (1.2 kg)[
+ (0.56 + 0.1)2 ] = 0.53 kg m2
6
6
6
IP = Icom + m(L + d/2)2 =
v
u
u
T = 2π t
0.53 kgm2
= 1.64 s
(1.2 kg)(9.8 m/s2 )(0.56 + 0.1)m
L
?
d
d