Richard Fitzpatrick
Oscillations and Waves:
An Introduction (2nd Edition):
Solutions to Exercises
Complete Chapter Solutions Manual
are included (Ch 1 to 11)
** Immediate Download
** Swift Response
** All Chapters included
, CHAPTER 1
Simple Harmonic Oscillation
1.1 The mass will fly off the platform when the platforms’s maximum downward acceleration
exceeds the acceleration, g, due to gravity. That is, when
ω 2 a = g. (1.1)
Here, ω = 2π f is the platform’s angular frequency of oscillation, f the same frequency in
hertz, and a the amplitude of the oscillation. It follows that
g
a= . (1.2)
4π 2 f 2
However, f = 5 Hz and g = 9.8 m s−1, so a = 0.0099 m.
1.2 The body will fly off the diaphragm when the diaphragm’s downward acceleration exceeds
the acceleration, g, due to gravity. Hence, as the frequency increases, the body will first fly
off the diaphragm when the maximum downward acceleration becomes equal to g: that is,
when
ω 2 a = g. (1.3)
Here, ω = 2π f is the diaphragm’s angular frequency of oscillation, f the same frequency in
hertz, and a the amplitude of the oscillation. It follows that
r
1 g
f = . (1.4)
2π a
However, a = 1 × 10−5 m and g = 9.8 m s−1 , so f = 157.6 Hz.
1.3 Referring to Fig. 1.1, the transverse restoring force is
f x = −2 T sin θ. (1.5)
m
T T
x
θ
l
FIGURE 1.1 Figure for Ex. 1.3
.
1
, 2 Oscillations and Waves: An Introduction (2nd Edition): Solutions to Exercises
However, from trigonometry,
x
.
tan θ = (1.6)
l
Now, assuming that |x| ≪ 1, it follows that tan θ is small. Hence, θ is small, and we can use
the small-angle approximations
sin θ ≃ tan θ ≃ θ. (1.7)
These approximations imply that
x
sin θ ≃ , (1.8)
l
and !
2T
fx ≃ − x. (1.9)
l
The transverse equation of motion of the mass is
..
m x = fx , (1.10)
which reduces to !
..x = − 2 T x. (1.11)
l
Comparing this equation to the standard form of the simple harmonic oscillator equation,
we deduce that the angular frequency of oscillation is
r
2T
ω= . (1.12)
l
1.4 Let x1 and x2 be the extensions of the first and second springs, respectively. The forces
exerted by these springs are f1 = −k1 x1 and f2 = −k2 x2 , respectively. If the springs are
connected in parallel then x = x1 = x2 , where x is the downward displacement of the mass
with respect to its equilibrium point. However, if the springs are connected in series then
x = x1 + x2 . If the springs are connected in parallel then f = f1 + f2 , where f is the net
downward force acting on the mass due to the springs. However, if the springs are connected
in series then f = f1 = f2 . ( f1 equals f2 because of Newton’s third law of motion.) The
generic equation of motion is
..
m x = f + m g. (1.13)
If the springs are connected in parallel then
..
m x = −(k1 + k2 ) x + m g. (1.14)
Let x = m g/(k1 + k2 ) + δx. It follows that
!
..
δx = −
k1 + k2
δx. (1.15)
m
This is the simple harmonic oscillator equation, so the angular frequency of oscillation be-
comes !1/2
k1 + k2
ω= . (1.16)
m
If the springs are connected in series then f = −k1 x1 = −k2 x2 , so that x2 /x1 = k1 /k2 . Also,
Oscillations and Waves:
An Introduction (2nd Edition):
Solutions to Exercises
Complete Chapter Solutions Manual
are included (Ch 1 to 11)
** Immediate Download
** Swift Response
** All Chapters included
, CHAPTER 1
Simple Harmonic Oscillation
1.1 The mass will fly off the platform when the platforms’s maximum downward acceleration
exceeds the acceleration, g, due to gravity. That is, when
ω 2 a = g. (1.1)
Here, ω = 2π f is the platform’s angular frequency of oscillation, f the same frequency in
hertz, and a the amplitude of the oscillation. It follows that
g
a= . (1.2)
4π 2 f 2
However, f = 5 Hz and g = 9.8 m s−1, so a = 0.0099 m.
1.2 The body will fly off the diaphragm when the diaphragm’s downward acceleration exceeds
the acceleration, g, due to gravity. Hence, as the frequency increases, the body will first fly
off the diaphragm when the maximum downward acceleration becomes equal to g: that is,
when
ω 2 a = g. (1.3)
Here, ω = 2π f is the diaphragm’s angular frequency of oscillation, f the same frequency in
hertz, and a the amplitude of the oscillation. It follows that
r
1 g
f = . (1.4)
2π a
However, a = 1 × 10−5 m and g = 9.8 m s−1 , so f = 157.6 Hz.
1.3 Referring to Fig. 1.1, the transverse restoring force is
f x = −2 T sin θ. (1.5)
m
T T
x
θ
l
FIGURE 1.1 Figure for Ex. 1.3
.
1
, 2 Oscillations and Waves: An Introduction (2nd Edition): Solutions to Exercises
However, from trigonometry,
x
.
tan θ = (1.6)
l
Now, assuming that |x| ≪ 1, it follows that tan θ is small. Hence, θ is small, and we can use
the small-angle approximations
sin θ ≃ tan θ ≃ θ. (1.7)
These approximations imply that
x
sin θ ≃ , (1.8)
l
and !
2T
fx ≃ − x. (1.9)
l
The transverse equation of motion of the mass is
..
m x = fx , (1.10)
which reduces to !
..x = − 2 T x. (1.11)
l
Comparing this equation to the standard form of the simple harmonic oscillator equation,
we deduce that the angular frequency of oscillation is
r
2T
ω= . (1.12)
l
1.4 Let x1 and x2 be the extensions of the first and second springs, respectively. The forces
exerted by these springs are f1 = −k1 x1 and f2 = −k2 x2 , respectively. If the springs are
connected in parallel then x = x1 = x2 , where x is the downward displacement of the mass
with respect to its equilibrium point. However, if the springs are connected in series then
x = x1 + x2 . If the springs are connected in parallel then f = f1 + f2 , where f is the net
downward force acting on the mass due to the springs. However, if the springs are connected
in series then f = f1 = f2 . ( f1 equals f2 because of Newton’s third law of motion.) The
generic equation of motion is
..
m x = f + m g. (1.13)
If the springs are connected in parallel then
..
m x = −(k1 + k2 ) x + m g. (1.14)
Let x = m g/(k1 + k2 ) + δx. It follows that
!
..
δx = −
k1 + k2
δx. (1.15)
m
This is the simple harmonic oscillator equation, so the angular frequency of oscillation be-
comes !1/2
k1 + k2
ω= . (1.16)
m
If the springs are connected in series then f = −k1 x1 = −k2 x2 , so that x2 /x1 = k1 /k2 . Also,
