Solutions 2.4-Page 140
Problem 3
A mass of 3 kg is attached to the end of a spring that is stretched 20 cm by a force of
15N. It is set in motion with initial position x0 = 0 and initial velocity v0 = −10 m/s.
Find the amplitude, period, and frequency of the resulting motion.
The spring constants, k = 15 N/ 0.2 m = 75 N/m. There is no mention of damping in the
problem statement, and no outside forces acting on the system. The equation that
governs the motion of the mass is 3 x ′′ + 75 x = 0 . Dividing through by the mass
x ′′ + 25 x = 0
ω 0 , the circular frequency, is calculated as
k
= 5 rad / s.
m
From eq.10 on pg.135, the solution of the above differential equation is of the following
form:
x(t ) = A cos 5t + B sin 5t
Two equations will be necessary to solve for the two unknowns, A and B. Differentiating
the position function to get the velocity function will provide the second equation.
x ′(t ) = v(t ) = 5B cos 5t − 5 A sin 5t
The initial conditions are now substituted
x(0) = 0 = A
v(0) = −10 = 5 B
B = −2
x(t ) = −2 sin 5t
The amplitude is 2 m.
ω
5
= 0.80 Hz
Frequency = 0 =
2π 2π
1
= 1.26 s
Period =
Frequency
Problem 8
Most grandfather clocks have pendulums with adjustable lengths. One such clock loses
10 min per day when the length of its pendulum is 30 in. With what length pendulum
will this clock keep perfect time?
A relationship between the period and length of the pendulum must be developed for the
two situations. According to eq.6 on pg.134, the circular frequency, ω , of a pendulum is
given by ω 2 = g / L = GM / R 2 L . Therefore the period, p = 2π / ω = 2πR L / GM .
Dividing this equation for period by another period with another length gives the
necessary relationship for this problem:
p1
=
p2
L1
L2
(This result is also given in Problem 5.)
The given information is as follows:
L1 = 30
Assuming the pendulum executes n cycles per day (1440 minutes in a day), its period
when it’s keeping perfect time is p 2 = 1440 / n minutes.
Since the pendulum takes 10 min longer to carry out the same number of cycles, the
period when the pendulum loses 10 min is p1 = 1450 / n minutes.
Solving for L2 gives L2 =
L2 = 29.6 in
L1 p 2
p1
2
2
=
30(1440) 2
(1450) 2
Problem 10
Consider a floating cylindrical buoy with radius r , height h , and uniform density ρ ≤ 0.5
(recall that the density of water is 1 g/cm3). The buoy is initially suspended at rest with
its bottom at the top surface of the water and is released at time t = 0. Thereafter it is
acted on by two forces: a downward gravitational force equal to its weight mg = ρπr 2 hg
and an upward force of buoyancy equal to the weight πr 2 xg of water displaced, where
x = x(t ) is the depth of the bottom of the buoy beneath the surface at time t (Fig.2.4.9).
Conclude that the buoy undergoes simple harmonic motion around its equilibrium
position xe = ρh with period p = 2π ρh / g . Compute p and the amplitude of the
motion if ρ = 0.5 g/cm3, h = 200 cm, and g = 980 cm/s2.
The F = ma is ρπr 2 hx ′′ = ρπr 2 hg − πr 2 xg , if the downward direction is taken to be
positive.
(Note: The upward buoyancy force should be ρ water πr 2 xg in order to have the proper
units of force. ρ is left out because ρ = 1 for water.)
The above equation simplifies to x ′′ + ( g / ρh )x = g . This is a nonhomogeneous equation.
Recall from pg.118 that the solution of a nonhomogeneous equation is a particular
solution plus the complimentary solution.
xc = A cos ω o t + B sin ω 0 t
The particular solution is the equilibrium position. (Substitute xe = ρh into the
differential equation and you will see that it works.) Thus the general solution is:
x(t ) = A cos ω o t + B sin ω 0 t + ρh
ω o = g / ρh
Differentiating to get a second equation gives x ′ = Bω o cos ω o t − Aω o sin ω o t .
Substituting the initial conditions, x(0) = x ′(0) = 0 , and simplifying yields:
x(t ) = ρh(1 − cos ω o t ) This is an equation for harmonic
motion.
With the given values for ρ , h, and g, the period is easily solved for
using p = 2π ρh / g .
p = 2.01 sec
Problem 14
Suppose that the mass in a mass-spring-dashpot system with m = 25, c = 10, and k = 226
is set in motion with x(0) = 20 and x ′(0) = 41 . (a) Find the position function x(t ) and
show that its graph looks as indicated in Fig.2.4.12. (b) Find the pseudoperiod of the
oscillations and the equations of the “envelope curves” that are dashed in the figure.
(a) From the given information, the differential equation governing the motion of the
mass is
25 x ′′ + 10 x ′ + 226 x = 0
x ′′ + 2(0.2) x ′ + 9.04 x = 0
From pg.137, p =
c
= 0.2 and ω 0 = k / m = 3.01
2m
From eq.17 on pg.137, the characteristic equation corresponding to this differential
2
2
equation has roots r1 , r2 = − p ± ( p 2 − ω 0 )1 / 2 which depend on the sign of p 2 − ω 0 .
From pg.137,
c 2 − 4km 0.04 − 4(3.01)25
2
p2 − ω0 =
=
< 0 and thus the system is underdamped.
4m 2
4(25) 2
For the underdamped case (pg.138), the characteristic equation has complex roots
− p ± i ω 0 − p 2 = −0.2 ± 3i
2
Therefore the solution of the governing differential equation is
x(t ) = e −0.2t ( A cos 3t + B sin 3t )
Differentiating the position function will give the necessary second equation so that the
initial conditions can be used to solve for A and B.
x ′(t ) = −0.2e −0.2t ( A cos 3t + B sin 3t ) + e −0.2t (3B cos 3t − 3 A sin 3t )
Substituting the initial conditions gives the following system of equations:
x(0) = 20 = A
x ′(0) = 41 = −0.2(20) + (3B)
B = 15
C = 20 2 + 15 2 = 25
From eq.22, x(t ) = 25e −0.2t cos(3t − α ) where α = tan − 1(15 / 20) = .6435 rad ( α is in the
first quadrant.)
x (t ) = 25e −0.2 t cos(3t − 0.64)
The graph is given below. The computer program Maple was used to graph the position
equation.
(b) The pseudoperiod is T = 2π / 3
From fig.2.4.8, the envelope curves are x = ±25e −0.2t
Problem 23
This problem deals with a highly simplified model of a car of weight 3200 lb (mass
m = 100 slugs in fps units). Assume that the suspension system acts like a single spring
and its shock absorbers like a single dashpot, so that its vertical vibrations satisfy Eq. (4)
with appropriate values of the coefficients. (a) Find the stiffness coefficients k of the
spring if the car undergoes free vibrations at 80 cycles per minute (cycles/min) when its
shock absorbers are disconnected. (b) With the shock absorbers connected the car is set
into vibration by driving it over a bump, and the resulting damped vibrations have a
frequency of 78 cycles/min. After how long will the time-varying amplitude be 1% of its
initial value?
With m = 100 , we have ω = k / 100 . We are also given information about the
frequency. Recall from Eq.14 on pg.135 that the relationship between frequency and
circular frequency is ω = 2πν .
ω = 2π (80 cycles/min)(1 min/60 sec) = 8π / 3
Equating the known information on circular frequency yields:
8π / 3 = k / 100
k = 7018 lb/ft
(b) Now ω = 2π (78 cycles/min)(1 min/60 sec) = 8.16814 sec −1 .
4km − c 2
Eq.21 on pg.138 states that ω1 = ω 0 − p =
. Substituting the known
2m
values and solving for the damping constant yields c = 372.31
2
Therefore p = c / 2m = 1.8615 .
Finally e − pt = 0.01 gives t = 2.47 sec
2
Problem 24
Problems 24-34 deal with a mass-spring-dashpot system having position function x(t)
satisfying Eq.(4). We write xo = x(0) and v0 = x ′(0) and recall that p = c /(2m) ,
ω 0 2 = k / m , and ω1 2 = ω 0 2 − p 2 . The system is critically damped, overdamped, or
underdamped, as specified in each problem.
(Critically damped) Show that in this case that
x(t ) = ( x0 + v0 t + px0 t )e − pt
From
eq.19
on
pg.138,
x(t ) = e − pt (c1 + c 2 t ) .
Differentiating
yields
− pt
− pt
x ′ = − pe (c1 + c 2 t ) + c 2 e . Now the differential equations can be used to solve for the
constants.
x0 = c1
v0 = − p( x0 ) + c2
c2 = v0 + x0 p
Substituting the for the constants gives
x(t ) = ( x0 + v0 t + px0 t )e − pt
Problem 34
Problems 24-34 deal with a mass-spring-dashpot system having position function x(t)
satisfying Eq.(4). We write xo = x(0) and v0 = x ′(0) and recall that p = c /(2m) ,
ω 0 2 = k / m , and ω1 2 = ω 0 2 − p 2 . The system is critically damped, overdamped, or
underdamped, as specified in each problem.
(Underdamped) A body weighing 100 lb (mass m = 3.125 slugs in fps units) is
oscillating attached to a spring and a dashpot. Its first two maximum displacements of
6.73 in. and 1.46 in. are observed to occur at times 0.34 s and 1.17 s, respectively.
Compute the damping constant (in pound-seconds per foot) and spring constant (in
pounds per foot).
Note that the problem does not state whether the maximum displacements occur
approximately a pseudoperiod apart or approximately a half pseudoperiod apart. If they
are approximately a full pseudoperiod apart then the displacements are positive.
However, if they are a half pseudoperiod apart then one of the displacements would be
negative.
For an underdamped system x(t ) = Ce − pt cos(ω 1t − α ) , then
x ′(t ) = − pCe − pt cos(ω 1t − α ) + Cω 1e − pt sin(ω1t − α ) . The derivative is set to zero because
maximums (or minimums) occur when the derivative is zero. Simplifying yields
tan(ω 1t − α ) = − p / ω 1 . Thus at a maxima, the tan function has a constant value. The
tangent of an angle is the same for another angle only if the angles are 2 π apart.
Thus ω1t 2 − ω 1t1 = 2π (see the result of Problem 32). This equation can be used along
with the given times when the maximums occur to find that ω1 = 7.57 .
The result of Problem 33 is ln
x1 2πp
. With x1 = 6.73 and x 2 = 1.46 , this leads to
=
ω1
x2
p = 1.84
Eq.16 on pg.137 gives c = 2mp = 2(100 / 32)(1.84) = 11.51 lb-sec/ft
And Eq.21 yields k = (4m 2ω1 + c 2 ) / 4m = 189.68 lb/ft
2