Physics 8
Spring 2012
TA:
NAME:
Quiz 13 - Solutions
Make sure your name is on your quiz, and please box your final answer. Because we
will be giving partial credit, be sure to attempt all the problems, even if you don’t finish them!
1. Consider a mass, m, attached to a spring with spring constant k. The other end of the
spring is attached to a motor that exerts a driving force F (t) = F0 cos (ωt), where F0
is a constant and ω is the driving frequency. When the motor is turned on the mass
oscillates back and forth with the same frequency ω, but is also subject to a resistive
(drag) force proportional to its velocity FD = −mγv, where m is the mass and γ is a
constant.
(a) Write down the sum of the forces to determine the acceleration, ẍ, of the mass
(Note: don’t forget about the spring force!).
(b) By substitution, show that the equation you found in part (a) is solved by
x (t) = A cos (ωt − δ) ,
and determine the constants A and δ in terms of the natural spring frequency
ω0 , driving frequency, ω, F0 , m, and γ. Hints: Recall that cos (A − B) =
cos (A) cos (B)+sin (A) sin (B), and sin (A − B) = sin (A) cos (B)−cos (A) sin (B),
and that the sine and cosine terms all need to satisfy the expression in part (a).
Finally, note that if tan δ = A/B, then sin δ = √A2A+B 2 , and cos δ = √A2B+B 2 .
————————————————————————————————————
Solution
1. (a) The mass is subject to three forces: the spring force (FS = −kx), the drag force
(FD = −mγv), and the driving force (F (t) = F0 cos (ωt)). The sum of the forces
is
−kx − mγv + F0 cos (ωt) = ma.
Writing the velocity as ẋ, and the acceleration as ẍ we find
ẍ + γ ẋ + ω02 x =
F0
cos (ωt) ,
m
where we have written the natural frequency as ω02 = k/m. The acceleration is
found simply by solving for ẍ, but we’ll leave it in this form.
1
(b) We simply plug in our solution for x (t). We need to take two derivatives,
ẋ (t) = −ωA sin (ωt − δ)
ẍ (t) = −ω 2 A cos (ωt − δ) .
Now, we plug in these expressions to our equation,
−ω 2 A cos (ωt − δ) − γωA sin (ωt − δ) + ω02 A cos (ωt − δ) =
F0
cos (ωt) .
m
Grouping together the cosine terms gives
F0
cos (ωt) .
ω02 − ω 2 A cos (ωt − δ) − γωA sin (ωt − δ) =
m
Expanding the cosine and sine terms we have
[ω02 − ω 2 ] A (cos (ωt) cos δ + sin (ωt) sin δ)
−γωA (sin (ωt) cos δ − cos (ωt) sin δ) =
F0
m
cos (ωt) .
Now, we need the sine terms to cancel off since the driving force only involves
cosines. So, we need
2
ω0 − ω 2 A sin (ωt) sin δ − γωA sin (ωt) cos δ = 0,
or
tan δ =
ω02
γω
,
− ω2
which determines the phase shift δ.
Once we get rid of the sine terms we are left with
2
F0
ω0 − ω 2 cos δ + γω sin δ A cos (ωt) =
cos (ωt) .
m
Canceling off the cos (ωt) terms and writing
γω
,
sin δ = q
2 2
2
2
2
(ω − ω0 ) + γ ω
we find

2
ω02 − ω 2
cos δ = q
,
2 2
2
2
2
(ω − ω0 ) + γ ω
and
2
ω02 )
2
2

(ω −
+γ ω 
F0
q
A=
.
m
2 2
2
2
2
(ω − ω0 ) + γ ω
q
2
But, the term in brackets is just (ω 2 − ω02 ) + γ 2 ω 2 , which gives for A,
F0 /m
A= q
,
2 2
2
2
2
(ω − ω0 ) + γ ω
which solves the problem.
2