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Physics 18 Spring 2011
Homework 11 - Solutions
Wednesday April 6, 2011
Make sure your name is on your homework, 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. The homework is due at the beginning of class on Wednesday, April 13th. Because
the solutions will be posted immediately after class, no late homeworks can be accepted! You
are welcome to ask questions during the discussion session or during office hours.
1. An ultrasonic transducer, of the type used in medical ultrasound imaging, is a very thin
disk (m = 0.10 grams) driven back and forth in SHM at 1.0 MHz by an electromagnetic
coil.
(a) The maximum restoring force that can be applied to the disk without breaking it
is 40,000 N. What is the maximum oscillation amplitude that won’t rupture the
disk?
(b) What is the disk’s maximum speed at this amplitude?
————————————————————————————————————
Solution
(a) Since the displacement of the oscillator obeys the expression x(t) = A cos (ωt + δ),
then the acceleration is a = ẍ = −ω 2 A cos (ωt + δ), and so the maximum acceleration occurs when the cosine is -1. Thus, amax = Aω 2 . The force associated with
this acceleration is F = ma, and so the maximum amplitude is
A=
F
F
= 2 2,
2
mω
4π mf
where f is the ordinary frequency. Plugging in the numbers gives
A=
F
4π 2 mf 2
=
4π 2
40000
= 10−5 meters, or 10 microns.
× 10−4 × (106 )2
(b) The maximum speed is found in the same way as the acceleration and gives
vmax = Aω = 2πf A, and so
vmax = 2πf A = 2π × 106 × 10−5 = 20π ≈ 63 m/s.
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2. An object of mass m is suspended from a vertical spring of force constant 1800 N/m.
When the object is pulled down 2.50 cm from equilibrium and released from rest, the
object oscillates at 5.50 Hz.
(a) Find m.
(b) Find the amount the spring is stretched from its unstretched length when the
object is in equilibrium.
(c) Write expressions for the displacement x, the velocity vx , and the acceleration ax
as functions of time.
————————————————————————————————————
Solution
(a) The oscillation (angular) frequency of a spring ω0 =
quency is f = ω0 /2π. Thus, the mass is
m=
p
k/m. The ordinary fre-
k
1800
= 2
= 1.51 kg.
2
2
4π f
4π (5.50)
(b) Attaching the mass to the spring stretches it by applying a gravitational force,
mg, which is compensated by the string force kx. Setting these two equal gives
x = mg/k, so
1.51 × 9.8
mg
=
= 8.2 mm.
x=
k
1800
(c) Again, the general solution of the position of a function of time is x(t) = A cos (ωt + δ),
where δ is the phase shift. Since the particle starts being pulled down, it’s initial
position is at x(t = 0) = −A, and so δ = π. The amplitude is A = 2.5 cm, and
ω = 2πf = 11π. So, we can immediately write
x(t) = 2.5 cos(11πt + π) = −2.5 cos(11πt).
We’ll take the last expression for x(t) = −2.5 cos(11πt) for convenience.
Now, the velocity is v = ẋ = 27.5π sin(11πt) ≈ 86.4 sin(11πt). Finally, the acceleration is a = v̇ = 86.4(11π) cos(11πt) ≈ 2985 cos(11πt) ≈ 30 m/s2 cos(11πt).
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3. You have designed a cat door that consists of a square piece of plywood that is 1.0 in
thick and 6.0 in on a side, and is hinged at its top. To make sure the cat has enough
time to get through it safely, the door should have a natural period of at least 1.0 s.
Will your design work? If not, explain qualitatively what you would need to do to
make it meet your requirements.
————————————————————————————————————
Solution
This is a physical pendulum, where we have to worry about the moment of inertia, I.
The period of a physical pendulum is
s
I
,
T = 2π
M gD
where I is the moment of inertia, M is the mass of the door, and D is the distance of
the center of mass from the pivot point. We need to determine the moment of inertia
of the door. Consulting the moments in Table 9-1 we see that the moment of inertia
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M (a2 + b2 ), where a is the distance on the side
of a solid box about the center is I = 12
(6 inches in our case), and b is the thickness (1 inch). To make the door pivot about
the edge we use the parallel-axis theorem, adding M (a/2)2 + M (b/2)2 to I to find
1
Idoor = M (a2 + b2 ).
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So we just now need to plug in everything, noting that D = a/2 is the distance to the
center of mass. So,
s
s
2M (a2 + b2 )
2(a2 + b2 )
T = 2π
= 2π
.
3M ga
3ga
Now, recalling that there are 2.54 centimeters (0.0254 meters) per inch, then b = 0.0254
meters, and a = 6 × 0.0254 = 0.152 meters. So, we can determine the period,
s
s
2(a2 + b2 )
2((0.0254)2 + (0.154)2 )
= 2π
= 0.652 seconds.
T = 2π
3ga
3(9.8)(0.154)
So, your design won’t work! You should make the height at least 9 1/2 inches in order
for the period to be one second.
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4. In one of your chemistry labs, you determine that one of the vibrational modes of
the HCl molecule has a frequency of 8.969 × 1013 Hz. Using the result of Problem 8,
find the “effective spring constant” between the H atom and the Cl atom in the HCl
molecule.
————————————————————————————————————
Solution
From problem 8, we see that the angular frequency of the molecule is
s
s
k
k(mH + mCl )
=
.
ω=
µ
mH mCl
Solving for the spring constant gives
k=
mH mCl 2
ω .
mH + mCl
Now,
mH = 1 amu = 1.67 × 10−27 kg
mCl = 35 amu = 5.92 × 10−26 kg.
So, plugging in all the numbers gives
k=
mH mCl 2 (1.67 × 5.92) × 10−53
13 2
ω =
8.969
×
10
= 13.1 N/m,
mH + mCl
(.167 + 5.92) × 10−26
which is huge for a molecule!
4
5. A straight tunnel is dug through Earth, as
shown in the figure. Assume that the walls
of the tunnel are frictionless.
(a) The gravitational force exerted by
Earth on a particle of mass m at
a distance r from the center of
Earth when r < RE is Fr =
3
− (GmME /RE
) r, where ME is the
mass of Earth and RE is its radius.
Show that the net force on a particle of mass m a distance x from
the middle of the tunnel is given by
3
Fx = − (GmME /RE
) x and that the
motion of the particle is therefore
simple harmonic motion.
(b) Show that the period of the motion is
independent of the length ofpthe tunnel and is given by T = 2π RE /g.
(c) Find its numerical value in minutes.
————————————————————————————————————
Solution
E
r sin θ, where sin θ = x/r. So,
(a) From the figure we see that Fx = Fr sin θ = − GmM
R2
E
Fx = mẍ = −
GmME
x.
2
RE
So, the force is proportional to the distance along the tunnel (the force is F ∝ −x),
so the motion is simple harmonic.
p
2
E
(b) We see that ẍ = − GM
x,
which
is
gives
ω
=
GME /RE
, and so the period is
2
0
RE
s
2
2π
RE
T =
= 2π
.
ω0
GME
q
2
Recalling that g = GME /RE , then the period is T = 2π RgE .
(c) Plugging in the values RE = 6.37 × 106 meters, and g = 9.8 m/s2 gives
s
r
RE
6.37 × 106
T = 2π
= 2π
= 84.4 minutes.
g
9.8
This means that if you dug a tunnel from one spot on Earth, to any other spot
on Earth you could travel between them in 42 minutes, just by falling!
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