Download A Pivoting Rod on a Spring

skiladæmi 12
Due: 11:59pm on Wednesday, November 25, 2015
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A Pivoting Rod on a Spring
A slender, uniform metal rod of mass M and length l is pivoted without friction about an axis through its midpoint and
perpendicular to the rod. A horizontal spring, assumed massless and with force constant k, is attached to the lower end of
the rod, with the other end of the spring attached to a rigid support.
Part A
We start by analyzing the torques acting on the rod when it is deflected by a small angle θ from the vertical. Consider
first the torque due to gravity. Which of the following statements most accurately describes the effect of gravity on the
rod?
Choose the best answer.
ANSWER:
Under the action of gravity alone the rod would move to a horizontal position. But for small deflections from
the vertical the torque due to gravity is sufficiently small to be ignored.
Under the action of gravity alone the rod would move to a vertical position. But for small deflections from the
vertical the restoring force due to gravity is sufficiently small to be ignored.
There is no torque due to gravity on the rod.
Correct
Assume that the spring is relaxed (exerts no torque on the rod) when the rod is vertical. The rod is displaced by a small
angle θ from the vertical.
Part B
Find the torque τ due to the spring. Assume that θ is small enough that the spring remains effectively horizontal and
you can approximate sin (θ) ≈ θ (and cos(θ) ≈ 1).
Express the torque as a function of θ and other parameters of the problem.
Hint 1. Find the change in spring length
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Deflecting the rod will stretch or compress the spring by a length δ. The spring will react with a restoring force
given by Hooke's law:
.
F = −kδ
What is δ?
Express your answer in terms of l and θ.
Hint 1. Relation between displacement and rotation angle
Consider an object at a distance r from the pivot it is rotating about. If it rotates through an angle ϕ, then
its displacement Δx is given by
.
Δx = rϕ
ANSWER:
δ
l
= 2
θ
Hint 2. Find the moment arm
The torque τ about a point is defined as the product of the force F acting on a body times the moment arm
(perpendicular distance d from the line of action of the force to the center point):
τ = F d.
What is d for the given situation?
Express your answer in terms of quantities given in the problem introduction.
ANSWER:
l
= d
2
ANSWER:
τ
= 2
−
kl θ
4
Correct
Since the torque is opposed to the deflection θ and increases linearly with it, the system will undergo angular
simple harmonic motion.
Part C
What is the angular frequency ω of oscillations of the rod?
Express the angular frequency in terms of parameters given in the introduction.
Hint 1. How to find the oscillation frequency
ω
can be found from the equation of motion by comparing it to the standard form
d
2
θ(t)
dt
2
2
= −ω θ(t).
This equation describes any angular simple harmonic motion. Bring your equation of motion into this standard
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form, and hence extract the expression for ω in terms of the parameters of this specific problem.
Hint 2. Solve the angular equation of motion
The angular equation of motion for this problem relates the change in angular momentum L to the torque τ :
τ =
dL(t)
dt
=I
where I is the moment of inertia of the rod about its center.
Solve this equation for d
2
θ(t)
dt
2
d
2
θ(t)
dt
2
,
.
You should already have found the expression for the torque in Part B.
Express your answer in terms of k, l, I , and θ.
ANSWER:
d
2
θ(t)
dt
2
= 2
−kl θ
4I
Correct
Hint 3. Determine the moment of inertia of the rod
What is the moment of inertia I of a rod of length l about its midpoint?
Express your answer in terms of l and M .
ANSWER:
I
= Ml
2
12
Correct
Compare the equation you found in the second hint to that given in the first to determine ω.
ANSWER:
ω
= −
−
−
√
3k
M
Correct
Note that if the spring were simply attached to a mass m, or if the mass of the rod were concentrated at its ends, −
−
−
−
ω would be √ k/m . The frequency is greater in this case because mass near the pivot point doesn't move as
much as the end of the spring. What do you suppose the frequency of oscillation would be if the spring were
attached near the pivot point?
Measuring the Acceleration Due to Gravity with a Speaker
To measure the magnitude of the acceleration due to gravity g in an unorthodox manner, a student places a ball bearing on
the concave side of a flexible speaker cone . The speaker cone acts as a simple harmonic oscillator whose amplitude is A
and whose frequency f can be varied. The student can measure both A and f with a strobe light. Take the equation of
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motion of the oscillator as
( ) =
cos (
+
,
)
where ω
= 2πf
y(t) = A cos (ωt + ϕ),
and the y axis points upward.
Part A
If the ball bearing has mass m, find N (t) , the magnitude of the normal force exerted by the speaker cone on the ball
bearing as a function of time.
Your result should be in terms of A, either f (or ω), m, g, a phase angle ϕ, and the constant π.
Hint 1. Determine the total force on the ball bearing
What is the net force ∑ Fy on the ball bearing?
Your answer should include N (t) , the normal force as it varies with time.
ANSWER:
∑ F y = may
= N (t) − mg
Hint 2. Find the acceleration of the ball bearing
What is a(t), the acceleration of the ball bearing as a function of time?
Express your answer in terms of given variables. Your answer should not contain N .
Hint 1. How to approach the problem
Acceleration is the second derivative of the displacement:
a(t) =
d
2
dt
y
2
.
Hint 2. Find y(t)
What is y(t), the vertical displacement as a function of time?
Give your answer in terms of A, f , and ϕ.
ANSWER:
= y(t)
Acos(2πf t + ϕ)
ANSWER:
= a(t)
2
−A(2πf ) cos(2πf t + ϕ)
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ANSWER:
N (t)
= m(g − A4π
2
f
2
cos(2πf t + ϕ))
Correct
Part B
The frequency is slowly increased. Once it passes the critical value f b , the student hears the ball bounce. There is
now enough information to calculate g. What is g?
Express the magnitude of the acceleration due to gravity in terms of f b and A.
Hint 1. Determine the force on the ball bearing when it loses contact
What is the net force ∑ Fy on the ball bearing the instant it loses contact with the speaker and is thrown into
the air?
Hint 1. Specify the possible forces
What forces are acting on the ball bearing at that moment?
ANSWER:
There are no forces acting on the ball bearing.
weight only
the normal force only
the normal force and weight
Correct
ANSWER:
∑ Fy
= −mg
Correct
Hint 2. Find the value of cos (2πfb t + ϕ) when the ball loses contact
In the first part of this problem you obtained an equation for the normal force:
2
N (t) = mg − mA(2πf )
.
cos(2πf t + ϕ)
At the critical value of f b , the ball just loses contact with the speaker when the speaker is completely convex
(i.e., when it is at the top of its oscillation). What is the value of cos (2πf b t + ϕ) at this moment?
ANSWER:
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−1 0 1 Correct
As the frequency continues to increase, the normal force will go to zero sooner in the ball's oscillation.
Hint 3. Relation between ω and f
Recall that ω
= 2πf
.
ANSWER:
= g
4Aπ
2
f
2
b
Correct
Exercise 15.8
A certain transverse wave is described by
y(x, t) = B cos[2π(
where B = 5.70 mm , L = 30.0 cm , and τ = 3.40×10−2 s .
Part A
Determine the wave's amplitude.
ANSWER:
A
= 5.70×10−3 m Correct
Part B
Determine the wave's wavelength.
ANSWER:
= 0.300 m λ
Correct
Part C
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Determine the wave's frequency.
x
L
−
t
τ
)] ,
ANSWER:
f
= 29.4 Hz Correct
Part D
Determine the wave's speed of propagation.
ANSWER:
v
= 8.82 m/s Correct
Part E
Determine the wave's direction of propagation.
ANSWER:
+x direction
­x direction
Correct
Exercise 15.16
Part A
With what tension must a rope with length 2.10 m and mass 0.115 kg be stretched for transverse waves of frequency
45.0 Hz to have a wavelength of 0.710 m ?
ANSWER:
T
= 55.9 N Correct
Ant on a Tightrope
A large ant is standing on the middle of a circus tightrope that is stretched with tension Ts . The rope has mass per unit
length μ. Wanting to shake the ant off the rope, a tightrope walker moves her foot up and down near the end of the
tightrope, generating a sinusoidal transverse wave of wavelength λ and amplitude A. Assume that the magnitude of the
acceleration due to gravity is g.
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Part A
What is the minimum wave amplitude Amin such that the ant will become momentarily "weightless" at some point as
the wave passes underneath it? Assume that the mass of the ant is too small to have any effect on the wave
propagation.
Express the minimum wave amplitude in terms of Ts , μ, λ, and g.
Hint 1. Weight and weightless
Weight is generally defined as being the equal and opposite force to the normal force. On a flat surface in a
static situation, the weight is equal to the force due to gravity acting on a mass.
"Weightless" is a more colloquial term meaning that if you stepped on a scale (e.g., in a falling elevator) it would
read zero. Think about what happens to the normal force in this situation.
Note that the force due to gravity does not change and would still be the same as when the elevator was static.
Hint 2. How to approach the problem
In the context of this problem, when will the ant become "weightless"?
ANSWER:
When it has no net force acting on it
When the normal force of the string equals its weight
When the normal force of the string equals twice its weight
When the string has a downward acceleration of magnitude g
Hint 3. Find the maximum acceleration of the string
Assume that the wave propagates as y(x, t)
amax of a point on the string?
. What is the maximum downward acceleration = A sin(ωt − kx)
Express the maximum downward acceleration in terms of π and any quantities given in the problem
introduction.
Hint 1. How to approach the problem
Use the formula given for the displacement of the string to find the acceleration of the string as a function
of position and time. Then determine what the maximum value of this acceleration is. (At some time, the
bit of rope underneath the ant will have this maximum downward acceleration.)
Hint 2. Acceleration of a point on the string
Find the vertical acceleration ay (x, t) of an arbitrary point on the string as a function of time.
Express your answer in terms of A, ω, t, k, and x.
Hint 1. How to find the acceleration
Differentiate the expression given for y(x, t), the displacement of a point on the string, twice.
Hint 2. The first derivative
Differentiate the given equation for the displacement of the string y(x, t) to find the vertical
velocity v y (x, t) of the rope.
Express your answer in terms of A, ω, t, k, and x.
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ANSWER:
= v y (x, t)
ωAcos(ωt − kx)
ANSWER:
= 2
ay (x, t)
−ω Asin(ωt − kx)
Hint 3. Find the maximum downward acceleration
The maximum downward acceleration amax is the most negative possible value of ay (x, t). As the wave
passes beneath the ant, at some time or another the ant will be at a point where the acceleration of the
string has this most negative value.
What is amax ?
Express your answer in terms of ω and quantities given in the problem introduction.
Hint 1. When will the acceleration reach its most negative value?
The most negative acceleration occurs when sin(ωt − kx)
= 1
.
ANSWER:
amax
= 2
−ω A
Hint 4. Determine ω in terms of given quantities
The angular frequency ω of the wave in the string was not given in the problem introduction. To solve the
problem, find an expression for ω in terms of given quantities.
Express the angular frequency in terms of Ts , μ, λ, and π.
Hint 1. How to approach this question
Combine a general formula for ω, a relationship among frequency, wavelength, and velocity, and a
formula for the velocity of a wave on a string to find an expression for ω in terms of quantities
given in the problem introduction.
Hint 2. General formula for ω
The angular frequency of a wave is equal to 2π times the normal frequency: ω
= 2πf
Hint 3. Relationship among frequency, wavelength, and velocity
The frequency, wavelength, and velocity of a wave are related by v
= λf
.
Hint 4. Speed of a wave on a string
What is the speed v of any wave on the string described in the problem introduction?
ANSWER:
−−
v
= √
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Ts
μ
.
ANSWER:
−−
ω
= √
Ts
2π
μ
λ
ANSWER:
amax
= −
Ts
μ
2
(
2π
λ
) A
Hint 4. Putting it all together
Once you have an expression for the maximum acceleration of a point on the string amax , determine what
amplitude is required such that amax = −g. This will be the minimum amplitude Amin for which the ant
becomes weightless.
ANSWER:
A min
= μgλ
2
4π 2 Ts
Correct
Two Velocities in a Traveling Wave
Wave motion is characterized by two velocities: the velocity with which the wave moves in the medium (e.g., air or a string)
and the velocity of the medium (the air or the string itself).
Consider a transverse wave traveling in a string. The mathematical form of the wave is
.
y(x, t) = A sin(kx − ωt)
Part A
Find the speed of propagation v p of this wave.
Express the velocity of propagation in terms of some or all of the variables A, k, and ω.
Hint 1. Perform an intermediate step
Note that the phase of the wave (kx − ωt), and therefore the displacement of the string, is equal to zero at (x, t) = (0, 0).
At what position x
= Δx
is the phase equal to zero a short time t
Express your answer in terms of Δt, ω, and k.
ANSWER:
Δx
= ωΔt
k
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later?
= Δt
ANSWER:
vp
= ω
k
Correct
Part B
Find the y velocity v y (x, t) of a point on the string as a function of x and t.
Express the y velocity in terms of ω, A, k, x, and t.
Hint 1. How to approach the problem
In the problem introduction, you are given an expression for y(x, t), the displacement of the string as a function
of x and t. To find the y velocity, take the partial derivative of y(x, t) with respect to time. That is, take the time
derivative of y(x, t) while treating x as a constant.
Hint 2. A helpful derivative
d
dt
sin(at + b) = a cos(at + b)
ANSWER:
= v y (x, t)
−Aωcos(kx − ωt)
Correct
Part C
Which of the following statements about v x (x, t), the x component of the velocity of the string, is true?
Hint 1. How to approach this question
You are given a form for y(x, t). You are not given any information about Δx(x, t), but it is assumed that each
point on the string only moves in the y direction, i.e. Δx(x, t) = 0.
ANSWER:
v x (x, t)
= vp
v x (x, t)
= v y (x, t)
v x (x, t) has the same mathematical form as v y (x, t) but is 180∘ out of phase.
v x (x, t)
= 0
Correct
So the wave moves in the x direction, even though the string does not. What this means is that even though
individual points on the string only move up and down, a given shape or pattern of points (in this sinusoidal) will
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move to the right as time progresses.
Part D
Find the slope of the string ∂y(x,t)
∂x
as a function of position x and time t.
Express your answer in terms of A, k, ω, x, and t.
Hint 1. A helpful derivative
d
dx
cos(ax + b) = −a sin(ax + b)
ANSWER:
∂y(x,t)
∂x
= Akcos(kx − ωt)
Correct
Part E
Find the ratio of the y velocity of the string to the slope of the string calculated in the previous part.
Express your answer as a suitable combination of some of the variables ω, k, and v p .
ANSWER:
vy (x,t)
∂y(x,t)
= −v p
∂x
Correct
To understand why the ratio of the y velocity of the string to its slope is constant, draw the string with a wave
running along it at time t = 0 . In the vicinity of x = 0, the string is sloped upward. The bit of string at position x = 0 moves downward as the wave moves forward. One­half cycle later, the string in the vicinity of x = 0 will be
sloped downward, and the string at position x = 0 will move upward as the wave moves forward.
In general, if at some particular (x, t) the slope of the string is positive (∂y(x, t)/∂x > 0), that bit of string will be
moving downward (v y (x, t) < 0). If the slope at (x, t) is negative, that bit of string will be moving upward. This
explains why the sign of the ratio of string velocity to slope is always negative.
One way of understanding why the ratio has a constant magnitude is to observe that the more steeply the string is
sloped, the more quickly it will move up or down.
Problem 15.54
You are designing a two­string instrument with metal string 35.0 cm long, as shown in the figure . Both strings are under the
same tension. String S1 has a mass of 8.35 g and produces the note middle C (frequency 262 Hz ) in its fundamental
mode.
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Part A
What should be the tension in the string?
ANSWER:
T
= 802 N Correct
Part B
What should be the mass of string S2 so that it will produce A# (frequency 466 Hz ) as its fundamental?
ANSWER:
M2
= 2.64 g Correct
Part C
To extend the range of your instrument, you include a fret located just under the strings, but not normally touching
them. How far from the upper end should you put this fret so that when you press S1 tightly against it, this string will
produce C# (frequency 277 Hz ) in its fundamental? That is, what is x in the figure?
ANSWER:
x
= 1.90 cm Correct
Part D
If you press S2 against the fret, what frequency of sound will it produce in its fundamental?
ANSWER:
f
= 493 Hz Typesetting math: 100%
Correct
Score Summary:
Your score on this assignment is 99.5%.
You received 6.96 out of a possible total of 7 points.
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