A Pivoting Rod on a Springisa13/Efni/Einn/F.hl/Edlisfraedi 1 V/skiladaemi 12.pdf · What is ?...

skiladæmi 12Due: 11:59pm on Wednesday, November 25, 2015

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A Pivoting Rod on a Spring

A slender, uniform metal rod of mass and length is pivoted without friction about an axis through its midpoint andperpendicular to the rod. A horizontal spring, assumed massless and with force constant , is attached to the lower end ofthe 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 from the vertical. Considerfirst the torque due to gravity. Which of the following statements most accurately describes the effect of gravity on therod?

Choose the best answer.

ANSWER:

Correct

Assume that the spring is relaxed (exerts no torque on the rod) when the rod is vertical. The rod is displaced by a smallangle from the vertical.

Part B

Find the torque due to the spring. Assume that is small enough that the spring remains effectively horizontal andyou can approximate (and ).

Express the torque as a function of and other parameters of the problem.

Hint 1. Find the change in spring length

Deflecting the rod will stretch or compress the spring by a length . The spring will react with a restoring forcegiven by Hooke's law:

M lk

θ

Under the action of gravity alone the rod would move to a horizontal position. But for small deflections fromthe 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 thevertical the restoring force due to gravity is sufficiently small to be ignored.

There is no torque due to gravity on the rod.

θ

τ θsin (θ) ≈ θ cos(θ) ≈ 1

θ

δTypesetting math: 100%

.What is ?

Express your answer in terms of and .

Hint 1. Relation between displacement and rotation angle

Consider an object at a distance from the pivot it is rotating about. If it rotates through an angle , thenits displacement is given by

.

ANSWER:

Hint 2. Find the moment arm

The torque about a point is defined as the product of the force acting on a body times the moment arm(perpendicular distance from the line of action of the force to the center point):

What is for the given situation?

Express your answer in terms of quantities given in the problem introduction.

ANSWER:

ANSWER:

Correct

Since the torque is opposed to the deflection and increases linearly with it, the system will undergo angularsimple harmonic motion.

Part C

What is the angular frequency 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

can be found from the equation of motion by comparing it to the standard form

.

This equation describes any angular simple harmonic motion. Bring your equation of motion into this standardform, and hence extract the expression for in terms of the parameters of this specific problem.

F = −kδδ

l θ

r ϕΔx

Δx = rϕ

= δ θl2

τ Fd

τ = Fd.d

= dl2

= τ −k θl2

4

θ

ω

ω

= − θ(t)θ(t)d2

dt2 ω2

ωTypesetting math: 100%

Hint 2. Solve the angular equation of motion

The angular equation of motion for this problem relates the change in angular momentum to the torque :

,

where is the moment of inertia of the rod about its center.

Solve this equation for .

You should already have found the expression for the torque in Part B.

Express your answer in terms of , , , and .

ANSWER:

Correct

Hint 3. Determine the moment of inertia of the rod

What is the moment of inertia of a rod of length about its midpoint?

Express your answer in terms of and .

ANSWER:

Correct

Compare the equation you found in the second hint to that given in the first to determine .

ANSWER:

Correct

Note that if the spring were simply attached to a mass , or if the mass of the rod were concentrated at its ends, would be . 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 wereattached near the pivot point?

Measuring the Acceleration Due to Gravity with a Speaker

To measure the magnitude of the acceleration due to gravity in an unorthodox manner, a student places a ball bearing onthe concave side of a flexible speaker cone . The speaker cone acts as a simple harmonic oscillator whose amplitude is and whose frequency can be varied. The student can measure both and with a strobe light. Take the equation ofmotion of the oscillator as

,

L τ

τ = = IdL(t)

dt

θ(t)d2

dt2

I

θ(t)d2

dt2

k l I θ

= θ(t)d2

dt2−k θl2

4I

I l

l M

= I Ml2

12

ω

= ω 3kM

−−−√

mω k/m

− −−−√

gA

f A f

( ) = cos ( + )

Typesetting math: 100%

,where and the y axis points upward.

Part A

If the ball bearing has mass , find , the magnitude of the normal force exerted by the speaker cone on the ballbearing as a function of time.

Your result should be in terms of , either (or ), , , a phase angle , and the constant .

Hint 1. Determine the total force on the ball bearing

What is the net force on the ball bearing?

Your answer should include , the normal force as it varies with time.

ANSWER:

Hint 2. Find the acceleration of the ball bearing

What is , the acceleration of the ball bearing as a function of time?

Express your answer in terms of given variables. Your answer should not contain .

Hint 1. How to approach the problem

Acceleration is the second derivative of the displacement:

.

Hint 2. Find

What is , the vertical displacement as a function of time?

Give your answer in terms of , , and .

ANSWER:

ANSWER:

y(t) = A cos (ωt + ϕ)ω = 2πf

m N(t)

A f ω m g ϕ π

∑ Fy

N(t)

= ∑ = mFy ay N(t) − mg

a(t)

N

a(t) = yd2

dt2

y(t)

y(t)

A f ϕ

= y(t) Acos(2πft + ϕ)

= a(t) −A cos(2πft + ϕ)(2πf)2


ANSWER:

Correct

Part B

The frequency is slowly increased. Once it passes the critical value , the student hears the ball bounce. There isnow enough information to calculate . What is ?

Express the magnitude of the acceleration due to gravity in terms of and .

Hint 1. Determine the force on the ball bearing when it loses contact

What is the net force on the ball bearing the instant it loses contact with the speaker and is thrown intothe air?

Hint 1. Specify the possible forces

What forces are acting on the ball bearing at that moment?

ANSWER:

Correct

ANSWER:

Correct

Hint 2. Find the value of when the ball loses contact

In the first part of this problem you obtained an equation for the normal force:

.

At the critical value of , 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 at this moment?

ANSWER:

= N(t) m(g − A4 cos(2πft + ϕ))π2 f 2

fbg g

fb A

∑ Fy

There are no forces acting on the ball bearing.

weight only

the normal force only

the normal force and weight

= ∑ Fy −mg

cos(2π t + ϕ)fb

N(t) = mg − mA(2πf cos(2πft + ϕ))2

fbcos (2π t + ϕ)fb


Correct

As the frequency continues to increase, the normal force will go to zero sooner in the ball's oscillation.

Hint 3. Relation between and

Recall that .

ANSWER:

Correct

Exercise 15.8

A certain transverse wave is described by

,

where = 5.70 , = 30.0 , and = 3.40×10−2 .

Part A

Determine the wave's amplitude.

ANSWER:

Correct

Part B

Determine the wave's wavelength.

ANSWER:

Correct

Part C

Determine the wave's frequency.

−101

ω f

ω = 2πf

= g 4Aπ2 fb2

y(x,t) = Bcos[2π( − )]xL

tτ

B mm L cm τ s

= 5.70×10−3 A m

= 0.300 λ m


ANSWER:

Correct

Part D

Determine the wave's speed of propagation.

ANSWER:

Correct

Part E

Determine the wave's direction of propagation.

ANSWER:

Correct

Exercise 15.16

Part A

With what tension must a rope with length 2.10 and mass 0.115 be stretched for transverse waves of frequency45.0 to have a wavelength of 0.710 ?

ANSWER:

Correct

Ant on a Tightrope

A large ant is standing on the middle of a circus tightrope that is stretched with tension . The rope has mass per unitlength . Wanting to shake the ant off the rope, a tightrope walker moves her foot up and down near the end of thetightrope, generating a sinusoidal transverse wave of wavelength and amplitude . Assume that the magnitude of theacceleration due to gravity is .

= 29.4 f Hz

= 8.82 v m/s

+x direction

x direction

m kgHz m

= 55.9 T N

Tsμ

λ Ag


Part A

What is the minimum wave amplitude such that the ant will become momentarily "weightless" at some point asthe wave passes underneath it? Assume that the mass of the ant is too small to have any effect on the wavepropagation.

Express the minimum wave amplitude in terms of , , , and .

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 astatic 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 wouldread 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.


In the context of this problem, when will the ant become "weightless"?

ANSWER:

Hint 3. Find the maximum acceleration of the string

Assume that the wave propagates as . What is the maximum downward acceleration of a point on the string?

Express the maximum downward acceleration in terms of and any quantities given in the problemintroduction.


Use the formula given for the displacement of the string to find the acceleration of the string as a functionof position and time. Then determine what the maximum value of this acceleration is. (At some time, thebit of rope underneath the ant will have this maximum downward acceleration.)

Hint 2. Acceleration of a point on the string

Find the vertical acceleration of an arbitrary point on the string as a function of time.

Express your answer in terms of , , , , and .

Hint 1. How to find the acceleration

Differentiate the expression given for , the displacement of a point on the string, twice.

Hint 2. The first derivative

Differentiate the given equation for the displacement of the string to find the verticalvelocity of the rope.


ANSWER:

Amin

Ts μ λ g

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

y(x, t) = A sin(ωt − kx)amax

π

(x, t)ay

A ω t k x

y(x, t)

y(x, t)(x, t)vy

A ω t k xTypesetting math: 100%

ANSWER:

Hint 3. Find the maximum downward acceleration

The maximum downward acceleration is the most negative possible value of . As the wavepasses beneath the ant, at some time or another the ant will be at a point where the acceleration of thestring has this most negative value.

What is ?

Express your answer in terms of and quantities given in the problem introduction.

Hint 1. When will the acceleration reach its most negative value?

The most negative acceleration occurs when .

ANSWER:

Hint 4. Determine in terms of given quantities

The angular frequency of the wave in the string was not given in the problem introduction. To solve theproblem, find an expression for in terms of given quantities.

Express the angular frequency in terms of , , , and .

Hint 1. How to approach this question

Combine a general formula for , a relationship among frequency, wavelength, and velocity, and aformula for the velocity of a wave on a string to find an expression for in terms of quantitiesgiven in the problem introduction.

Hint 2. General formula for

The angular frequency of a wave is equal to times the normal frequency: .

Hint 3. Relationship among frequency, wavelength, and velocity

The frequency, wavelength, and velocity of a wave are related by .

Hint 4. Speed of a wave on a string

What is the speed of any wave on the string described in the problem introduction?

ANSWER:

= (x, t)vy ωAcos(ωt − kx)

= (x, t)ay − Asin(ωt − kx)ω2

amax (x, t)ay

amax

ω

sin(ωt − kx) = 1

= amax − Aω2

ω

ωω

Ts μ λ π

ωω

ω

2π ω = 2πf

v = λf

v

= vTs

μ

−−√


ANSWER:

ANSWER:

Hint 4. Putting it all together

Once you have an expression for the maximum acceleration of a point on the string , determine whatamplitude is required such that . This will be the minimum amplitude for which the antbecomes weightless.

ANSWER:

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

.

Part A

Find the speed of propagation of this wave.

Express the velocity of propagation in terms of some or all of the variables , , and .

Hint 1. Perform an intermediate step

Note that the phase of the wave , and therefore the displacement of the string, is equal to zero at .

At what position is the phase equal to zero a short time later?

Express your answer in terms of , , and .

ANSWER:

= ωTs

μ

−−√ 2π

λ

= amax − ATs

μ ( )2πλ

2

amax= −gamax Amin

= Aminμgλ2

4π2Ts

y(x, t) = A sin(kx − ωt)

vp

A k ω

(kx − ωt)(x, t) = (0, 0)

x = Δx t = Δt

Δt ω k

= Δx ωΔtk


ANSWER:

Correct

Part B

Find the y velocity of a point on the string as a function of and .

Express the y velocity in terms of , , , , and .


In the problem introduction, you are given an expression for , the displacement of the string as a functionof and . To find the y velocity, take the partial derivative of with respect to time. That is, take the timederivative of while treating as a constant.

Hint 2. A helpful derivative

ANSWER:

Correct

Part C

Which of the following statements about , the x component of the velocity of the string, is true?

Hint 1. How to approach this question

You are given a form for . You are not given any information about , but it is assumed that eachpoint on the string only moves in the y direction, i.e. .

ANSWER:

Correct

So the wave moves in the x direction, even though the string does not. What this means is that even thoughindividual points on the string only move up and down, a given shape or pattern of points (in this sinusoidal) willmove to the right as time progresses.

= vpωk

(x, t)vy x t

ω A k x t

y(x, t)x t y(x, t)

y(x, t) x

sin(at+b) = acos(at+b)ddt

= (x, t)vy −Aωcos(kx − ωt)

(x, t)vx

y(x, t) Δx(x, t)Δx(x, t) = 0

has the same mathematical form as but is out of phase.

(x, t) =vx vp

(x, t) = (x, t)vx vy

(x, t)vx (x, t)vy 180∘

(x, t) = 0vx


Part D

Find the slope of the string as a function of position and time .


Hint 1. A helpful derivative

ANSWER:

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 , , and .

ANSWER:

Correct

To understand why the ratio of the y velocity of the string to its slope is constant, draw the string with a waverunning along it at time . In the vicinity of , the string is sloped upward. The bit of string at position

moves downward as the wave moves forward. Onehalf cycle later, the string in the vicinity of will besloped downward, and the string at position will move upward as the wave moves forward.

In general, if at some particular the slope of the string is positive ( ), that bit of string will bemoving downward ( ). If the slope at is negative, that bit of string will be moving upward. Thisexplains 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 issloped, the more quickly it will move up or down.

Problem 15.54

You are designing a twostring instrument with metal string 35.0 long, as shown in the figure . Both strings are under thesame tension. String has a mass of 8.35 and produces the note middle C (frequency 262 ) in its fundamentalmode.

∂y(x,t)∂x

x t

A k ω x t

cos(ax+b) = −asin(ax+b)ddx

= ∂y(x,t)

∂xAkcos(kx − ωt)

ω k vp

= (x,t)vy

∂y(x,t)

∂x

−vp

t = 0 x = 0x = 0 x = 0

x = 0

(x, t) ∂y(x, t)/∂x > 0(x, t) < 0vy (x, t)

cmS1 g Hz


Part A

What should be the tension in the string?

ANSWER:

Correct

Part B

What should be the mass of string so that it will produce A# (frequency 466 ) as its fundamental?

ANSWER:

Correct

Part C

To extend the range of your instrument, you include a fret located just under the strings, but not normally touchingthem. How far from the upper end should you put this fret so that when you press tightly against it, this string willproduce C# (frequency 277 ) in its fundamental? That is, what is in the figure?

ANSWER:

Correct

Part D

If you press against the fret, what frequency of sound will it produce in its fundamental?

ANSWER:

= 802 T N

S2 Hz

= 2.64 M2 g

S1Hz x

= 1.90 x cm

S2

= 493 f Hz


Correct

Score Summary:Your score on this assignment is 99.5%.You received 6.96 out of a possible total of 7 points.


A Pivoting Rod on a Springisa13/Efni/Einn/F.hl/Edlisfraedi 1 V/skiladaemi 12.pdf · What is ?...

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