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Consider a traveling wave described by the formula

$y_1(x,t) = A \sin(k x - \omega t)$ .

This function might represent the lateral displacement of a string, a local electric field, the position of the surface of a body of water, or any of a number of other physical manifestations of waves.

Part A

Which one of the following statements about the wave described in the problem introduction is correct?

ANSWER:

The wave is traveling in the

direction.
The wave is traveling in the

direction.
The wave is oscillating but not traveling.
The wave is traveling but not oscillating.

Part B

Which of the expressions given is a mathematical expression for a wave of the same amplitude that is traveling in the opposite direction? At time t=0

this new wave should have the same displacement as y_1(x,t)

, the wave described in the problem introduction.

ANSWER:

$A \cos (k x - \omega t)$
$A \cos (k x + \omega t)$
$A \sin (k x - \omega t)$
$A \sin (k x + \omega t)$

The principle of superposition states that if two functions each separately satisfy the wave equation, then the sum (or difference) also satisfies the wave equation. This principle follows from the fact that every term in the wave equation is linear in the amplitude of the wave.

Consider the sum of two waves , where is the wave described in Part A and is the wave described in Part B. These waves have been chosen so that their sum can be written as follows:

$y_{\rm s}(x,t) = y_{\rm e}(x) y_{\rm t}(t)$ .

This form is significant because , called the envelope, depends only on position, and depends only on time. Traditionally, the time function is taken to be a trigonometric function with unit amplitude; that is, the overall amplitude of the wave is written as part of .

Part C
Find and . Keep in mind that should be a trigonometric function of unit amplitude.

erinhyena808

Could you please proofread and correct scientifically andgrammatically my work, thanks ^_^

Here itis:

The percentage error was found to be 0.186% and 6.66%respectively.

This implies that the predicted value of the freefall accelerationis fairly accurate while the predicated value of deceleration isnot as much. The encountered errors were due to severalexperimental and human limitation was that the experiment involved.The experimental limitation was due to the uncertainty of theclinometer which was Â± 0.5o.

I.e. In our calculation, a difference of 0.5o can lead to adifference of 40 - 80 cm in height as shown below:

The human limitations was due to the human eye being unable todetect the exact point on the track at which the gondoladecelerates. This is because the braking happened very quickly.This adds an uncertainty to the location of deceleration when theangels are being measured.

In addition, the human ability cannot decide on the exact period oftime at which the gondola freefalls or decelerates. This is due tothe human reaction speed being unable to exactly determine thestart time of the momentary freefall and deceleration. The timetaken for humans to react cannot be measured thus it cannot besubtracted from the findings to get an accurate timing.

All of these limitations that have affected the accuracy of theresults could be avoided to some extent by marking the exact zonesrequired on the track of the giant drop as well as using anelectronic angle measurer to measure the angles . Additionally, theusage of an electronic timer to measure periods of the freefall anddeceleration is also suggested.

During the analysis, finding an average to the deceleration in theaccelerometer reading (using Logger Pro) was only estimated usingthe human naked eye. This is because an average could not be takenelectronically. Thus, there is also some uncertainty involved inthe theoretical value.

NOTE:

The value for freefall acceleration obtained from thevideo-analysis technique was 9.754 Â± 0.3237 m/s2. 9.754 m/s2 has apercentage error of :
0.469%

This value indicates it is fairly accurate and can be used as aconforming result to the theoretical value. The error involved isdue to the fact that the video was shaking, thus plotting thevelocity/ time graph of the gondola at different times wasnegatively affected; although the axes was roughly re-set at everyframe.

In a circuit containing only resistors, the basic (though notnecessarily explicit) assumption is that the current reaches itssteady-state value instantly. This is not the case for a circuitcontaining inductors. Due to a fundamental property of an inductorto mitigate any "externally imposed" change in current, the currentin such a circuit changes gradually when a switch is closed oropened.
Consider a series circuit containing a resistor of resistance andan inductor of inductance connected to a source of emf andnegligible internal resistance. The wires (including the ones thatmake up the inductor) are also assumed to have negligibleresistance.
Let us start by analyzing the process that takes place after switchis closed (switch remains open). In our further analysis, lowercaseletters will denote the instantaneous values of various quantities,whereas capital letters will denote the maximum values of therespective quantities.
Note that at any time during the process, Kirchhoff's loop ruleholds and is, indeed, helpful:
.
Part A
Immediately after the switch is closed, what is the current in thecircuit?
Hint A.1
How to approach the problem
Hint not displayed
ANSWER:

zero
Answer not displayed
Part B
Immediately after the switch is closed, what is the voltage acrossthe resistor?
Hint B.1
Ohm's law
Hint not displayed
ANSWER:

zero
Answer not displayed
Part C
Immediately after the switch is closed, what is the voltage acrossthe inductor?
Hint C.1
A formula for voltage across an inductor
Hint not displayed
ANSWER:

zero
Answer not displayed
Part D
Shortly after the switch is closed, what is the direction of thecurrent in the circuit?
ANSWER:
clockwise
counterclockwise
There is no current because the inductor does not allow the currentto increase from its initial zero value.
Answer not displayed
Part E
Shortly after the switch is closed, what is the direction of theinduced EMF in the inductor?
ANSWER:
clockwise
counterclockwise
There is no induced EMF because the initial value of the current iszero.
Answer not displayed
Part F
Eventually, the process approaches a steady state. What is thecurrent in the circuit in the steady state?
ANSWER:

Answer not displayed
Part G
What is the voltage across the inductor in the steady state?
Hint G.1
Induced EMF and current
Hint not displayed
ANSWER:
zero

Answer not displayed
Part H
What is the voltage across the resistor in the steady state?
ANSWER:
zero

Answer not displayed
Part I
Now that we have a feel for the state of the circuit in its steadystate, let us obtain the expression for the current in the circuitas a function of time. Note that we can use the loop rule (goingaround counterclockwise):
.
Note as well that and . Using these equations, we can get, aftersome rearranging of the variables and making the subsitution,
.
Integrating both sides of this equation yields
.
Use this last expression to obtain an expression for . Rememberthat and that .
Express your answer in terms of , , and . You may or may not needall these variables. Use the notation exp(x) for .
ANSWER:
=
Answer not displayed
Part J
Just as in the case of R-C circuits, the steady state here is neveractually reached: The exponential functions approach their limitsasymptotically as . However, it usually does not take very long forthe value of to get very close to its presumed limiting value. Thenext several questions illustrate this point.
Note that the quantity has dimensions of time and is called thetime constant (you may recall similar terminology applied to R-Ccircuits). The time constant is often denoted by . Using , one canwrite the expression

as
.
Find the ratio of the current at time to the maximum current.
Express your answer numerically, using three significantfigures.
ANSWER:
=
Answer not displayed
Part K
Find the time it takes the current to reach 99.999% of its maximumvalue.
Express your answer numerically, in units of . Use threesignificant figures.
ANSWER:
=
Answer not displayed

Part L
Find the time it takes the current to reach 99.999% of its maximumvalue. Assume that ohms and millihenrys.
Express your answer in seconds, using three significantfigures.
ANSWER:
=
Answer not displayed
seconds
Part M
Find the time it takes the current to reach 99.999% of its maximumvalue. Assume that ohms and henrys.
Express your answer in seconds, using three significantfigures.
ANSWER:
=
Answer not displayed
seconds
Part N
What fraction of the maximum value will be reached by the currentone minute after the switch is closed? Again, assume that ohms andhenrys.
Use three significant figures in your answer.
ANSWER:
=
Answer not displayed
Now consider a different situation. After switch has been closedfor a long time, it is opened; simultaneously, switch is closed, asshown in the figure. This effectively removes the battery from thecircuit.
The questions below refer to the time immediately after switch isopened and switch is closed.
Part O
What is the direction of the current in the circuit?
ANSWER:
clockwise
counterclockwise
The current is zero because there is no EMF in the circuit.
Answer not displayed
Part P
What is happening to the magnitude of the current?
ANSWER:
The current is increasing.
The current is decreasing.
The current remains constant.
Answer not displayed
Part Q
What is the direction of the EMF in the inductor?
ANSWER:
clockwise
counterclockwise
The EMF is zero because the current is zero.
The EMF is zero because the current is constant.
Answer not displayed
Part R
Which end of the inductor has higher voltage (i.e., to which end ofthe inductor should the positive terminal of a voltmeter beconnected in order to yield a positive reading)?
Hint R.1
How to approach the problem
Hint not displayed
ANSWER:
left
right
The potentials of both ends are the same.
The answer depends on the magnitude of the time constant.
Answer not displayed
Part S
For this circuit, Kirchhoff's loop rule gives
.
Note that , since the current is decreasing. Use this equation toobtain an expression for .
Hint S.1
Initial and final conditions
Hint not displayed
Hint S.2
Let us cheat a bit
Hint not displayed
Express your answer in terms of , , and . Use exp(x) for .
ANSWER:
=
Answer not displayed

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