🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →
Log in
Physics
1
answer
0
watching
420
views

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 +x direction.
The wave is traveling in the -x 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 y_1(x,t)+y_2(x,t), where y_1(x,t) is the wave described in Part A and y_2(x,t) 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 y_e(x), called the envelope, depends only on position, and y_t(t) 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 y_e(x).

Part C  
Find y_e(x) and y_t(t). Keep in mind that y_t(t) should be a trigonometric function of unit amplitude.

Watch
1
answer
0
watching
420
views

For unlimited access to Homework Help, a Homework+ subscription is required.

Supratim Pal
Supratim PalLv10
8 Nov 2020

Unlock all answers

Get 1 free homework help answer.
Already have an account? Log in
Share

Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

Huygens' principle (first described by the Dutch scientist Christiaan Huygens in 1678) uses geometry to determine the shape of a wavefront at a time t, given some initial wavefront at an earlier time. This can be constructed by imagining that the initial wavefront is a source of wavelets that propagate from each point at the speed of light. From this principle, one can determine the angles of reflection and refraction by considering the speed of light at each point on the wavefront. In this problem, we will explore some of these concepts.

Part A

First, let's look at a plane wave that is incident on a flat piece of material with an index of refraction n. Part of the light is transmitted through the material and part of it is reflected. For the moment, let's just look at the part that is transmitted. At time t=0, the wavefront is a distance d away from the surface of the material. At time t=t1, the wavefront is at the position of the material interface. Use the Huygens' principle to determine how far into the material (d') the wavefront has propagated by time t=2t1.

Express your answer in terms of the variables n, t1, and the speed of light in a vacuum c.

Note that there would be no change in the direction of the wavefront at any point because the wavefront encountered the material interface at the same time at all points. Thus, all of the individual wavefronts all propagated at the same speed, thereby maintaining the flat wavefront.

Now, instead of having a flat wavefront propagating normal to the material interface, we have a flat wavefront propagating toward the material at an angle of 55° to the axis perpendicular to the material interface. In this part, we will look at the relative positions of a few points—A, B, and C—on the wavefront to illustrate Huygens' principle. (Intro 1 figure) Point C touches the vacuum/material interface at time t=0 whereas point B is a distance d and point A is a distance 2d away from the interface.

Part B

What is the time tB it will take for point B of the wavefront to encounter the vacuum/material interface?

Express your answer numerically to two decimal places in units of t1 (the time it takes light to travel a distance d in a vacuum).

It should be noted that the accuracy of the method increases the smaller the distance is before making a new wavefront and redoing the wavelets. If you tried to just draw a large wavelet from point B, then it would hit the surface after only traveling a distance d. This large wavelet would make it difficult to visualize how to add up all of the other wavelets to make a coherent wavefront. As a result, one should just draw a line perpendicular to the wavefront at point B until it hits the surface, at which point the wavelets will change owing to the new index of refraction.

 

Part C

How far did point C go into the material interface in the time t=tB that it took for point B to get to the interface? For this part we are looking for the distance traversed by the point C in the material (dC).

Give your answer in terms of the time tB, c, and n.

 

Part D

What is the new angle θ at which point C of the wavefront is propagating (relative to a line perpendicular to the vacuum/material interface)? Try to use the fact that you have a spherical wavefront propagating from t=0 at the point where C met the vacuum/material interface until time tB when the wavefront at point B reached the interface.

Express your answer in terms of inverse trigonometric functions and n.

 

Goal Use the superposition principle to calculate the electricfield due to two point charges.

Problem Charge q1 = 7.00 µC is at the origin, and charge q2 = -5.35µC is on the x-axis, 0.300 m from the origin (Fig. 15.12).

(a) Find the magnitude and direction of the electric field at pointP, which has coordinates (0, 0.400) m.

(b) Find the force on a charge of 2.00 10-8 C placed at P.

(a) Calculate the electric field at P.
Find the magnitude of 1 with Equation 15.6.
E1 = 3.93 105 N/C
Vector 1 is vertical, making an angle of 90° with respect to thepositive x-axis. Use this fact to find its components. E1x = E1cos(90°) = 0
E1y = E1 sin(90°) = 3.93 105 N/C
Next, find the magnitude of 2, again with Equation 15.6.
E2 = 1.92 105 N/C
Obtain the x-component of 2, using the triangle in Figure 15.12 tofind cos ?.
E2x = E2 cos ? = (1.92 105 N/C)(0.600)
E2x = 1.15 105 N/C
Obtain the y-component in the same way, but a minus sign has to beprovided for sin ? because this component is directeddownward.
E2y = E2 sin ? = (1.92 105 N/C)(0.800)
E2y = -1.54 105 N/C
Sum the x-components to get the x-component of the resultantvector. Ex = E1x2x = 0 + 1.15 105 N/C
Ex = 1 N/C
+ E
Sum the y-components to get the y-component of the resultantvector. Ey = E1y2y = 3.93 105 N/C - 1.54 105 N/C
Ey = 2 N/C
+ E
Use the Pythagorean theorem to find the magnitude of the resultantvector.
E = 3 N/C

The inverse tangent function yields the direction of the resultantvector.
? = 4°

(b) Find the force on a charge of 2.00 10-8 C placed at P.
Calculate the magnitude of the force (the direction is the same asthat of because the charge is positive). F = Eq = E(2.00 10-8C)
F = 5 N




(a) Place a charge of -3.60 µC at point P and find the magnitudeand direction of the electric field at the location of q2.

E = N/C



? = ° counterclockwise from the positive x-axis



(b) Find the magnitude and direction of the force on q2.

F = N



? = ° counterclockwise from the positive x-axis



I Don't know how to do this Please help Me. Thank you. The top partis how i figured out the rest i think its the missing informationto solve the bottom part.

Weekly leaderboard

Home
Homework Help 3,900,000
Physics 110,000
Start filling in the gaps now
Log in