At what angle are the electronic and the magnetic wave related in an electromagnetic signal?

Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri

Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part $kx \pm \omega t$ , the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but here is an explanation of this:

For both cases, + and -, after a certain time $\delta t$ ( $\delta t >0$ ), the displacement y of the wave will be determined by the $kx\pm\omega (t+\delta t)$ term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply $\pm \omega (t+\delta t)$ .

To know which side, right or left of the origin, would go through the origin after a time $\delta t$ (and thus know the direction of propagation) we have to see how we can achieve that same displacement y not by a time variation but by a space variation $\delta x$ (we would be looking where in space is what we would have in the future in time). The term would be then $k(x+\delta x)\pm\omega t$ , which at the origin is $k \delta x \pm \omega t$ . This would mean that, when the original equation has $kx+\omega t$ , we must have that $\delta x>0$ for $k\delta x+\omega t$ to be equal to $kx+\omega\delta t$ , and when the original equation has $kx-\omega t$ , we must have that $\delta x$ for $k\delta x-\omega t$ to be equal to $kx-\omega \delta t$

Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .

In conclusion, when $kx+\omega t$ , the part of the wave on the positive side ( $\delta x>0$ ) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0

3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

2 years ago

Which of the following is a force acting at a distance

wariber [46]

D. gravitational force. is your answer hope this helps

8 0

2 years ago

Read 2 more answers

Balboa Park in San Diego has an outdoor organ. When the air temperature increases, the fundamental frequency of one of the organ

musickatia [10]

The fundamental frequency of one of the organ pipes will go up or increase.

When pressured air is forced into an organ pipe, it echoes at a particular pitch, generating the sound of the pipe organ. Each pipe has been adjusted to a particular pitch on the musical scale.

A musical instrument called an outdoor pipe organ is used to perform music. It produces some calming tones and has a really serene sound. The organ pipe produces the sound of the outdoor organ. The wavelength of the sound is also dependent on the length of the pipe. The fundamental frequency of one of the organ pipes will grow as the speed of the sound increases as the ambient air temperature rises.

The correct option is (c).

Learn more about frequency here:

brainly.com/question/254161

#SPJ4

6 0

1 year ago

PLEASE HELP ME!

PilotLPTM [1.2K]

Answer:

a) A = 3 cm, b) T = 0.4 s, f = 2.5 Hz,

2) A standing wave the displacement of the wave is canceled and only one oscillation remains

Explanation:

a) in an oscillatory movement the amplitude is the highest value of the signal in this case

A = 3 cm

b) the period of oscillation is the time it takes for the wave to repeat itself in this case

T = 0.4 s

the period is the inverse of the frequency

f = 1 /T

f = 1 /, 0.4

f = 2.5 Hz

2) a traveling wave is a wave for which as time increases the displacement increases, in the case of a transverse wave the oscillation is perpendicular to the displacement and in the case of a longitudinal wave the oscillation is in the same direction of the displacement.

A standing wave occurs when a traveling wave bounces off some object and there are two waves, one that travels in one direction and the other that travels in the opposite direction. In this case, the displacement of the wave is canceled and only one oscillation remains.

8 0

2 years ago