Max Interference at x=200: Waves & Interference HW

[TalonStriker]

Homework Statement

There are 2 in-phase radio wave generators at x = 100 and x = 200. The frequency of each wave is 3.2 MHz. Find spot closest to x = 200 such that the interference at that point is maximum.

Homework Equations

$$\Delta\Phi = \frac{2\pi}{\lambda} \Delta x + \Phi_{0}$$
$$v = c = f\lambda$$ c = speed of light

The Attempt at a Solution

I assumed that point d was close to x = 300. So that
$$\Delta x = (100-d) - d$$
$$\Delta x = (100 - 2d)$$

$$2\pi m = \frac{2\pi}{c/f} (100 - 2d) + 0$$ for some integer m; \Phi_{0} = 0 since the waves are in-phase

$$m = \frac{1}{c/f} (100 - 2d)$$ cancel out 2 * pi

$$c/f * m = (100 - 2d)$$

$$(c/f * m ) - 100= 2d$$

$$d = \frac{(c/f * m) - 100}{2}$$Ignoring d, I solved for m such that $$\frac{(c/f * m) - 100}{2}$$ was equal to 300. Using the m i found (rounding it to nearest integer), i plugged it back into $$d = \frac{(c/f * m) - 100}{2}$$ to find d.

Is the method I used correct? What did I do wrong? (Could you please provide me the correct answer & method...i have an exam in 5 hours).

Thanks!

 

[anathema]

Hello there,

Your method is on the right track, but there are a few errors in your calculations. Let's go through it step by step:

1. First, let's define our variables:
- x = 200 (position of one generator)
- d = unknown distance from x = 200 to the point of maximum interference
- \lambda = 3.2 MHz (wavelength of the radio waves)
- m = integer (used to represent the number of wavelengths between the two generators)

2. Next, let's set up our equation for the phase difference:
\Delta\Phi = \frac{2\pi}{\lambda} \Delta x + \Phi_{0}

Since the waves are in-phase, \Phi_{0} = 0. So our equation becomes:
\Delta\Phi = \frac{2\pi}{\lambda} \Delta x

3. Now, we can plug in our known values:
\Delta\Phi = \frac{2\pi}{3.2 \times 10^{6}} \Delta x

4. We also know that at the point of maximum interference, the phase difference should be a multiple of 2\pi. So we can set \Delta\Phi = 2\pi m, where m is an integer.

5. Combining these two equations, we get:
2\pi m = \frac{2\pi}{3.2 \times 10^{6}} \Delta x

6. Solving for \Delta x, we get:
\Delta x = \frac{3.2 \times 10^{6}}{m}

7. Now, we want to find the value of d, which is the distance from x = 200 to the point of maximum interference. We can do this by using the Pythagorean theorem:
d^{2} + \Delta x^{2} = (200 - x)^{2}

Substituting in our value for \Delta x, we get:
d^{2} + \left(\frac{3.2 \times 10^{6}}{m}\right)^{2} = (200 - x)^{2}

8. We can simplify this equation by substituting in our known values for x and \lambda:
d^{2} + \left(\frac{3.2 \times 10^{6}}{m}\right)^{2} = (200

 

[m2003]

I cannot provide you with the exact answer without knowing the specific values for c (speed of light) and f (frequency). However, your approach seems correct in using the equation \Delta\Phi = \frac{2\pi}{\lambda} \Delta x + \Phi_{0} to find the point of maximum interference. The value of \Delta\Phi should be equal to 2\pi or a multiple of 2\pi for maximum interference to occur.

To find the value of m, you can use the equation \Delta\Phi = \frac{2\pi}{\lambda} \Delta x + \Phi_{0} and plug in the values of \Delta\Phi and \Delta x (which is 100 - 2d) and solve for m. Once you find the value of m, you can plug it back into the equation d = \frac{(c/f * m) - 100}{2} to find the value of d.

In summary, your approach seems correct, but make sure to use the correct values of c and f and solve for m before finding the value of d. Good luck on your exam!