An interference pattern is set up by two point sources of the same frequency, vibrating in phase with one another. A point on the second nodal line is 50.0 cm from one source and 59 cm from the other. The speed of the waves is 35 cm/s. Calculate the frequency of the sources. I know you do path difference (like Pms-Pms) and then do the universal wave equation, I am confused where it says in phase with one another and the 2nd nodal line

The "first nodal line" is a line along which the path difference is 1/2 wave, a point of destructive interference. In this case, 9 cm is the wavelength, since there would be one extra wavelength in path difference along the second nodal line.

The frequency is then f = (35 cm/s)/9 cm = 3.9 Hz. Those are very slow waves, by the way. Are you sure you got the units right?

It does not matter whether the two sources are in phase with one another or not. It affects where the nodes are, but not the separation between the nodes.

For more background about this, see
http://www.glenbrook.k12.il.us/gbssci/phys/Class/light/u12l3a.html

mmmh are you sure it is 9 cm? It is destructive interference which means a minimum, therefore PmS1-PmS2= (m+1/2) lambda...I get 9 cm when I don't account for the m + 1/2

You are correct. 9 cm is (3/2) wavelength along a second nodal line. The wavelength is therefore 6 cm.

Thanks for all of your help, I talked to my other friend who had originally posted, and we both appreciate this, thanks!

Sorry about the error. I'm glad you and BobPursley were on the ball. I was thinking of the path length difference between the first and second nodal line.

To determine the frequency of the sources, we need to consider the interference pattern formed by the two point sources. Let's break down the problem step by step:

1. "In phase with one another": When two point sources are vibrating in phase, it means that they have the same phase difference and the crests and troughs of their waves align perfectly. This results in constructive interference, where the waves reinforce each other.

2. "Second nodal line": In an interference pattern, nodal lines are regions where the waves from the two sources perfectly cancel each other out, resulting in destructive interference. The second nodal line refers to the second line parallel to the two sources along which destructive interference occurs.

3. Finding the path difference: For a point on the second nodal line, the path difference (Pms - Pms) is given by the distance between the point and one source (d1) minus the distance between the point and the other source (d2).

4. Applying the universal wave equation: The universal wave equation relates the frequency (f), speed (v), and wavelength (λ) of a wave and is given by the formula: v = f * λ. We can rearrange the equation to solve for frequency: f = v / λ.

Now, let's apply this information to solve the problem:

Given:
- Distance from the point on the second nodal line to one source (d1) = 50.0 cm
- Distance from the point on the second nodal line to the other source (d2) = 59 cm
- Speed of the waves (v) = 35 cm/s

1. Calculate the path difference: Pms-Pms = d1 - d2 = 50.0 cm - 59 cm = -9.0 cm (negative since it's on the second nodal line)

2. Calculate the wavelength: The path difference corresponds to an integral number of wavelengths. Since we are dealing with a nodal line, the path difference is equal to (m + 0.5) * λ, where m is an integer. In this case, since it's the second nodal line, m = 1. So, -9.0 cm = (1 + 0.5) * λ.
Simplifying, we find that the wavelength (λ) = -9.0 cm / 1.5 = -6.0 cm (negative because the wave is inverted due to the destructive interference).

3. Calculate the frequency: Using the universal wave equation, f = v / λ. Substituting the given values, we have f = 35 cm/s / -6.0 cm.

Calculating the frequency, we find that f ≈ -5.83 Hz (rounded to two decimal places).

Note: The negative sign indicates that the wave is inverted due to the destructive interference, but we take the absolute value for the frequency since it represents the number of cycles per second.