Finding mass attached to the string

Question

I'm given the following problem:

One end of a string with a linear mass density of 7.60* 10^-4 is connected to an oscillator with a frequency of 50.0 Hz. The other end is connected to a hanging variable mass. The string passes over a pulley, and the string between the oscillator and the pulley can vibrate freely.
The length of the vibrating section of the string is 0.500 m. What mass must be attached so that the string vibrates with

i) one antinode between the pulley and oscillator and

ii) three antinodes between the pulley and the oscillator?

I'm given solution for this problem and the part that confuses me a lot is this:
i) In this case, the fundamental frequency is excited, so m = 1. We can calculate tensions as follows...
And then they calculate tension and from there T = mg they find mass.
For ii) they are doing the same, except it is third fundamental frequency that is excited, so m = 3.

So, my questions are:

1. What does it mean for a frequency to be excited?

2. Why is m = 1 for the fundamental frequency and m = 3 for the third fundamental frequency? And why are they finding another mass if they have these 1 and 3.

Answer 1

1) What does it mean for a frequency to be excited?

That's unfortunate wording. I would say that the string was vibrated sinusoidally at a frequency. The string is being "excited", not the frequency.

2) Why is m = 1 for the fundamental frequency and m = 3 for the third fundamental frequency? And why are they finding another mass if they have these 1 and 3.

Another unfortunate choice. The m=1 and m=3 items are specific integers referring to the number of antinodes in the standing wave (or resonant wave), not masses. It would be better to use a different letter, such as n or k. n=1 yields the lowest resonant frequency for the string and is defined to be the fundamental frequency.

There is no such thing as the third fundamental. It is properly called the 3rd partial or second overtone. Because the standing wave wavelengths of a string fixed at both ends form a harmonic series ($\lambda,\frac{\lambda }{2},\frac{\lambda }{3},\frac{\lambda }{4}, ...$), the frequencies are referred to as harmonic frequencies, and n=3 (m=3 for your problem) is the third harmonic.