Answer:
0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz
Explanation:
The fundamental frequency of a standing wave on a string is given by

where
L is the length of the string
T is the tension in the string
is the mass per unit length
For the string in the problem,
L = 30.0 m

T = 20.0 N
Substituting into the equation, we find the fundamental frequency:

The next frequencies (harmonics) are given by

with n being an integer number and f being the fundamental frequency.
So we get:


