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
[tex]f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}[/tex]
where
L is the length of the string
T is the tension in the string
[tex]\mu[/tex] is the mass per unit length
For the string in the problem,
L = 30.0 m
[tex]\mu=9.00\cdot 10^{-3} kg/m[/tex]
T = 20.0 N
Substituting into the equation, we find the fundamental frequency:
[tex]f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz[/tex]
The next frequencies (harmonics) are given by
[tex]f_n = nf[/tex]
with n being an integer number and f being the fundamental frequency.
So we get:
[tex]f_2 = 2 (0.786 Hz)=1.572 Hz[/tex]
[tex]f_3 = 3 (0.786 Hz)=2.358 Hz[/tex]
[tex]f_4 = 4 (0.786 Hz)=3.144 Hz[/tex]
