Answer:
The frequency of the second harmonic (
) is 11.97 Hz.
Step-by-step explanation:
Given;
mass of the string, m = 25 g = 0.025kg
tension on the string, T = 43 N
length of the string, L = 12 m
The speed of wave on the string is given as;
![v = \sqrt{(T)/(\mu) }](https://img.qammunity.org/2021/formulas/physics/college/bodk7tthgkdap293fnpjwyj1hlhb2ybmjg.png)
where;
μ is mass per unit length = 0.025 / 12 = 0.002083 kg/m
![v = \sqrt{(43)/(0.002083) }\\\\v = 143.678 \ m/s](https://img.qammunity.org/2021/formulas/physics/college/cmvsxpar0mdg2qz3qcotpvcmk5m7qwo5vw.png)
The wavelength of the first harmonic wave is given as;
![L = (1)/(2) \lambda _o\\\\\lambda _o = 2L \\\\\lambda _o = 2 \ * \ 12\\\\\lambda _o = 24 \ m](https://img.qammunity.org/2021/formulas/physics/college/8bf6d2ph3cj68pq38b9odx89nk91zkk1e4.png)
The frequency of the first harmonic is given as;
![f_o = (v)/(\lambda _o) = (v)/(2L) = (143.678)/(24) = 5.99 \ Hz\\\\](https://img.qammunity.org/2021/formulas/physics/college/rl51hl2dyvidszzfwonbia4c4kp5125rsz.png)
The wavelength of the second harmonic wave is given as;
![L = \lambda_1 \\\\\lambda_1 = 12 \ m](https://img.qammunity.org/2021/formulas/physics/college/n425e0c6c7zpecuov59qlfjgljfn4hklov.png)
The frequency of the second harmonic is given as;
![f_1 = (v)/(\lambda _1) = (143.678)/(12) = 11.97 \ Hz = 2((v)/(\lambda _0)) = 2f_o](https://img.qammunity.org/2021/formulas/physics/college/uaspkzb12etojxdr354rhdmbf2uue1n8fg.png)
Therefore, the frequency of the second harmonic (
) is 11.97 Hz.