Answer
$\approx 180\;\rm Hz$
Work Step by Step
We need to treat the Tendons here as strings.
We know for standing waves in a string, the frequency is given by
$$f_m=\dfrac{mv}{2L}$$
And the fundamental frequency ($m=1$) is then given by
$$f_1=\dfrac{v}{2L}$$
where $v$ in a string is given by $v=\sqrt{T/\mu}$ where $T$ is the tension and $\mu$ is the linear density of the string.
$$f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}\tag 1$$
Noting that $\rho=\dfrac{m}{V}=\dfrac{m}{AL}$ where $m$ is the mass of the string, $L$ is the length of the strung, and $A$ is its cross-sectional area. And we know that $m/L=\mu$
Thus,
$$\mu=\rho A$$
Plugging into (1):
$$f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{\rho A}}$$
Plugging the known;
$$f_1=\dfrac{1}{2(0.20)}\sqrt{\dfrac{500}{(1100)(90\times 10^{-6})}}$$
$$f_1= 177.7\;\rm Hz\approx\color{red}{\bf180}\;\rm Hz$$
