To solve this problem it is necessary to apply the concepts related to the period based on variables such as gravity, distance and frequency.
By definition, know that the Period is
[tex]T_1 = 2\pi \sqrt{\frac{L}{g}}[/tex]
Where,
L = Length
g = Gravity
At the same time, frequency can be defined as,
[tex]f_1 = \frac{1}{T_1}[/tex]
So using this for [tex]10\mu[/tex] m we have that,
[tex]T_1 = 2\pi \sqrt{\frac{L}{g}}[/tex]
[tex]T_1 = 2\pi \sqrt{\frac{10*10^{-6}}{9.8}}[/tex]
[tex]T_1 = 0.635*10^{-2}s[/tex]
Then the frequency is
[tex]f_1 = \frac{1}{T_1}[/tex]
[tex]f_1 = \frac{1}{0.635*10^{-2}}[/tex]
[tex]f_1 = 157.6\approx 160Hz[/tex]
For the second length of 50\mu m we have that
[tex]T_1 = 2\pi \sqrt{\frac{L}{g}}[/tex]
[tex]T_1 = 2\pi \sqrt{\frac{5*10^{-5}}{9.8}}[/tex]
[tex]T_1 = 1.4*10^{-2}s[/tex]
Then the frequency is
[tex]f_1 = \frac{1}{T_1}[/tex]
[tex]f_1 = \frac{1}{1.4*10^{-2}}[/tex]
[tex]f_1 = 70Hz[/tex]
Therefore the correct answer is A.