Answer to Question #253117 in Optics for Tom

Question

A string 50.0 cm long has a mass per unit length equal to 20.0x10^-5 kg/m. To what tension should this string be stretched if it's fundamental frequency is to be 20.0 Hz.

Answer 1

$l = 50 \;cm = 0.5 \;m \\ m = \frac{M}{l} = 20 \times 10^{-5} \; kg/m \\ l = 20 \;Hz \\ f = \frac{1}{2l} \sqrt{\frac{T}{m}} \\ f^2 = \frac{1}{4l^2} \times \frac{T}{m} \\ T = f^2 \times 4l^2 \times m \\ = 20^2 \times 4 \times 0.5^2 \times 20 \times 10^{-5} \\ = 800 \times 10^{-4} \\ = 8 \times 10^{-2} \;N$