Period of a vibrating stretched string

Length = 0.8m
Mass Density = 0.5kg/m
Vibrates at ground tone 259 Hz
Tensile force of the string = 85863.68N

Find period and angular frequency for the forth overtone.

Quick answer: fourth over tone has a frequency that is 4 times the ground tone
=> f4 = 1036/s

Therefore
w4 = 2 pi f4
T4 = 1/f4 = 0.965ms

ALTERNATIVELY:

First you need to calculate the speed of wave propagation:

c = sqrt(T/rho)
= sqrt(85863.68 kg m/s^2 / (0.5kg/m))
= 414.4 m/s

Another way to obtain this velocity is:

for the ground tone we have that the wave length is : lambda = 2 L

Also, c = lambda * f,

so

c = 2 * 0.8 m * 259 /s = 414.4 m/s

The fourth overtone has wavelength lambda = L/2, so

f4 = c/(L/2) = 414.4 m/s / ( 0.4 m) =1036 /s