The fundamental frequency is f. The first overtone is the second harmonic and will be at a frequency 2·f, thus the wavelength will be
λ=2 L/2=L=100 cm.
The same tone will be if the frequencies’ ratio will be 1/√(6&2), thus to emit the same tone the rod might be clamped at the points of
x=λ/6=100/6=16.67 cm.
To amid the same tone in the second overtone the frequency must be 3·f, thus the wavelength of the third harmonic is
λ=2 L/3=2 100/3=66.67 cm.
x=λ/(20/3)=66.67/(20/3)=10 cm.
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