No, that's backwards. The harmonics would be over submultiples, i.e. divisors, of ##2\pi##, i.e. ##2\pi/2,\, 2\pi/3,\, 2\pi/4,\, 2\pi/5,\,\ldots## They have higher frequencies, hence shorter periods. Thus all of them would have ##2\pi## as a period, but not necessarily as a shortest period.