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Your solution is correct. Obviously, any periodic function with period $T$ is also periodic with period $kT$, $k\in\mathbb{Z}^+$. In the given example, the smallest possible period is $T=1/5$, which is equivalent to $\omega_0=10\pi$. There is no obvious reason why one should choose $T^\prime=kT$ and $\omega_0^{\prime}=\omega_0/k$, with $k>1$. Nevertheless,...