Where does time function of wave on string go when expressed in the Fourier series?
A Standing wave on string of length $L,$ fixed at its ends $x=0$ and $x=L$ is: $\quad y(x, t)=A \sin (k x) \cos \left(\omega t+\phi_{0}\right) \quad$
Where: $k=\frac{n \pi}{L}$
A periodic function $f(x)$ with period $P$ is represented by the Fourier series: $f(x)=\frac{1}{2} a_{0}+\sum_{n=1}^{\infty} a_{n} \cos \left(n \frac{2 \pi}{p} x\right)+\sum_{n=1}^{\infty} b_{n} \sin \left(n \frac{2 \pi}{p} x\right)$
Where: $$ a_{0}=\frac{2}{p} \int_{-P / 2}^{P / 2} f(x) d x \quad a_{n}=\frac{2}{p} \int_{-P / 2}^{P / 2} f(x) \cos \left(\frac{2 \pi}{p} n x\right) d x \quad b_{n}=\frac{2}{p} \int_{-P / 2}^{P / 2} f(x) \sin \left(\frac{2 \pi}{P} n x\right) d x $$ For question where a guitar is played and the string is put into motion by plucking it. If we want to write $y(x)$ as a sum of the basis function, $y_{n}(x)$ we write: $$ y(x, 0)=\sum_{n=1}^{\infty} a_{n} \sin \left(k_{n} x\right) \quad \rightarrow \quad y(x, t)=\sum_{n=1}^{\infty} a_{n} \sin \left(k_{n} x\right) \cos \left(\omega_{n} t\right) $$ [since the wave function is usually odd, so the $a_n$ function will be eliminated)
Also in the case where the wave is neither an odd or even function when we have values for $a_{0}, a_{n}, b_{n}$ (not just 0).
And the periodic function is given by $f(x)=\frac{1}{2} a_{0}+\sum_{n=1}^{\infty} a_{n} \cos \left(n \frac{2 \pi}{p} x\right)+\sum_{n=1}^{\infty} b_{n} \sin \left(n \frac{2 \pi}{p} x\right)$.
Where do we add the $\cos \left(\omega_{n} t\right)$ part?
Does the equation look like this: $y(x, t)=\frac{1}{2} a_{0} \cos \left(\omega_{n} t\right)+\sum_{n=1}^{\infty} a_{n} \cos \left(n \frac{2 \pi}{p} x\right) \cos \left(\omega_{n} t\right)+\sum_{n=1}^{\infty} b_{n} \sin \left(n \frac{2 \pi}{p} x\right) \cos \left(\omega_{n} t\right)$