The answer is to the negative according to the following counterexample for $n=2\;$ :
If $x_1\left(t\right),x_2\left(t\right)$ are the vertical displacements of the top and bottom masses $m_1,m_2$ from their equilibrium positions respectively (and not from the ceiling), the equations of motion are
\begin{align}
m_1\ddot{x}_1 & =+k\left(x_2-2x_1\right)
\tag{01.1}\\
m_2\ddot{x}_2 & =-k\left(x_2-x_1\right)
\tag{01.2}
\end{align}
and by matrices
\begin{equation}
\begin{bmatrix}
m_1 & 0 \\
0 & m_2
\end{bmatrix}
\begin{bmatrix}
\ddot{x}_1 \\
\ddot{x}_2
\end{bmatrix}
+ k\begin{bmatrix}
+2 & -1 \\
-1 & +1
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0
\end{bmatrix}
\tag{02}
\end{equation}
that is
\begin{equation}
\mathrm{M}\ddot{\mathbf{x}}+\mathrm{K}\mathbf{x}=\mathbf{0}
\tag{03}
\end{equation}
where
\begin{equation}
\mathbf{x} \equiv
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
, \quad
\mathrm{M} \equiv
\begin{bmatrix}
m_1 & 0 \\
0 & m_2
\end{bmatrix}
, \quad
\mathrm{K} \equiv
k\begin{bmatrix}
+2 & -1 \\
-1 & +1
\end{bmatrix}
\tag{04}
\end{equation}
Since $\mathrm{M}$ is invertible $(m_1m_2 \ne 0)$, multiplying equation (03) by $\mathrm{M}^{-1}$ we have
\begin{equation}
\ddot{\mathbf{x}}+\left(\mathrm{M}^{-1}\mathrm{K}\right)\mathbf{x}=\ddot{\mathbf{x}}+\mathrm{S}\mathbf{x}=\mathbf{0}
\tag{05}
\end{equation}
where
\begin{equation}
\mathrm{S}\equiv \mathrm{M}^{-1}\mathrm{K}=
\begin{bmatrix}
m_1^{-1} & 0 \\
0 & m_2^{-1}
\end{bmatrix}
k
\begin{bmatrix}
+2 & -1 \\
-1 & +1
\end{bmatrix}
=\dfrac{k}{m_2}
\begin{bmatrix}
2\dfrac{m_2}{m_1} & -\dfrac{m_2}{m_1} \\
-1 & +1
\end{bmatrix}
\tag{06}
\end{equation}
or
\begin{equation}
\mathrm{S}
=\omega_{0}^2
\begin{bmatrix}
2\mu & -\mu\\
-1 & +1
\end{bmatrix}
\;, \quad \omega_{0}\equiv\sqrt{\dfrac{k}{m_2}}\; , \quad \mu \equiv \dfrac{m_2}{m_1} =\text{mass ratio}
\tag{07}
\end{equation}
For the eigenvalues $\xi$ of the matrix
\begin{equation}
\dfrac{\mathrm{S}}{\omega_{0}^2}=
\begin{bmatrix}
2\mu & -\mu\\
-1 & +1
\end{bmatrix}
\tag{08}
\end{equation}
we have
\begin{equation}
\begin{vmatrix}
2\mu-\xi & -\mu\\
-1 & 1-\xi
\end{vmatrix}
=0 \quad \Longrightarrow \quad
\xi^2-\left(2\mu+1\right)\xi +\mu=0
\tag{09}
\end{equation}
So, the eigenvalues are the following positive real numbers
\begin{align}
\xi_1 & =\dfrac{\left(2\mu+1\right)+\sqrt{4\mu^2+1}}{2}\in \mathbb{R}^{+}
\tag{10.1}\\
\xi_2 & =\dfrac{\left(2\mu+1\right)-\sqrt{4\mu^2+1}}{2}\in \mathbb{R}^{+}
\tag{10.2}
\end{align}
Now if
\begin{equation}
\mathrm{P}=
\begin{bmatrix}
p_{11} & p_{12} \\
p_{21} & p_{22}
\end{bmatrix}
\tag{11}
\end{equation}
is an invertible real matrix which diagonalizes the matrix of equation (08), then
\begin{equation}
\mathrm{P}^{-1}
\begin{bmatrix}
2\mu & -\mu\\
-1 & +1
\end{bmatrix}
\mathrm{P}
=
\begin{bmatrix}
\xi_1 & 0 \\
0 & \xi_2
\end{bmatrix}
\tag{12}
\end{equation}
and consequently
\begin{equation}
\mathrm{P}^{-1}\mathrm{S}\mathrm{P}
= \omega_{0}^2
\begin{bmatrix}
\xi_1 & 0 \\
0 & \xi_2
\end{bmatrix}
\tag{13}
\end{equation}
Defining
\begin{equation}
\mathbf{y}\equiv\mathrm{P}^{-1}\mathbf{x}
\tag{14}
\end{equation}
equation (05) gives after multiplication by $\mathrm{P}^{-1}$
\begin{equation}
\ddot{\mathbf{y}}+\left(\mathrm{P}^{-1}\mathrm{S} \mathrm{P}\right)\mathbf{y}=\mathbf{0}
\tag{15}
\end{equation}
or
\begin{equation}
\begin{bmatrix}
\ddot{y}_1 \\
\ddot{y}_2
\end{bmatrix}
+ \omega_{0}^2
\begin{bmatrix}
\xi_1 & 0 \\
0 & \xi_2
\end{bmatrix}
\begin{bmatrix}
y_1 \\
y_2
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0
\end{bmatrix}
\tag{16}
\end{equation}
that is 2 independent differential equations
\begin{align}
\ddot{y}_1+\xi_1\omega_{0}^{2} y_{1} & =0
\tag{17.1}\\
\ddot{y}_2+\xi_2\omega_{0}^{2} y_{2} & =0
\tag{17.2}
\end{align}
with general solutions respectively
\begin{align}
y_{1}\left(t\right) & = A_{1}\sin\left(\omega_1 t+\phi_{1}\right)
\tag{18.1}\\
y_{2}\left(t\right) & = A_{2}\sin\left(\omega_2 t+\phi_{2}\right)
\tag{18.2}
\end{align}
where
\begin{align}
\omega_1 & =\sqrt{\xi_1}\,\omega_0 =\sqrt{\dfrac{\left(2\mu+1\right)+\sqrt{4\mu^2+1}}{2}}\,\omega_0
\tag{19.1}\\
\omega_2 & =\sqrt{\xi_2}\,\omega_0 =\sqrt{\dfrac{\left(2\mu+1\right)-\sqrt{4\mu^2+1}}{2}}\,\omega_0
\tag{19.2}
\end{align}
The motion of the system is the composition of two independent harmonic(sinusoid) oscillations of angular frequencies $\omega_1,\omega_2$
\begin{align}
x_{1}\left(t\right)& = p_{11}\cdot y_{1}\left(t\right)+ p_{12}\cdot y_{2}\left(t\right)=p_{11} A_{1}\sin\left(\omega_1 t+\phi_{1}\right)+p_{12} A_{2}\sin\left(\omega_2 t+\phi_{2}\right)
\tag{20.1}\\
x_{2}\left(t\right)& = p_{21}\cdot y_{1}\left(t\right)+ p_{22}\cdot y_{2}\left(t\right)=p_{21} A_{1}\sin\left(\omega_1 t+\phi_{1}\right)+p_{22} A_{2}\sin\left(\omega_2 t+\phi_{2}\right)
\tag{20.2}
\end{align}
The constants $ A_1,\phi_1,A_2,\phi_2 $ are determined from the initials conditions, for example from the positions and velocities of the two masses at time $t=0$ : $x_1\left(0\right),\dot{x}_1\left(0\right),x_2\left(0\right),\dot{x}_2\left(0\right)$.
As a final conclusion the motion of the system, equations (20), is never sinusoid and in general not even periodic.
