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Your Answer
\par The natural period of oscillation is\begin{equation T=\frac{2 \pi}{\omega_{0}}=2 \pi \sqrt{\frac{M}{k_{1}+k_{2}}}\end{equation\par Since the mass of spring cannot be ignored, we take the effective mass of the oscillator into account.It consists of the mass of the object and the effective mass of the spring, which is 1/3 of the actual mass of it. Then the angular frequency can be expressed a\begin{equation \omega_{0}=\sqrt{\frac{k_{1}+k_{2}}{M+m_{0}}}\end{equationwhere $m_0=\frac{1}{3}m_s$, representing the effective and actual masses of the spring\par In harmonic motion in a spring-mass system, the elastic potential energy is $U=kx^2/2$ and the kinetic energy of the oscillating mass $m$ is $K=mv^2/2$.In the absence of non-conservative for\qdelces, the total mechanical energy is conserved, where $$E=K_{max}=U_{max}=K+U,$$ implyin\begin{equation} k=\frac{m v_{\max }^{2}}{A^{2}}\end{equation
JOJ Answer
\par The natural period of oscillation is
\begin{equation}
T=\frac{2 \pi}{\omega_{0}}=2 \pi \sqrt{\frac{M}{k_{1}+k_{2}}}.
\end{equation}
\par Since the mass of spring cannot be ignored, we take the effective mass of the oscillator into account.
It consists of the mass of the object and the effective mass of the spring, which is 1/3 of the actual mass of it. Then the angular frequency can be expressed as
\begin{equation}
\omega_{0}=\sqrt{\frac{k_{1}+k_{2}}{M+m_{0}}},
\end{equation}
where $m_0=\frac{1}{3}m_s$, representing the effective and actual masses of the spring.
\par In harmonic motion in a spring-mass system, the elastic potential energy is $U=kx^2/2$ and the kinetic energy of the oscillating mass $m$ is $K=mv^2/2$.
In the absence of non-conservative for
k=\frac{m v_{\max }^{2}}{A^{2}}.
\end{equation}