Calculating maximum velocity of a mass tethered between two springs

Question

I have the question "The diagram shows a mass tethered between two springs. It is displaced by 10cm then released. The mass oscillates with SHM with a frequency of 0.55Hz. Calculate the maximum velocity of the mass."

I have used the equation:

V = wASin(wt)

Therefore, V = (2$\pi$ 0.55) x 0.1 x Sin(2$\pi$ 0.55 x 1.82)

Therefore, V = 0.04 ms$^-1$.

Is this correct ?

Answer 1

Simple harmonic motion results in a position in general given by $x=A\sin(\omega t+\phi)$ where $A$ and $\phi$ are constants depending on the initial conditions and $\omega$ is the angular frequency. You appear to have correctly deduced $A$ in this case will be equal to the maximum displacement, $10\,\mathrm{cm}$. By taking the time derivative of position, you can determine velocity, $v=\omega A\cos(\omega t+\phi)$. You appear to have correctly deduced $\omega=2\pi f=2\pi\cdot 0.55\,\mathrm{Hz}$. The maximum velocity must occur when $\cos(\cdot)=1$, so $v_{\mathrm{max}}=\omega A$.

Answer 2

You can compute the maximum speed of the mass via conservation of energy: the sum of its kinetic and potential energies $E=\frac12kx^2+\frac12m\dot x^2$ is constant. The spring constant $k$ is related to the angular frequency of the oscillation by the equation $\omega_0^2=k/m$. Substituting for $k$ gives $E=\frac12m\omega_0^2x^2+\frac12m\dot x^2$. At the maximum displacement $\pm A$, $\dot x=0$, and maximum velocity is achieved when the potential energy is zero. This means that $\frac12m\omega_0^2x^2=\frac12m\dot x^2$, from which $|\dot x|=\omega_0A$.