Question

An object vibrates at the end of a vertical spring ( k = 23N/m). The speed of the object is 0.12m/s when the displacement is 0.15m. If the maximum displacement is 0.20m, what is the maximum speed of the object?

Answer 1

Total energy of the system is $E = \frac{k a^2}{2}$, where $k$ is the spring constant, and $a$ is the amplitude. This quantity is conserved, and is equal to the sum of kinetic and potential energies at any moment: $E = \frac{m v^2}{2} + \frac{k x^2}{2}$.

For given displacement and velocity, the equation $E = \frac{m v_1^2}{2} + \frac{k x_1^2}{2}$ holds.

The object reaches its maximum speed when $x = 0$, then $E = \frac{m v_{max}^2}{2}$, from where $v_{max} = \sqrt \frac{2 E}{m}$.

Since we know the total energy in terms of amplitude $E = \frac{k a^2}{2}$ , let us express the mass from the equation of energy for given position and velocity $m = \frac{ 2 E - k x_1^2}{v_1^2}$

, and substitute it into the expression for maximum velocity: $v_{max} = v_1 \sqrt{\frac{2 E}{2 E - k x_1^2}}$ . Expressing energy in terms of the amplitude, and simplifying, obtain $v_{max} = v_1 \sqrt{\frac{a^2}{a^2 - x_1^2}}$ .

Substituting $v_1 = 0.12 \frac{m}{s}$, $a = 0.2 m$, $x_1 = 0.15 m$, and calculating, obtain $v_{max} \approx 0.18 \frac{m}{s}$.