A mass of 0.40 kg,hanging from a spring with a spring constant of 75.0 N/m,is set into an up-and-down simple harmonic motion.What is the speed of the mass when moving through the equilibrium point? The starting displacement from equilibrium is 0.20 m.
A)zero
B)2.7 m/s
C)3.9 m/s
D)3.4 m/s
E)0.55 m/s

Question

Quizplus · Accepted Answer

The speed of the mass at the equilibrium point can be found using the conservation of mechanical energy principle. At the equilibrium point, all the energy is kinetic, and at the starting displacement, all the energy is potential. The potential energy at the starting point is given by $PE = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the displacement from equilibrium. Substituting the given values, $PE = \frac{1}{2} \times 75.0 \, \text{N/m} \times (0.20 \, \text{m})^2 = 1.5 \, \text{J}$. This potential energy is converted into kinetic energy $KE = \frac{1}{2}mv^2$ at the equilibrium point, where $m$ is the mass and $v$ is the velocity. Solving for $v$, we get $v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 1.5 \, \text{J}}{0.40 \, \text{kg}}} = 2.7 \, \text{m/s}$.

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