The energy of a spring-mass system is E₀ when the maximum displacement is x₀. If the total mechanical energy of the system is held constant and the mass is increased by a factor of two, the maximum velocity of the system is A) (1/2) v₀. B) (0.707) v₀. C) v₀. D) (1.414) v₀.

Question

Quizplus · Accepted Answer

The total mechanical energy in a spring-mass system is given by $E = \frac{1}{2}kx^2 = \frac{1}{2}mv^2$, where $k$ is the spring constant, $x$ is the maximum displacement, $m$ is the mass, and $v$ is the maximum velocity. If the mass is doubled, to keep the energy constant, the velocity must be reduced by a factor of $\sqrt{2}$, which is approximately 0.707. This is because the kinetic energy part of the equation $\frac{1}{2}mv^2$ now has a doubled mass, so to keep $E$ constant, $v^2$ must be halved, and thus $v$ must be multiplied by $\frac{1}{\sqrt{2}}$.

The Energy of a Spring-Mass System Is E0 When the Maximum

The Energy of a Spring-Mass System Is E₀ When the Maximum