dwsmith said:$\mathcal{E} = \frac{1}{2}m\dot{x} + \frac{1}{2}kx^2$
The derivative is
$$
\frac{d\mathcal{E}}{dt} = \frac{1}{2}m\ddot{x} + kx\dot{x}
$$
but the solution is suppose to be
$$
\frac{d\mathcal{E}}{dt} = \dot{x}(m\ddot{x} + kx).
$$
How?
I like Serena said:Hi dwsmith! :)
The first part of your energy is the kinetic energy.
However, kinetic energy contains speed squared instead of just speed.
So your energy formula should be:
$\qquad \mathcal{E} = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2$
Differentiating $\mathcal{E}$ allows us to find the equation of motion for a mass-spring system under the influence of an external force. This equation, $\dot{x}(m\ddot{x} + kx)$, can then be used to analyze the behavior and stability of the system.
To differentiate $\mathcal{E}$, we first identify the terms that contain the variables of interest (in this case, $x$ and $\dot{x}$). Then, we use the rules of differentiation to find the derivative of each term. Finally, we combine the derivatives to reach the equation of motion, $\dot{x}(m\ddot{x} + kx)$.
The term $m\ddot{x}$ represents the acceleration of the mass in the system. It takes into account both the mass of the object and the net force acting on it. This term is crucial in understanding the dynamics of the system and how it responds to external forces.
The constant $k$ represents the stiffness of the spring in the system. A higher value of $k$ indicates a stiffer spring, which results in a faster response and higher frequency oscillations. On the other hand, a lower value of $k$ indicates a less stiff spring, resulting in a slower response and lower frequency oscillations.
No, this equation of motion is specific to a mass-spring system under the influence of an external force. Other factors, such as damping and non-linearity, may need to be considered for more complex systems. Additionally, the equation may need to be modified for different types of forces acting on the system.