I have a question that says
"A Particle $P$ of mass $m$ kg is attached to one end of a spring. When the displacement of $P$ from its equilibrium poisition is $x$ metres, the magnitude of tension in the spring is $4nx N$ and the resistance force on P is $5c\frac{dx}{dt}$
Write a differential equation which models the motion of the particle.
My answer:
$$m\frac{d^2x}{dt^2}=4nx-5c\frac{dx}{dt}$$
$$m\frac{d^2x}{dt^2}+5c\frac{dx}{dt}-4nx$$
The textbook answer:
$$m\frac{d^2x}{dt^2}+5c\frac{dx}{dt}+4nx$$
I don't get why $4nx$ is positive? Wouldn't that require the resistant force to act in the same direction as the tension?
Best Answer
You need to get the signs correct. The elastic force is always in the opposite direction to displacement. The resistance is always opposite to the velocity. Let's suppose that the equilibrium is at $x=0$. When the object is at $x=1$, the elastic force is in the negative direction. But the object might be moving towards either right or left. If it's moving in the positive direction, the resistive force is in the negative direction. If the object is moving towards the origin, the resistive force is towards the positive direction