Derivation of equation of motion for mechanical seismograph

Consider a simple seismograph consisting of a mass $$M$$ hung from a spring on a rigid framework attached to the earth, as shown in the picture.

The motion of the mass is apparently described by the equation $$\frac{d^2y}{dt^2}+\gamma\frac{dy}{dt}+\omega_o^2y=-\frac{d^2\eta}{dt^2} \tag{1}$$

where $$y$$ is used to denote the displacement of $$M$$ relative to the earth and $$\eta$$ to denote the displacement of the earth's surface itself. I want to know how to derive this equation.

I know that damped oscillations are described by $$\frac{d^2y}{dt^2}+\gamma\frac{dy}{dt}+\omega_o^2y=0 \tag{2}$$ and that the oscillations of the earth's surface act as an external force on the system (replace the framework holding the spring with a hand). Therefore the equation becomes $$\frac{d^2y}{dt^2}+\gamma\frac{dy}{dt}+\omega_o^2y=\frac{F_o}{M}cos(\omega t) \tag{3}$$ Remembering that $$M\frac{d^2\eta}{dt^2}=F_ocos(\omega t) \tag{4}$$ I get a final equation $$\frac{d^2y}{dt^2}+\gamma\frac{dy}{dt}+\omega_o^2y=\frac{d^2\eta}{dt^2} \tag{5}$$ The question is, where have I lost the sign for the $$\frac{d^2\eta}{dt^2}$$ term?

1 Answer

The question is, where have I lost the sign for the $$\frac{d^2\eta}{dt^2}$$ term?

Since positive direction is up and for

$$\eta''>0$$

you get external force pushing down (in the coordinate system connected to Earth's surface), equation $$(4)$$ should've been

$$M\frac{d^2\eta}{dt^2}=\color{red}{-}F_ocos(\omega t)$$

letting us get the correct result.