17.1, 17.2 - Velocity and energy in oscillations

Velocity, phase and two possible directions

If an oscillator's displacement is zero, has it run out of energy? In ideal SHM the opposite can be true: at equilibrium it has its greatest speed. Its position is zero, while the slope of its displacement graph is greatest in magnitude.

For the phase choice $x=x_0\sin\omega t$, the object crosses equilibrium moving positively at $t=0$. Its velocity must therefore begin at a positive maximum, fall to zero at the positive turning point, and then become negative. A cosine describes exactly that timing.

The maximum-speed relation is $v_0=\omega x_0$. It follows by setting $x=0$ in the supplied displacement-dependent velocity equation below, rather than being printed as a separate formula.

The velocity is the gradient of the displacement curve. On a graph, check that its sign agrees with the slope at every instant, rather than choosing sine and cosine independently. Since $\cos\omega t=\sin(\omega t+\pi/2)$, velocity leads displacement by $\pi/2$ radians for this phase convention.

[DIAGRAM: asset_name: Three phase-consistent SHM graphs; asset_slug: oscillations-three-phase-graphs; caption: Velocity has the sign of the displacement gradient while acceleration opposes displacement; description: Three aligned graphs plot displacement, velocity and acceleration for a sine-phase oscillator with amplitude 0.050 metres and angular frequency 10 radians per second. Displacement starts at zero rising, velocity at its positive maximum, and acceleration at zero becoming negative. Vertical guides mark quarter periods.]

Eliminating time with a trigonometric identity

On a unit circle, the horizontal and vertical coordinates are cosine and sine. Pythagoras gives the identity

It is exact for every angle $\phi$, unlike the small-angle approximations. It is not on the data sheet: learn it. Using $x/x_0=\sin\omega t$ and $v/(\omega x_0)=\cos\omega t$,

Taking the square root introduces two possible signs.

Kinetic and potential energy exchange

A restoring interaction can store potential energy as the oscillator is displaced. When the object moves back towards equilibrium, that potential energy decreases and kinetic energy increases. With negligible resistance and no driving force, total mechanical energy stays constant.

For a horizontal Hooke's-law spring, the area under its force–extension graph gives elastic potential energy $E_P=\tfrac12kx^2$, taking zero at equilibrium, where the spring has its natural length. Since $k=m\omega^2$ for that system,

Mass $m$ is in kilograms, displacement $x$ in metres, angular frequency $\omega$ in $\mathrm{rad\,s^{-1}}$ and velocity $v$ in $\mathrm{m\,s^{-1}}$, giving both energies in joules. These energy relations are not on the data sheet: learn the kinetic relation and understand the potential-energy derivation. For general ideal SHM, the oscillation potential energy is measured relative to its equilibrium minimum and has the same quadratic form.

At $x=0$, this potential energy is minimum and speed is maximum. At either turning point, speed and kinetic energy are zero while potential energy is maximum. The object is not “out of energy” there; its energy is in potential rather than kinetic form. Acceleration is greatest in magnitude, ready to change the direction of motion.