Physics · Ch 10 — Oscillations

Displacement, velocity, acceleration and its graphical representation - SHM

10.2.2

Displacement, velocity, acceleration and its graphical representation - SHM

Let $P$ be the particle's position on the reference circle of radius $A$ at time $t$ ; its displacement from the mean position is $y=A\sin\omega t$ . The amplitude $A$ is the maximum value taken by $y$ (when $\sin\omega t=1$ ); for SHM the amplitude stays constant throughout the motion. Velocity is the rate of change of displacement: differentiating gives $v=dy/dt=A\omega\cos\omega t$ , which, using $\cos\omega t=\sqrt{1-\sin^2\omega t}$ and $\sin\omega t=y/A$ , can be re-written entirely in terms of the instantaneous displacement as $v=\omega\sqrt{A^2-y^2}$ . This shows the two extremes directly: at the mean position ( $y=0$ ) the speed is maximum, $v=\omega A$ ; at the extreme positions ( $y=\pm A$ ) the speed is zero. Acceleration is the rate of change of velocity: $a=dv/dt=-A\omega^2\sin\omega t=-\omega^2y$ , i.e. $\dfrac{d^2y}{dt^2}=-\omega^2y$ -- this second-order differential equation is the defining equation of SHM. At the mean position the acceleration is zero even though the speed is maximum there, and at the extreme positions the acceleration has its maximum magnitude $A\om …

Figure 10.9Displacement, velocity and acceleration of a particle at some instant of time

What this figure shows. A particle P sits on the reference circle at angle theta from a fixed axis, with its position vector, and the geometric construction used to read off displacement (the projection y = A sin theta), velocity (the component of the tangential speed v along the projection axis, v cos theta), and acceleration (the component of the centripetal acceleration along the same axis) all drawn on one diagram. It visually ties together why displacement, velocity and acceleration in SHM are simply different trigonometric projections of the same uniform circular motion, each shifted from the others b …

Table 10.1Displacement, velocity and acceleration at different instants of time

Time	0	T/4	T/2	3T/4	T
omega t	0	pi/2	pi	3pi/2	2pi
Displacement, y = A sin(omega t)	0	A	0	-A	0
Velocity, v = A omega cos(omega t)	A omega	0	-A omega	0	A omega

Figure 10.10Variation of displacement, velocity and acceleration at different instants of time

What this figure shows. Three sine/cosine curves are stacked one above the other, all plotted against time t over one full period T: the displacement curve x = A sin(omega t), the velocity curve v = omega A cos(omega t) drawn one quarter-period ahead of displacement, and the acceleration curve a = -omega^2 A sin(omega t) drawn a further quarter-period ahead (i.e. exactly inverted relative to displacement). Lining the three curves up vertically makes the 90-degree phase lead of velocity over displacement, and the 180-degree phase difference between acceleration and displacement, visual …

Misc Example 10.4Ratio of time period to amplitude from two velocity-position pairs

Worked out. A particle in SHM has velocity v1 at position x1 and velocity v2 at position x2; the question asks to show that T/(2 pi A) works out to a formula in terms of x1, x2, v1 and v2. Starting from v^2 = omega^2(A^2 - x^2), writing this once for each point gives v1^2 = omega^2(A^2-x1^2) and v2^2 = omega^2(A^2-x2^2). Subtracting the two equations eliminates A^2 and isolates omega^2 in terms of (x2^2-x1^2)/(v1^2-v2^2); dividing the two original equations instead eliminates omega^2 and isolates A^2 in terms of the same four quantities. Combining the ratio A/omega obtained this way, and using T = 2 pi/omega, gives the required result T/A = 2 pi sqrt[(x2^2-x1^2)/(v1^2 x2^2 - v2^2 x1^2)], purely algebraic mani …

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