Velocity, Acceleration and Time Between Positions

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. In simple harmonic motion, SHM for short, a particle takes a quarter of a period, T over four, to go from the mean position to an extreme. So how long for just the first half of that distance? Half the distance, half the time, so T over eight, right? Hold that answer for now. Such timing questions do appear in JEE papers. So, let's begin.

0:28Velocity and acceleration in SHM

We start from the equation of SHM: x equals capital A times sine of, omega t plus phi. Here x is the displacement, capital A the amplitude, the Greek letter omega the angular frequency, and phi, another Greek letter, the initial phase. Okay? So how do we get v, the velocity, from this displacement? Yes, exactly, we differentiate with respect to time! v equals d x over d t, which is capital A times omega, times cosine of, omega t plus phi. With me? But problems usually give a position, not a time, right? The sine of the phase is x over capital A, so what is its cosine, in terms of x? It is plus or minus the root of, one minus x squared over capital A squared. So v equals plus or minus omega times the root of, capital A squared minus x squared. The plus-minus sign is there since the particle passes each point twice, once each way. Clear? Differentiate once more. The acceleration, small a, equals d v over d t, which is minus omega squared times x. The minus sign means it always points back towards the mean position. Got it?

1:44Where is it fastest? Where is the pull largest?

Both formulas are on screen now. So, where is the particle fastest? At the mean position, correct, where x is zero and the root is largest. We write it v max, with max as a small subscript, and v max equals capital A times omega. Okay? The size of the acceleration grows with the distance from the mean position, so it is largest at the extremes, x equals plus or minus capital A, where small a max equals capital A times omega squared. Right? Now, at the extremes the particle stops for an instant. Zero velocity, so zero acceleration too, right? Oh no, careful, this is exactly the trap! At the extremes, v is zero, yet the acceleration is the largest, since the pull is hardest there. Clear?

2:33Graphs and phase differences

Now, square the velocity formula and rearrange: x squared over capital A squared, plus v squared over capital A squared omega squared, equals one. So the v x graph, velocity against displacement, is an ellipse, and in suitable units, a circle. Okay? And the small a, x graph, acceleration against displacement, is a straight line, since small a equals minus omega squared times x. Its slope is minus omega squared. With me? Now the phases. Velocity has a cosine where displacement has a sine, and a cosine runs a quarter cycle ahead, so velocity leads by pi over two. The extra minus sign in acceleration is half a cycle, so it leads by pi. Got it?

3:16Time between positions: follow the phase

Back to our puzzle, with the reference circle. Point P goes round a circle of radius capital A at a steady rate omega, and its shadow N, the red dot, moves in SHM, with y sub N equals capital A times sine of, omega t plus phi. The red arrow at P is its velocity, of size capital A times omega, and the marked angle is the phase, growing steadily at the rate omega. Okay? The little triangle is the Greek capital delta, meaning change in, and theta, another Greek letter, is the phase. So the time between two positions, delta t, equals delta theta over omega, which is delta theta over two pi, times the period T, clear? Starting from phase zero at the centre, at what phase does x reach half the amplitude? Sine theta is one half, so theta is pi over six, one twelfth of a full two pi. The arrow on screen means from zero to half the amplitude, and that takes T over twelve, not T over eight! That is just a third of T over four, because the particle is fastest near the centre. So our puzzle is solved. The same way, capital A over root two takes T over eight, root three over two times capital A takes T over six, and the full amplitude takes T over four. Just find the phase from sine theta, and divide by omega. Right?

4:36Example 1: v max, a max, speed and acceleration at x = 2 cm

Example one, an easy one. An SHM has an amplitude of four centimetres and a period of two seconds. Find v max, a max, and the speed and acceleration at x equals two centimetres. Pause the video, and try it first. Done? Okay, let's check. Omega is two pi over the period, two pi over two, which is pi radians per second; the little minus one on the s means per second. Okay? So v max is capital A times omega, four pi centimetres per second, and small a max is capital A times omega squared, four pi squared centimetres per second squared. With me? Now, x equals two is half the amplitude. So the speed there is half of v max, two pi, right? Oh no, careful, that is the trap, because speed is not proportional to x. The formula gives pi times the root of, sixteen minus four, which is pi root twelve, or two root three pi centimetres per second, much more than half. Clear? And the acceleration is minus omega squared times x, minus pi squared times two, so minus two pi squared centimetres per second squared. It is proportional to x, so its size is exactly half of small a max here. Got it?

5:52Example 2: amplitude and period from two speeds in SHM

Now a harder one. The speed of a particle in SHM is eight centimetres per second at x equals three centimetres, and six centimetres per second at x equals four centimetres. Find the amplitude and the period. Pause the video, and try it first. Done? Okay, let's check. We square the velocity formula: v squared equals omega squared times, capital A squared minus x squared. Right? Put in the two readings. Eight squared is sixty four, so sixty four equals omega squared times, capital A squared minus nine. And thirty six equals omega squared times, capital A squared minus sixteen. Okay? Two equations, two unknowns. Which single step removes capital A, the amplitude? Yes, exactly, we subtract the two equations! Sixty four minus thirty six gives twenty eight, which equals seven omega squared, so omega squared is four; the double arrow on screen means which gives. Careful now, omega is not four; take the root, so omega is two radians per second. With me? Put omega squared equals four into the first equation. Sixty four over four gives sixteen, so capital A squared is twenty five, and the amplitude is five centimetres. Got it? And the period is two pi over omega, two pi over two, which is pi seconds. Clear? And here is a topper's shortcut. With subscripts one and two for the two readings, omega squared equals, v one squared minus v two squared, over, x two squared minus x one squared. Sixty four minus thirty six, over, sixteen minus nine, gives four in one line!

7:40Quick revision: v max, a max and why 0 to A/2 takes T/12

Okay? So, quick revision. Velocity is plus or minus omega times the root of, capital A squared minus x squared, and v max equals capital A times omega, at the centre. Acceleration is minus omega squared times x, and small a max equals capital A times omega squared, at the extremes. For timing, use delta theta over omega: the centre to half the amplitude takes T over twelve, not T over eight. Now try the practice problems on this page. All the best!