Mixed Problems: Set 1 (SHM)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You carry a pendulum clock to the Moon, where gravity is one sixth of the Earth's. So the pendulum swings six times slower, right? Hold on to that answer; we will test it soon. Today is our first set of mixed problems on simple harmonic motion, SHM for short, and they come up often in JEE Main. So, let's begin.

0:27Idea 1: the largest speed and acceleration

Let us start with idea number one. Here, x equals zero point one times the sine of, ten times t, plus pi over four, in S I units, so x in metres and the time t in seconds. Find the largest speed and acceleration, okay? Any SHM is x equals A sine of, omega t plus phi, with A the amplitude; omega, the Greek curly w, the angular frequency; and phi, the Greek letter with a line through it, the initial phase. The speed is largest at the mean position: v max, the v with a small max below it, equals A times omega. Clear? And the acceleration is largest at the extremes, x equals plus or minus A, where a max equals A times omega squared. Got it? Now, there is a pi over four inside the bracket. So tell me, does it change the largest speed, or the largest acceleration? No, it does not change them at all! Both answers use only A and omega; the phase just decides where the particle starts. So here, A is zero point one metre, and omega is ten radians per second, written rad s to the power minus one. With me? Now multiply. v max is zero point one times ten, which gives one metre per second, written m s to the power minus one. And a max is zero point one times ten squared, which gives ten metres per second squared, written m s to the power minus two. Okay?

1:59Idea 2: a pendulum clock on the Moon

Now, idea number two: our puzzle. The clock goes to the Moon, where g sub M, the g with a small M for Moon, is g over six. Gravity is six times smaller, so the period is six times longer, right? No, no, careful, that is the trap! For a small amplitude, a simple pendulum has T equals two pi times the square root of l over g, with l the length, g the acceleration due to gravity, and the tick, the square root sign. And the sign like a stretched alpha says it: T is proportional to one over root g. Clear? So take the ratio. T sub M, the period on the Moon, over T, equals the square root of g over g sub M, and that is root six. With me? Root six is about two point four five; the wavy equals sign means approximately. So each swing takes about two point four five times longer, not six times. The square root softens the change. Got it?

3:00Idea 3: what fraction is potential energy?

Now, idea number three: energy. In the figure, the orange curve is the potential energy, U equals half of k times x squared, with k the spring constant. The blue curve is the kinetic energy, K, and the dashed line is the total, E equals K plus U, fixed at half of k times A squared. They cross at A over root two, where K equals U, okay? So at A over three, the potential energy is one third of the total, right? Oh no, careful, look at the orange curve: it is a parabola, not a straight line. U goes as x squared and E as A squared, so U over E equals x squared over A squared. Clear? So U over E is one third, squared, which is one ninth. Only one ninth is potential; the other eight ninths, K over E, is kinetic. Square the fraction; never just copy it. Got it?

3:58Example 1: period of springs 2k and 3k in series

Alright, example one, on springs. Springs of spring constants two k and three k are joined in series and carry a mass m. Find the period. Pause the video and try it first. Done? Okay, let's check. Two springs, so just add them: two k plus three k gives five k, right? No, no, wait, adding is for springs in parallel, like the first picture, a block between two springs. Ours are in series, joined end to end, like the second picture; the third, a physical pendulum, we skip today. In series, the reciprocals add: one over k eff, the k with a small eff for effective, equals one over k one plus one over k two, where k one and k two are the two spring constants. With me? So, one over two k, plus one over three k. With six k as the common denominator, that is three over six k, plus two over six k, which gives five over six k. Okay? Now flip it: k eff equals six k over five, which is smaller than even the softer spring, two k. Right? Now the period: T equals two pi times the square root of m over k eff. With k eff as six k over five, T equals two pi times the square root of, five m over six k. Very good!

5:24Example 2: shortest time from x = -A/2 to A/root 2

Now a harder one, JEE level. Find the time for an SHM to go from x equals minus A over two, to x equals A over root two, by the shortest route. Pause the video and try it first. Done? Okay, let's check. Many of you know these times: from the mean position, reaching A over two takes T over twelve, and reaching A over root two takes T over eight. So the answer is T over eight minus T over twelve, which is T over twenty four, right? Oh no, careful, minus A over two is on the other side of the mean position. So let us use the reference circle: point P goes round a circle of radius A, and N, its shadow on the vertical axis along the red dashed line, moves in simple harmonic motion. The orange arrow is the velocity of P, along the tangent, of size A times omega. Here y sub N equals A sine of, omega t plus phi, the angle at the centre. Calling the displacement x and the phase theta, x equals A sine theta. With me? At minus A over two, sine theta is minus a half, so theta is minus pi over six, thirty degrees below the horizontal. At A over root two, sine theta is one over root two, so theta is pi over four, forty five degrees above. Clear? So on the shortest route, P sweeps up across the horizontal axis, and N passes through the mean position. Delta theta, the change in phase, written with the Greek triangle, is pi over four plus pi over six, which gives five pi over twelve, or seventy five degrees. Got it? The phase grows at the rate omega, so delta t equals delta theta over omega. With omega as two pi over T, that is five pi over twelve, times T over two pi. The pi cancels, giving five times T, over twenty four. Okay? And a quick check with those times, done the right way. The points are on opposite sides of the mean, so add them: T over twelve plus T over eight gives five times T, over twenty four. We get the same answer, perfect. Very good!

7:41Quick revision: max speed, energy, pendulum, springs, timing

So, quick revision. The largest speed is A omega, the largest acceleration A omega squared, and U over E is x squared over A squared. For a pendulum, T is proportional to one over root g; for springs in series, the reciprocals add. For the time between positions, use the phase: delta t equals delta theta over omega. Now try the practice problem on this page. All the best!