A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is 0.090 m, it takes the block 2.70 s to travel from x = 0.090 m to x = -0.090 m. If the amplitude is doubled, to 0.180 m, how long does it take the block to travel (b) from x = 0.090 m to x = -0.090 m?

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Hey everyone in this problem, we have a wood wooden cube fixed on one end of a perfect spring that lies on a horizontal smooth surface where it performs simple harmonic motion. The motion has an amplitude a The mass covers the distance positive a too negative a in 2.64 seconds. The amplitude is then doubled to a one equals to A. And were asked how much time will the cube take when traveling from X one equals a 1/4, 2 X two equals negative A 1/4. No, Let's start with. What is the initial period when we're talking about how much time something takes? We're looking at um periods or fractions of periods. So the initial period while the mass covers a distance positive a too negative a in 2.64 seconds. So from positive a two negative a represents half a period. And you can imagine that if you were to draw, say a sine wave or cosine wave you draw a cosine wave to go from positive A The amplitude up here down to the bottom, negative a. That's half of your period halfway through one entire cycle. So we have half a period which means that t over two a half the period is equal to the time. It takes 2.64 seconds. This tells us that that initial period is going to be T equals 5.28 seconds. Alright, so we have our initial period. Now the amplitude is going to be doubled. So what happens to the period Once the amplitude is doubled. Well let's recall some equations relating to period, we know that the angular frequency omega is equal to two pi times the frequency F, Which is equal to two pi over the period T. Which is also equal to the square root of K over M. And what you'll notice is that none of these equations contain the amplitude. A. This means that the period does not depend on the amplitude. Now, if the period doesn't depend on the amplitude, then changing the period or sorry, changing the amplitude, doubling, the amplitude doesn't change the period. So the period t. In this second case when we're talking about the amplitude a one is still going to be 5.28 seconds. All right now we're looking at traveling from a 1/4 to negative a 1/4. And when you have positions or distances like these that aren't nice quarter multiples of your period, then what you need to do is you need to work from the equation instead of working like we did here, that's half a period. So we can divide by two, we can't do that all the time. That's a little shortcut that we can use in special cases, in this case, we're gonna need to work from the equation. So let's look at the general equation, we're talking about simple harmonic motion, we have the X of T is equal to a coast omega T plus five were five as a phase shift. Now we're going to say that at time T equals zero. We're gonna let the mass B at its amplitude A one. Now if we do that what happens is that A one is equal to a 1? Coast T equals zero. So we have zero plus five. Okay so we're just plugging in this 0.2 equals zero. X. Equals a one. We have that A one is equal to a one. Cose if I so divide by a one, we get that one is equal to coast five. Now when does coast equal 1? We know that it equals one at five equals zero. Okay that's one option. So we're going to say that five is equal to zero and now we don't have to worry about that face shift. So from now on our equation is going to be X. Of T. Is equal to a one. Cose of omega T. What is omega well above, we wrote all of these possible equations for omega. So we know that we can write omega as two pi over the period T. This is gonna be two pi Over 5.28 seconds. Remember the period we found initially is the same period that we have here when the amplitude has doubled because the period does not depend on the amplitude. This is going to give us an angular frequency of 1. radiance per second. Now we want to look at our two time points. So remember what we're trying to find. We're trying to find how much time it will take the queue to travel from. X one equals a 1/4 2 X two equals negative A 1/4. So with the equation we found now we have omega what we can do is figure out the time point that the cube is at X. One. The time point that the cube is at X. Two. And then we can take the difference to figure out how much time it will take. Okay so that's where we're at. So four X one Is equal to a 1/4. Our equation is going to be X. Which is a 1/4 is equal to our amplitude times coast omega. 1.19 radiance per second times a time. And we're going to call this T. One to correspond with X one. Now we can divide both sides by a one. Get that 1/4 is equal to cose of 1.19 radiance per second times T. One. And to solve for T. One we're gonna take arcos 1.2 Whoops not 1. 1/4. And we're gonna divide that by 1.19 Radiance per second in order to isolate T one. And this is going to give us AT one value Of 1.10766 seconds. Alright so at T one equals 1.11 seconds approximately. Our cuba is going to be a 1/4. Now let's do our second point we have X two is equal to negative A 1/4. And we're gonna do the same thing negative A 1/4 is equal to a one coast 1.19 radiance per second times T two KT two to correspond with X two. We divide by 81 we get negative 1/4 is equal to coast 1.19 radiance for a second times T two. And in order to isolate T two We get T two post inverse of minus 1/4 divided by 1.19 radiance per second and we get a T. Two value Of 1.53233 seconds approx. And now we want to figure out the time it takes to go from X one to X two. So what we're gonna do is we're gonna subtract these two times and figure out that difference. And so the time taken Is going to be equal to T 2 - which is equal to 1. seconds minus 1.10766 seconds. Which gives us a time difference Of 0.42467 seconds approx. And if we go up to our answer choices you see that these have Two decimal places. And if we compare with our answer and we round we see that we have answer choice b. The time taken for the Cube to go from a 1/4 to negative a 1/4 is going to be .42 seconds. Thanks everyone for watching. I hope this video helped you in the next one.

Verified Solution

Key Concepts

Simple Harmonic Motion (SHM)

Amplitude

Period of Oscillation