An object is undergoing SHM with period 0.900 s and amplitude 0.3...

Question

An object is undergoing SHM with period 0.900 s and amplitude 0.320 m. At t = 0 the object is at x = 0.320 m and is instantaneously at rest. Calculate the time it takes the object to go (a) from x = 0.320 m to x = 0.160 m. (b) from x = 0.160 m to x = 0.

Accepted Answer

Everyone in this problem, a crane executes simple harmonic motion at the end of a spring with an amplitude of 168 centimeters in a period of 0. seconds. When T equals zero, the crate has maximum displacement. Okay. And is momentarily at rest and were asked, what is the time taken by the block to move from in part one, cm to 62 cm and in part to 62 cm - zero cm. Alright. So we're giving information about the period and the amplitude, we don't have information about the wavelength. Okay. So we can't use our trick. We don't know if 100 and 68 62 centimeters or quarter multiples of that wavelength or not. So, what we need to do is we need to look at the equation for positions a function of time. Okay. So, recall we have that X of T is equal to a coast omega T. Alright. So in order to use this, we need to find a and we need to find Omega. Well, we know that A Is equal to cm. What about omega? Well, Omega is our angular frequency and recall that this can be related to our period. Okay. Through the following Omega is equal to two pi over the period T we're given a period T of 0.419 seconds. And so we get two pi divided by 0. seconds. And we get an angular frequency of 14.9, 9, 6 ingredients per second. So we have our a value, we have our omega value. So in general, the equation that relates position and time, It's going to be X of T is equal to cm. Name's cose Of 14.996 Iranians per second times time team. Alright. So if we want to figure out the time taken to move from cm to 60, cm, what we need to do is figure out the time That we are at 168 cm. Figure out the time that we are at 62 cm and subtracted to take the difference. Alright, so let's start with 100 and 68 centimeters. So this is going to be part one When our position is 168 cm our equation becomes 100 and 68 centimeters is equal to centimeters times Kassian Of 14.996 radiance per second times time T And let's call this T1 Okay time where we're at that first position. Alright. Well, we can divide by 168. We have 100 and 68 centimeters divided by 100 and 68 centimeters. This is just going to give us one And on the right hand side, we're left with coasts of 14.996 radiance per second times time. T alright. So we have coast of something is equal to one. We know that coast of zero is equal to one. And so that tells us that 14. radiance per second times T is equal to zero. Now, if you didn't remember that coast of zero is equal to one. Okay, you can always take the inverse coast coast inverse of one and you will get zero. Alright. So we have 14.996 ratings per second. T is equal to zero which tells us that T one. Sorry, I dropped the one. This is T1 must be equal to zero seconds. All right. So when T zero, the crate has maximum displacement, which makes sense. Okay. This checks out at T equals zero. Crate has maximum displacement which means that it's at its amplitude, its amplitude is 168 centimeters. Okay. And so we're at the position 168 centimeters when T one is equal to zero seconds. Okay. That matches with the information we were getting. Now let's do the second part. Okay. When are we at 62 cm? Well, our position is 62 cm. Okay. So we get 62 cm is equal to 168 cm times cosign of 14.996 radiance per second times teeth. Now notice when we do this, we divide centimeters by centimeters. Okay. So our position it is okay to be in centimeters as long as our amplitude is also in centimeters. Okay. We need the position and the amplitude to be the same unit here. Alright. So we divide, We get 62 cm divided by cm is equal to cosign 14.996 radiance per second times the time T And let's label this T two cheat you. Okay. The second time point and this tells us that T2 is going to be equal to what? Well, we take the inverse coast Of the left hand side 62 cm divided by 168 cm. And then once we do that, we're gonna have 14.996 radiance per second T two. And so we have to divide by 14.996 radiance per second. And this is going to leave us with a T two value of 0. seconds. Alright. So the time taken to go from cm to 62 cm is going to be the difference in these times. Okay. So the time taken Is going to be equal to T 2 -11 which is equal to 0. seconds minus zero seconds. Okay? Which is just that second time we found t to 0.79542 seconds. Alright. So that is going to be our answer for part one. Now Let's move on to part two. So for part two, we're asked to find the time it takes to move from 62 centimeters to zero centimeters. Okay. We already know the time when we are at 62 centimeters because we found it in the previous question. Okay. So we already know that T two is equal to 0.79542 seconds. Okay. The time at 62 centimeters, let's now find the time that we are at when we are at position zero centimeters. So when we are at zero centimeters, our equation is zero centimeters is equal to centimeters times co sign 14.996 radiance per second times T and we're gonna call it T three this time. Okay. This is the third time point we've looked at This tells us that zero is equal to Kassian Of 14.996 radiance per second times T three. When does Kassian of something equal zero? Okay. Kassian of something equal zero When the Angle is Pi over two. So we get pi over two is equal to 14. radiance per second times T three. And we can divide by 14.996 to get our value of T three. And we get that T three is equal to approximately 0.1047 seconds. Alright. So the time taken here to go from X equals 62 centimeters to X equals zero centimeters is going to be the time where we're at zero centimeters. T three minus the time when we were at centimeters T two. This is going to be 0.1047 seconds minus 0.79542 seconds four A time of 0.025158 seconds approx. Alright. So if we look at our answer choices, we go back up, We see that part one. We had a time of approximately 0. seconds. And for part two, we had a time of approximately 0.0252 seconds. And so this corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

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An object is undergoing SHM with period 0.900 s and amplitude 0.320 m. At t = 0 the object is at x = 0.320 m and is instantaneously at rest. Calculate the time it takes the object to go (a) from x = 0.320 m to x = 0.160 m. (b) from x = 0.160 m to x = 0.

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