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Mass of 280 kg is attached spring and set into oscillation on horizontal frictionless surface The simple harmonic motion of the mass described bY x(t) (0.800 mJcos[...

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Mass of 280 kg is attached spring and set into oscillation on horizontal frictionless surface The simple harmonic motion of the mass described bY x(t) (0.800 mJcos[(10.0 rads)t]- Determine the following.amplitude 800oscillation for the oscillating Massforce constant for the spring Nmposition of the mass after has been oscillating for one half periodposition of the mass one-sixth of period after it has been releasedHow can You determine the position of the object at specific time from an expressi

mass of 280 kg is attached spring and set into oscillation on horizontal frictionless surface The simple harmonic motion of the mass described bY x(t) (0.800 mJcos[(10.0 rads)t]- Determine the following. amplitude 800 oscillation for the oscillating Mass force constant for the spring Nm position of the mass after has been oscillating for one half period position of the mass one-sixth of period after it has been released How can You determine the position of the object at specific time from an expression for the position af the object at ay time? m time takes the mass t0 get t0 the position 0.100 m after has been released



Answers

An object of mass $0.2 \mathrm{~kg}$ is hung from a spring whose spring constant is $80 \mathrm{~N} / \mathrm{m}$. The body is subject to a resistive force given by $-b v$, where $v$ is its velocity $(\mathrm{m} / \mathrm{sec})$ and $b=4 \mathrm{~N}-\mathrm{m}^{-1} \mathrm{sec}$. (a) Set up the differential equation of motion for free oscillations of the system, and find the period of such oscillations. (b) The object is subjected to a sinusoidal driving force given by $F(t)=F_{0} \sin \omega t$, where $F_{0}=2 \mathrm{~N}$ and $\omega=30 \mathrm{sec}^{-1}$. In the steady state, what is the amplitude of the forced oscillation?

Hello and welcome to this video solution of numerous. This question is based on the principles of simple harmonic motion. So this situation based on a spring my system and we have got the mass of the object as two kg. So let me tabulate the given parameters right away. We have mass M equal to two cages. Now it is acted upon by an horizontal force of 20 newtons That is required to hold the object at rest when it's pulled .2 m from the equilibrium position. So there is a extension of this mask which is equal to 0.2 m from the equilibrium point. And this is held by a force that is equal to Np Newton's. Now here we have to calculate the force consent of the spring. So party forced consent has given us F equal to from the basic equation for situation of spring, which is difficult to cakes. Now here we have forced us 20, which is equal to K. We have to calculate time. Success. It opened to great. From here we have two equal to 100 newtons per meter. So this is party part B. We have to calculate the frequency of oscillation. So omega is the angular frequency of oscillation which is calculated as but overall cable. Yeah, if you put this value of K and then you will get uh route 50 dads two radiance first again, omega is actually if you know maybe I called Jeff. So if you want to convert this into read in hearts, then you have to calculate if So if we will get to pay by omega, omega by to pay Which is 1.13 heads. So there's an answer to part B. Now let's move on to part C. Here we have to calculate the maximum velocity we have v max equal to oh my God, Now Omega, we already know it's route 50 Times is 0.2. The one for me. Days later this gives us meters per second per second. So this gives us 1.14 meters per second. There's a maximum velocity of population part D we have to calculate. Okay, so here it is asking that where is this maximum speed? Oprah Okay, so maximum speed walkers have the equilibrium point. That is that X equal to zero. We have caught maximum speed max speed or velocity whatever. It's a maximum velocity question. E now you two for fire. Okay. Now the question is to calibrate the maximum acceleration. So acceleration of maximum acceleration is equal to omega square Mhm. So it's equal to 50 times. Point to this is equal to 10 m past. Against where? Okay, so now we have to find out the that the distance where this equilibrium where this acceleration is maximum. Right? So the acceleration is maximum attics equal to plus my nancy. That is at the extreme ends. He's X equal to plus minus of 0.00.2. We do straight we have to we have in these positions we have caught we have the maximum accelerations fine. Now we have part G. Okay so we have to calculate the total energy of oscillation of the system. So total energy is given us half. Okay is square. This is equal to half times How much we got GSK 800 Times 0.2 Squared this is jules which just gives us regions. Next we have got part age. Now it is given that what is the speed of oscillation? When the position is 1/3 of the maximum value, That means we have X equal to a way through Which is equal to 0.2 by three m. The speed is given by an expression equal to omega route over over a squared minus X squared. Now you have to plug in the values of A and X. Here and omega, we know this gives us the velocity At x equal to a by three. Call to 1.33 m. again. Similarly, we have to calculate the acceleration i at that same point I acceleration equal to omega's critics here. You have to plug in the value of omega. Yes, it took 50 So 50 times of and so Excess. He had 2.0.2 x three, two by three meters per second squared. This gives us 3.33. I hope this is clear to you and have a very good rest of the they.

Harris Trans lectured or discussion? Supposed on a holy joint. Also face for two KZ object is stretched at the position off on two meter from its equilibrium position and the force required to hold all objectors when Dean you did. Okay, Mass is located. You of the object? No. Firstly, we have to find out the force constraint off the object. Okay, Uh, you got that? If this is the mean position, then we are stretching the our object and to hold the object the force required this physical Quentin union and that this force or object is that fresh? OK, eso basically, we have to find our fourth question. So we have the formula, our physicals tok ex gay will be forced by ex forces 20 Newton access 200.2 meter. So we will get the values 100 mutual or meter or force question. No, we have to find out the frequency off oscillation when we released this object. So it will start start emotion that Gordon simple harmonic motion. So now we have to find out the frequency of that motion so we'll make Isaac was to find you new is a questionable no makeup on to buy. And we know the form loaf Omega, that is Gabe. I am under route one of 12 by now putting the values here one upon to into 3.14 on the room 100. By goingto we will get the value off one point blank. Hutch. Okay, this is the frequents. No, we have to find out the maximum speed. Okay, so basically maximum, we have the form love. Maximum speed is the omega, so I am pretty sure is. So for maximum speed, we have to find out and little first for amplitude. Uh, we have the position. A also make a t. No ad is it goes to geo acts. Is that 0.2 meter? Because it already stretched. So course DJ Brokaw 01 So it will come to me. A temperature remains constant. Okay, so from here, going to under root 100 up on 1000.2. And by solving this, we will get the maximum speed off 1.41 major per second. Ok, Now, for maximum acceleration, we have the formula. You make us sweat. So why do in 200 by way to that the sea by Omega Square is acquitted to Steve. I am. So we will get the exploration off that Anitra per second. Also, we have to find out that Veer does this maximum spear and Mexican aspiration. So basically, if we take a access occurs toe a also omega T So villus people comes to be a omega sine omega T, and this is in minus, so V will be maximum. Then we get the value off omega for a omega sine omega T value of sine omega T must be minus would. So omega T is minus one at the angle off to 70 degrees, so omega t will be keep by a bite. Do OK, so because at home I got to you is it goes to treat bye bye to we will get the value off a omega. And this is a maximum value of spirit. Okay, No. By solving this to five i p cheap by bike to So we will get the value off the so he by 40. No, at this time we have to find out our position. So position will be access because toe a course to five i t into treaty by a four. He will be cancel out here from to cancer. So coarse three bye bye to mean scores to 70 and course Pose. I went Is Jiro so at excessive? Goes to Jiro. We have the Mexican velocity. Similarly, for acceleration, we have the form. Love s oration minus a Omega Square goes Omega T. Okay. No, it will be maximum at the value if also Omega T values minus one. Okay, so it will be at the angle of 1 80 degree. Okay. So Oh my God. He will be by So from here to buy by t Do you buy? Evil Comes Toby. Do you might do now at T by two. We have the value exegesis A cause by whitey into t by two people begins alert close by. So accessible B minus a. So at minus a position, the acceleration will exploration will be maximum. No, we have to find out the total energy off the system. So total energy is We're toe Kaine Antic energy. Let's put and she'll energy and it is Have I m If I will right there. A clear. So it will be half force Constant Amplitude square. This is the total energy. Okay, Half in 200 in tow, 2000.2 square. So we will get the value of total energy to chew. Okay, Now we have to find out the value off velocity value of speed and acceleration when the position is 1/3 off its maximum value. Okay, so if access the position course Omega T. So its maximum value is a on Dvir entitled off it is a by three. Okay. So also, if I have to write the that is spirits, it will be a omega sine omega T. But an acceleration will be minus Iomega's. Where goes Omega de? This is the velocity and acceleration at the position off one by tree over its them maximally. Okay, well, this is all for me for this video. I hope you will like the review. Thank you.

So this question belongs to the isolation in which for the previous problem, let us take the position of the mass. When the spring is his test at X equals to zero. So and the direction from left to right is positive X axis. So we have to determine the equations so forth apart a at remain position. Okay, at the main position when the masses at main position. So from the previous problem, the angular speed omega. This is equal to K by ems okay, it is equal to Okay, it is equal to 1200 mass m is three and this under rule. So from here we get 20 radian per second. Okay, so we can write that X. This is equal to amplitude was already given so this is equal to 2.0 centimeter and sine omega T plus five. Okay, five is the initial phase. So are the main position. This fire will be equals to zero. Okay. And omega can be taken as 20. So from here the equation obtained will be X. This is equal to 2.0 centimeter and signed 20. So this become the answer for the but a Okay no. Moving to the part B in which we have to determine the evacuation at the maximum stress position at the maximum stretched positions where the maximum stretched position, the fire angle this will be equals two by by two because it is at the right end so we can right here that the situation will be converted into sine omega T plus by by two. And this can be converted into cause so we can write directly here so 2.0 centimeter and of course 20 T. So this will be the equation for the second be part Okay, no moving to the last see part in which we have to determine the same equation at the maximum at maximum compressed position, compressed position. So maximum compressed position will be towards the left so the left the angle will be three by by two. Or we can take minus by by two. So five here becomes three by by two. So the situation will be converted into 2.0 centimeter and of course 20 T. And this minus because sign off three by by two plus omega T. This will be equals two minus course omega t. Okay so this will be the answer for the part C of the problem. Now next we have to determine in which way these three equations are same or different. So hence we can stay here. The phase angle is different for the each problem, amplitude and frequency are same in the each problem. Okay so face is different and remaining values are same. Okay. Mhm. Thank you.

Mm in this problem. I'm going to be working Simple harmonic motion. Example of a magic on the horizontal spring. And what we I'm velocity and acceleration of the mask that All right. So, what we have adjusted bring here Mhm Okay. Zero, that's well the Librium station what we're going to do Yeah, for any help. Two X most 20. All right. So, we want to find what the acceleration logically hard to the point once we stretched the spring and let him So I'll start with the once we have our allegiance from books, last force equals mass times acceleration. Okay, Same time X we're X here. We're going to from its equilibrium. So it's pretty easy to see that zero acceleration that equals zero. We have to give any applications for that one right now, for velocity we're gonna need to put in some numbers um and we'll use this equation for velocity at any point. All right. Three head. Well, Square root of our amplitude which is going to be our 20 cm problem minus hard position, which is equilibrium given to us by the problem. Little easier. Very sorry, that's maybe my zero. And that's all times hes for Amber and constant and Emma Carmax usually and this simplifies to be equal amplitude times the square root of Bernie. Hey, over and right. When the values given to us in the problem. Right, Okay. Is there 27 years. Okay, give it to us eight. You will fix them meters and every mhm uh to killing them. And right. So putting in our numbers we get the most Right. You mean years? Yeah. Only english Over two. I'm sorry. It's not 8 80 m for the spring content. Right? So I spent three years eight m divided by two. That gives us 0.2 liters trying to square you up 40 minute. Mhm mm hmm. Yeah. Mm mm hmm, mm. Mhm mm. Uh huh, mm mm mm. Mhm. Mhm. Uh huh, mm. Mm. In this problem, I mean.


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