(II) The graph of displacement vs. time for a small mass m at ...

Question

(II) The graph of displacement vs. time for a small mass m at the end of a spring is shown in Fig. 14–30. At t = 0 , 𝓍 = 0.43 cm.

(a) If m= 7.7 g , find the spring constant, k.

(b) Write the equation for displacement 𝓍 as a function of time.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Imagine that you are testing out a prototype spring system and that you have recorded the displacement versus time graph for a small mass that is attached to the end of the spring. The figure below shows the graph that is produced from this motion note that at T equals zero, the displacement will be X equals 0.79 centimeters. I, if the mass of the suspended component is 8.2 kg, determine what the spring constant K of your prototype spring system will be. I I express what the equation for the displacement X as a function of time will be. So it appears our end goal. What we're ultimately trying to solve for is we're trying to solve for two separate answers. Part I is asking us to determine what the spring constant K is for our prototype spring system. And I I is asking us to figure out what the equation for the displacement X as a function of time will be. So those are the two answers that we're trying to solve more. Ultimately. So looking at the figure that's provided to us for our graph, as we can see, we have our oscillating graph, which is denoted in red. And as you can see, it's a cosine graph, it appears to be because it's going up. And as you can see, we're given that the amplitude of our waves is 0.95 centimeters. And we're also told that the period is 0.72 seconds between, you know, our, our period is given to us, shown us by shown to us by our black arrows pointing to the left and right. And then we're also told that 0.79 is our displacement X. So 0.79 centimeters is our displacement X Awesome. So X is Ry AX and T is our X is our X axis. So X is our displacement for the Y axis and then T is our, is our horizontal X axis. Awesome. So now that we have a visualization of what's going on for this particular problem, let's look at our multiple choice answers to see what our final answer might be. Let us note that for our spring constant K, all of our units are gonna be in newtons per meter for part I first spring constant and for part I I, which is X of T equals some equation. So let's read them off to see what our final answer pair might be. So A is 4.5 multiplied by 10 to the power of two and 0.79 centimeters multiplied by cosine of 4.5 T minus 34. B is 4.5 multiplied by 10 to the power of two and 0.95 centimeters multiplied by cosine of 8.7 T 8.7 T minus 34 C is 6.2 multiplied by 10 to the power of two and 0.79 centimeters multiplied by cosine of 4.5 T minus 0.59. Finally D is 6.2 multiplied by 10 to the power of two and 0.95 centimeters and cosine of 8.7 T minus 0.59. Awesome. So now, first off, we need to write down all of our given variables. So from the provided graph, we know that the period T, which we're gonna use capital T to denote period is 0.72 seconds, we also know that the amplitude, which we're gonna use capital A to denote amplitude is equal to 0.95 centimeters. And we also know that M the mass is equal to 8.2 kg. So now we can start solving for part I, so to start solving for part I, we need to note that in order to solve for the spring constant K, we will need to recall and use the equation to help us sol for the period of an oscillating spring, which states that the period is equal to one divided by two multiplied by pi multiplied by the square root of M divided by K where M is the mass and K is our spring constant. So now we need to rearrange this equation to isolate and solve for K. So when we do that, we will get that K is equal to four multiplied by pi squared, multiplied by M all divided by T squared, which is equal to or multiplied by pi squared, multiplied by 8.2 kg is now we're plugging in all of our known variables to solve or our value of K divided by our period squared, which is 0.72 seconds squared. So when we plug that into our calculator, we will find that K is equal to 624 point 4657 and its units will be newtons per meter. Awesome. So now we can say that this is approximately equal to since we have to write our answer in scientific notation because all of our multiple choice answers for part I are written in scientific notation. So rounding to one decimal place in scientific notation, our final answer will be 6.2 multiplied by 10 to the power of two newtons per meter. And that's it. That's our first answer for part I hooray, we did it. So now we could start solving for part I I. So focusing on part I I, so we will first need to recall and use the equation to help us solver the displacement of an oscillating spring which is undergoing simple harmonic motion, which we can write this equation as X of T is equal to a multiplied by cosine of omega multiplied by T plus five. Awesome where Phi is our phase cha. So Phi in this case will be our phase angle representing the initial phase of oscillation. So the initial phase of the oscillation. So this is the phase angle is Phi and Omega represents our angular frequency of oscillation which let's take a moment here to pause and make a little note here that Omega which is our angular frequency of oscillation is equal to two multiplied by pi. So I need not that it's two multiplied by pi multiplied by F which F is the frequency. And then T in this case is the time. OK. Moving right along here. So now we need to solve for the value of omega first. So Omega is equal to two multiplied by pi multiplied by F which is also equal to two multiplied by pi divided by T which when we plug in our known variables will take two multiplied by pi divided by 0.72 seconds, which is gonna be equal to rounding the one decimal place, 8.7 radians per second. Fantastic. So now, at this point, we need to note that the initial conditions of the graph are, and let's write these in blue, T equals zero seconds and X of zero is equal to zero 0.79 centimeters. Thus, we can write that X of zero. So X of zero is equal to 0.95 centimeters multiplied by cosine A five, which we can also write ads, 0.79 centimeters is equal to 0.95 centimeters multiplied by cosine A five. So thus, we can rearrange this equation to isolate and solve for cosine of phi. So when we do that, we will get that cosine A five is equal to zero 0.79 centimeters divided by 0.95 centimeters. OK. So when we plug that into our calculator, this will all equal zero point 8316. So now we can isolate and solve for Phi. So when we isolate and solve for Phi, we will get that phi by itself is equal to inverse cosine of 0.8316. So when we plug that into our calculator, we will find that P is equal to plus or minus. Need to note this is very important plus or minus 0.58 eight eight. So three eights and it's gonna be and units of radiance which is approximately equal to plus or minus 0.59. When we round to two decimal places. So we need to make sure this is very, very important that our calculator is in units of radiance or else you will not get the correct answer. So let us note that the graph is shifted to the right from the zero phase cosine. Thus, the phase co uh sorry, the phase constant must be subtracted. Therefore, we could write that XFT is equal to 0.95 centimeters. And we need to multiply this by cosine of 8.7. And we need to multiply 8.75 T and we need to subtract 0.59 ratings. And that's it. That's our final answer for part. I hooray, we did it. So it's boxer final answer. Awesome. And let's quickly note once again that this is our answer for part I I, so looking at our multiple choice answers, the correct answer has to be the letter D which states IK is equal to 6.2 multiplied by 10 to the power of two newtons per meter. And I I is X of T is equal to 0.95 centimeters multiplied by cosine of 8.7 T minus 0.59. And that's it we solved for this problem. Hooray. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

Struggling with Physics?

(II) The graph of displacement vs. time for a small mass m at the end of a spring is shown in Fig. 14–30. At t = 0 , 𝓍 = 0.43 cm.
(a) If m= 7.7 g , find the spring constant, k.
(b) Write the equation for displacement 𝓍 as a function of time.
<IMAGE>

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Struggling with Physics?

(II) The graph of displacement vs. time for a small mass m at the end of a spring is shown in Fig. 14–30. At t = 0 , 𝓍 = 0.43 cm.(a) If m= 7.7 g , find the spring constant, k.(b) Write the equation for displacement 𝓍 as a function of time.<IMAGE>

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Intro to Simple Harmonic Motion (Horizontal Springs) practice set

(II) The graph of displacement vs. time for a small mass m at the end of a spring is shown in Fig. 14–30. At t = 0 , 𝓍 = 0.43 cm.
(a) If m= 7.7 g , find the spring constant, k.
(b) Write the equation for displacement 𝓍 as a function of time.
<IMAGE>