\int_{\sqrt{7} }^{\infty }(8y^{3}-12y)^{2}e^{-y^{2}}dy=3.6363. For a better experience, please enable JavaScript in your browser before proceeding. endobj endstream How To Register A Security With Sec, probability of finding particle in classically forbidden region, Mississippi State President's List Spring 2021, krannert school of management supply chain management, desert foothills events and weddings cost, do you get a 1099 for life insurance proceeds, ping limited edition pld prime tyne 4 putter review, can i send medicine by mail within canada. [2] B. Thaller, Visual Quantum Mechanics: Selected Topics with Computer-Generated Animations of Quantum-Mechanical Phenomena, New York: Springer, 2000 p. 168. What is the probability of finding the particle in classically forbidden region in ground state of simple harmonic oscillator. Legal. Turning point is twice off radius be four one s state The probability that electron is it classical forward A region is probability p are greater than to wait Toby equal toe. Mesoscopic and microscopic dipole clusters: Structure and phase transitions A.I. 9 OCSH`;Mw=$8$/)d#}'&dRw+-3d-VUfLj22y$JesVv]*dvAimjc0FN$}>CpQly . It may not display this or other websites correctly. h 1=4 e m!x2=2h (1) The probability that the particle is found between two points aand bis P ab= Z b a 2 0(x)dx (2) so the probability that the particle is in the classical region is P . Correct answer is '0.18'. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. You can see the sequence of plots of probability densities, the classical limits, and the tunneling probability for each . Disconnect between goals and daily tasksIs it me, or the industry? The classically forbidden region is given by the radial turning points beyond which the particle does not have enough kinetic energy to be there (the kinetic energy would have to be negative). That's interesting. This made sense to me but then if this is true, tunneling doesn't really seem as mysterious/mystifying as it was presented to be. June 5, 2022 . The calculation is done symbolically to minimize numerical errors. Published since 1866 continuously, Lehigh University course catalogs contain academic announcements, course descriptions, register of names of the instructors and administrators; information on buildings and grounds, and Lehigh history. In particular the square of the wavefunction tells you the probability of finding the particle as a function of position. You can't just arbitrarily "pick" it to be there, at least not in any "ordinary" cases of tunneling, because you don't control the particle's motion. Related terms: Classical Approach (Part - 2) - Probability, Math; Video | 09:06 min. The way this is done is by getting a conducting tip very close to the surface of the object. and as a result I know it's not in a classically forbidden region? Lehigh Course Catalog (1996-1997) Date Created . Is it possible to create a concave light? Mathematically this leads to an exponential decay of the probability of finding the particle in the classically forbidden region, i.e. You may assume that has been chosen so that is normalized. /Subtype/Link/A<> Although the potential outside of the well is due to electric repulsion, which has the 1/r dependence shown below. Possible alternatives to quantum theory that explain the double slit experiment? The wave function oscillates in the classically allowed region (blue) between and . We've added a "Necessary cookies only" option to the cookie consent popup. In classically forbidden region the wave function runs towards positive or negative infinity. << Making statements based on opinion; back them up with references or personal experience. endobj For a quantum oscillator, assuming units in which Planck's constant , the possible values of energy are no longer a continuum but are quantized with the possible values: . Confusion regarding the finite square well for a negative potential. If not, isn't that inconsistent with the idea that (x)^2dx gives us the probability of finding a particle in the region of x-x+dx? Probability of particle being in the classically forbidden region for the simple harmonic oscillator: a. Question: Probability of particle being in the classically forbidden region for the simple harmonic oscillator: a. Energy and position are incompatible measurements. The zero-centered form for an acceptable wave function for a forbidden region extending in the region x; SPMgt ;0 is where . Can you explain this answer? Probability distributions for the first four harmonic oscillator functions are shown in the first figure. Can a particle be physically observed inside a quantum barrier? +2qw-\
\_w"P)Wa:tNUutkS6DXq}a:jk cv "Quantum Harmonic Oscillator Tunneling into Classically Forbidden Regions" The probability of the particle to be found at position x at time t is calculated to be $\left|\psi\right|^2=\psi \psi^*$ which is $\sqrt {A^2 (\cos^2+\sin^2)}$. Probability 47 The Problem of Interpreting Probability Statements 48 Subjective and Objective Interpretations 49 The Fundamental Problem of the Theory of Chance 50 The Frequency Theory of von Mises 51 Plan for a New Theory of Probability 52 Relative Frequency within a Finite Class 53 Selection, Independence, Insensitiveness, Irrelevance 54 . Professor Leonard Susskind in his video lectures mentioned two things that sound relevant to tunneling. Download more important topics, notes, lectures and mock test series for Physics Exam by signing up for free. endobj The classically forbidden region is given by the radial turning points beyond which the particle does not have enough kinetic energy to be there (the kinetic energy would have to be negative). /D [5 0 R /XYZ 276.376 133.737 null] for 0 x L and zero otherwise. (4.303). 2003-2023 Chegg Inc. All rights reserved. In particular the square of the wavefunction tells you the probability of finding the particle as a function of position. Belousov and Yu.E. It can be seen that indeed, the tunneling probability, at first, decreases rather rapidly, but then its rate of decrease slows down at higher quantum numbers . /D [5 0 R /XYZ 234.09 432.207 null] How to match a specific column position till the end of line? I don't think it would be possible to detect a particle in the barrier even in principle. theory, EduRev gives you an
12 0 obj These regions are referred to as allowed regions because the kinetic energy of the particle (KE = E U) is a real, positive value. Can you explain this answer? << .1b[K*Tl&`E^,;zmH4(2FtS> xZDF4:mj mS%\klB4L8*H5%*@{N Como Quitar El Olor A Humo De La Madera, khloe kardashian hidden hills house address Danh mc h 1=4 e m!x2=2h (1) The probability that the particle is found between two points aand bis P ab= Z b a 2 0(x)dx (2) so the probability that the particle is in the classical region is P . /Font << /F85 13 0 R /F86 14 0 R /F55 15 0 R /F88 16 0 R /F92 17 0 R /F93 18 0 R /F56 20 0 R /F100 22 0 R >> Third, the probability density distributions for a quantum oscillator in the ground low-energy state, , is largest at the middle of the well . Jun >> WEBVTT 00:00:00.060 --> 00:00:02.430 The following content is provided under a Creative 00:00:02.430 --> 00:00:03.800 Commons license. My TA said that the act of measurement would impart energy to the particle (changing the in the process), thus allowing it to get over that barrier and be in the classically prohibited region and conserving energy in the process. (iv) Provide an argument to show that for the region is classically forbidden. H_{4}(y)=16y^{4}-48y^{2}-12y+12, H_{5}(y)=32y^{5}-160y^{3}+120y. stream (B) What is the expectation value of x for this particle? If the correspondence principle is correct the quantum and classical probability of finding a particle in a particular position should approach each other for very high energies. /D [5 0 R /XYZ 126.672 675.95 null] The same applies to quantum tunneling. << This Demonstration shows coordinate-space probability distributions for quantized energy states of the harmonic oscillator, scaled such that the classical turning points are always at . 06*T Y+i-a3"4 c A particle has a certain probability of being observed inside (or outside) the classically forbidden region, and any measurements we make . \int_{\sqrt{2n+1} }^{+\infty }e^{-y^{2}}H^{2}_{n}(x) dy, (4.298). You simply cannot follow a particle's trajectory because quite frankly such a thing does not exist in Quantum Mechanics. Quantum mechanics, with its revolutionary implications, has posed innumerable problems to philosophers of science. Or am I thinking about this wrong? The vertical axis is also scaled so that the total probability (the area under the probability densities) equals 1. 2. The probability of finding a ground-state quantum particle in the classically forbidden region is about 16%. For the particle to be found with greatest probability at the center of the well, we expect . ${{\int_{a}^{b}{\left| \psi \left( x,t \right) \right|}}^{2}}dx$. /Filter /FlateDecode /D [5 0 R /XYZ 188.079 304.683 null] /Type /Annot Classically, the particle is reflected by the barrier -Regions II and III would be forbidden According to quantum mechanics, all regions are accessible to the particle -The probability of the particle being in a classically forbidden region is low, but not zero -Amplitude of the wave is reduced in the barrier MUJ 11 11 AN INTERPRETATION OF QUANTUM MECHANICS A particle limited to the x axis has the wavefunction Q. Lehigh Course Catalog (1999-2000) Date Created . We have step-by-step solutions for your textbooks written by Bartleby experts! Solution: The classically forbidden region are the values of r for which V(r) > E - it is classically forbidden because classically the kinetic energy would be negative in this ca Harmonic . So, if we assign a probability P that the particle is at the slit with position d/2 and a probability 1 P that it is at the position of the slit at d/2 based on the observed outcome of the measurement, then the mean position of the electron is now (x) = Pd/ 2 (1 P)d/ 2 = (P 1 )d. and the standard deviation of this outcome is (a) Find the probability that the particle can be found between x=0.45 and x=0.55. classically forbidden region: Tunneling . Q) Calculate for the ground state of the hydrogen atom the probability of finding the electron in the classically forbidden region. Tunneling probabilities equal the areas under the curve beyond the classical turning points (vertical red lines). [3] P. W. Atkins, J. de Paula, and R. S. Friedman, Quanta, Matter and Change: A Molecular Approach to Physical Chemistry, New York: Oxford University Press, 2009 p. 66. accounting for llc member buyout; black barber shops chicago; otto ohlendorf descendants; 97 4runner brake bleeding; Freundschaft aufhoren: zu welchem Zeitpunkt sera Semantik Starke & genau so wie parece fair ist und bleibt Open content licensed under CC BY-NC-SA, Think about a classical oscillator, a swing, a weight on a spring, a pendulum in a clock. zero probability of nding the particle in a region that is classically forbidden, a region where the the total energy is less than the potential energy so that the kinetic energy is negative. Published:January262015. ncdu: What's going on with this second size column? ectrum of evenly spaced energy states(2) A potential energy function that is linear in the position coordinate(3) A ground state characterized by zero kinetic energy. %PDF-1.5 in the exponential fall-off regions) ? This dis- FIGURE 41.15 The wave function in the classically forbidden region. If the particle penetrates through the entire forbidden region, it can "appear" in the allowed region x > L. Thus, the particle can penetrate into the forbidden region. Free particle ("wavepacket") colliding with a potential barrier . daniel thomas peeweetoms 0 sn phm / 0 . 21 0 obj In general, we will also need a propagation factors for forbidden regions. Using Kolmogorov complexity to measure difficulty of problems? endobj /Length 1178 We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. The transmission probability or tunneling probability is the ratio of the transmitted intensity ( | F | 2) to the incident intensity ( | A | 2 ), written as T(L, E) = | tra(x) | 2 | in(x) | 2 = | F | 2 | A | 2 = |F A|2 where L is the width of the barrier and E is the total energy of the particle. endobj There is nothing special about the point a 2 = 0 corresponding to the "no-boundary proposal". The integral in (4.298) can be evaluated only numerically. Share Cite If you work out something that depends on the hydrogen electron doing this, for example, the polarizability of atomic hydrogen, you get the wrong answer if you truncate the probability distribution at 2a. What happens with a tunneling particle when its momentum is imaginary in QM? Okay, This is the the probability off finding the electron bill B minus four upon a cube eight to the power minus four to a Q plus a Q plus. Have you? What is the probability of finding the partic 1 Crore+ students have signed up on EduRev. To each energy level there corresponds a quantum eigenstate; the wavefunction is given by. Wavepacket may or may not . Peter, if a particle can be in a classically forbidden region (by your own admission) why can't we measure/detect it there? \[P(x) = A^2e^{-2aX}\] Classical Approach (Part - 2) - Probability, Math; Video | 09:06 min. Bulk update symbol size units from mm to map units in rule-based symbology, Recovering from a blunder I made while emailing a professor. ample number of questions to practice What is the probability of finding the particle in classically forbidden region in ground state of simple harmonic oscillatorCorrect answer is '0.18'. The bottom panel close up illustrates the evanescent wave penetrating the classically forbidden region and smoothly extending to the Euclidean section, a 2 < 0 (the orange vertical line represents a = a *). You may assume that has been chosen so that is normalized. I am not sure you could even describe it as being a particle when it's inside the barrier, the wavefunction is evanescent (decaying). Connect and share knowledge within a single location that is structured and easy to search. Correct answer is '0.18'. . From: Encyclopedia of Condensed Matter Physics, 2005. >> /Filter /FlateDecode Year . It is easy to see that a wave function of the type w = a cos (2 d A ) x fa2 zyxwvut 4 Principles of Photoelectric Conversion solves Equation (4-5). But for the quantum oscillator, there is always a nonzero probability of finding the point in a classically forbidden region; in other words, there is a nonzero tunneling probability. Why are Suriname, Belize, and Guinea-Bissau classified as "Small Island Developing States"? Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Wavepacket may or may not . HOME; EVENTS; ABOUT; CONTACT; FOR ADULTS; FOR KIDS; tonya francisco biography Can I tell police to wait and call a lawyer when served with a search warrant? .r#+_. Can I tell police to wait and call a lawyer when served with a search warrant? Energy eigenstates are therefore called stationary states . [1] J. L. Powell and B. Crasemann, Quantum Mechanics, Reading, MA: Addison-Wesley, 1961 p. 136. If so, why do we always detect it after tunneling. Find the Source, Textbook, Solution Manual that you are looking for in 1 click. This Demonstration calculates these tunneling probabilities for . For example, in a square well: has an experiment been able to find an electron outside the rectangular well (i.e. Solution: The classically forbidden region are the values of r for which V(r) > E - it is classically forbidden because classically the kinetic energy would be negative in this case. (v) Show that the probability that the particle is found in the classically forbidden region is and that the expectation value of the kinetic energy is . To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The potential barrier is illustrated in Figure 7.16.When the height U 0 U 0 of the barrier is infinite, the wave packet representing an incident quantum particle is unable to penetrate it, and the quantum particle bounces back from the barrier boundary, just like a classical particle. $x$-representation of half (truncated) harmonic oscillator? Each graph is scaled so that the classical turning points are always at and . A similar analysis can be done for x 0. ~! For the quantum mechanical case the probability of finding the oscillator in an interval D x is the square of the wavefunction, and that is very different for the lower energy states. Step by step explanation on how to find a particle in a 1D box. quantum mechanics; jee; jee mains; Share It On Facebook Twitter Email . Home / / probability of finding particle in classically forbidden region. /D [5 0 R /XYZ 125.672 698.868 null] represents a single particle then 2 called the probability density is the from PHY 1051 at Manipal Institute of Technology First, notice that the probability of tunneling out of the well is exactly equal to the probability of tunneling in, since all of the parameters of the barrier are exactly the same. For a quantum oscillator, we can work out the probability that the particle is found outside the classical region. The probability of finding the particle in an interval x about the position x is equal to (x) 2 x. Can you explain this answer? The difference between the phonemes /p/ and /b/ in Japanese, Difficulties with estimation of epsilon-delta limit proof. The connection of the two functions means that a particle starting out in the well on the left side has a finite probability of tunneling through the barrier and being found on the right side even though the energy of the particle is less than the barrier height. 11 0 obj A particle in an infinitely deep square well has a wave function given by ( ) = L x L x 2 2 sin. rev2023.3.3.43278. The classically forbidden region coresponds to the region in which. One idea that you can never find it in the classically forbidden region is that it does not spend any real time there. The classically forbidden region is where the energy is lower than the potential energy, which means r > 2a. For the n = 1 state calculate the probability that the particle will be found in the classically forbidden region. The oscillating wave function inside the potential well dr(x) 0.3711, The wave functions match at x = L Penetration distance Classically forbidden region tance is called the penetration distance: Year . Use MathJax to format equations. Now consider the region 0 < x < L. In this region, the wavefunction decreases exponentially, and takes the form So its wrong for me to say that since the particles total energy before the measurement is less than the barrier that post-measurement it's new energy is still less than the barrier which would seem to imply negative KE. So in the end it comes down to the uncertainty principle right? The Question and answers have been prepared according to the Physics exam syllabus. The turning points are thus given by En - V = 0. $\psi \left( x,\,t \right)=\frac{1}{2}\left( \sqrt{3}i{{\phi }_{1}}\left( x \right){{e}^{-i{{E}_{1}}t/\hbar }}+{{\phi }_{3}}\left( x \right){{e}^{-i{{E}_{3}}t/\hbar }} \right)$. Can you explain this answer? Calculate the classically allowed region for a particle being in a one-dimensional quantum simple harmonic energy eigenstate |n). Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. (a) Determine the probability of finding a particle in the classically forbidden region of a harmonic oscillator for the states n=0, 1, 2, 3, 4. Third, the probability density distributions | n (x) | 2 | n (x) | 2 for a quantum oscillator in the ground low-energy state, 0 (x) 0 (x), is largest at the middle of the well (x = 0) (x = 0). Textbook solution for Introduction To Quantum Mechanics 3rd Edition Griffiths Chapter 2.3 Problem 2.14P. 1999-01-01. Calculate the. So which is the forbidden region. Thus, the probability of finding a particle in the classically forbidden region for a state \psi _{n}(x) is, P_{n} =\int_{-\infty }^{-|x_{n}|}\left|\psi _{n}(x)\right| ^{2} dx+\int_{|x_{n}|}^{+\infty }\left|\psi _{n}(x)\right| ^{2}dx=2 \int_{|x_{n}|}^{+\infty }\left|\psi _{n}(x)\right| ^{2}dx, (4.297), \psi _{n}(x)=\frac{1}{\sqrt{\pi }2^{n}n!x_{0}} e^{-x^{2}/2 x^{2}_{0}} H_{n}\left(\frac{x}{x_{0} } \right) . Does a summoned creature play immediately after being summoned by a ready action? If we make a measurement of the particle's position and find it in a classically forbidden region, the measurement changes the state of the particle from what is was before the measurement and hence we cannot definitively say anything about it's total energy because it's no longer in an energy eigenstate. The probability is stationary, it does not change with time. MathJax reference. The best answers are voted up and rise to the top, Not the answer you're looking for? E.4). Can you explain this answer?, a detailed solution for What is the probability of finding the particle in classically forbidden region in ground state of simple harmonic oscillatorCorrect answer is '0.18'. H_{2}(y)=4y^{2} -2, H_{3}(y)=8y^{2}-12y. Mutually exclusive execution using std::atomic? /Border[0 0 1]/H/I/C[0 1 1] This is my understanding: Let's prepare a particle in an energy eigenstate with its total energy less than that of the barrier. 30 0 obj ,i V _"QQ xa0=0Zv-JH << If the correspondence principle is correct the quantum and classical probability of finding a particle in a particular position should approach each other for very high energies. The values of r for which V(r)= e 2 . >> Solution: The classically forbidden region are the values of r for which V(r) > E - it is classically forbidden because classically the kinetic energy would be negative in this ca 00:00:03.800 --> 00:00:06.060 . Note: Your message & contact information may be shared with the author of any specific Demonstration for which you give feedback. In fact, in the case of the ground state (i.e., the lowest energy symmetric state) it is possible to demonstrate that the probability of a measurement finding the particle outside the . we will approximate it by a rectangular barrier: The tunneling probability into the well was calculated above and found to be /Annots [ 6 0 R 7 0 R 8 0 R ] A few that pop in my mind right now are: Particles tunnel out of the nucleus of which they are bounded by a potential. We have step-by-step solutions for your textbooks written by Bartleby experts! Go through the barrier . In general, quantum mechanics is relevant when the de Broglie wavelength of the principle in question (h/p) is greater than the characteristic Size of the system (d). 24 0 obj Ela State Test 2019 Answer Key, Using the numerical values, \int_{1}^{\infty } e^{-y^{2}}dy=0.1394, \int_{\sqrt{3} }^{\infty }y^{2}e^{-y^{2}}dy=0.0495, (4.299), \int_{\sqrt{5} }^{\infty }(4y^{2}-2)^{2} e^{-y^{2}}dy=0.6740, \int_{\sqrt{7} }^{\infty }(8y^{3}-12y)^{2}e^{-y^{2}}dy=3.6363, (4.300), \int_{\sqrt{9} }^{\infty }(16y^{4}-48y^{2}+12)^{2}e^{-y^{2}}dy=26.86, (4.301), P_{0}=0.1573, P_{1}=0.1116, P_{2}=0.095 069, (4.302), P_{3}=0.085 48, P_{4}=0.078 93. Track your progress, build streaks, highlight & save important lessons and more! If so, how close was it? Can you explain this answer? . This is . 19 0 obj We have so far treated with the propagation factor across a classically allowed region, finding that whether the particle is moving to the left or the right, this factor is given by where a is the length of the region and k is the constant wave vector across the region. Free particle ("wavepacket") colliding with a potential barrier . Stahlhofen and Gnter Nimtz developed a mathematical approach and interpretation of the nature of evanescent modes as virtual particles, which confirms the theory of the Hartmann effect (transit times through the barrier being independent of the width of the barrier). /MediaBox [0 0 612 792] defined & explained in the simplest way possible. Description . Is it just hard experimentally or is it physically impossible? The turning points are thus given by En - V = 0. /Type /Annot /Type /Annot Get Instant Access to 1000+ FREE Docs, Videos & Tests, Select a course to view your unattempted tests. interaction that occurs entirely within a forbidden region. Turning point is twice off radius be four one s state The probability that electron is it classical forward A region is probability p are greater than to wait Toby equal toe. Why Do Dispensaries Scan Id Nevada, (4.172), \psi _{n}(x)=1/\sqrt{\sqrt{\pi }2^{n}n!x_{0} } e^{-x^{2} /2x^{2}_{0}}H_{n}(x/x_{0}), where x_{0} is given by x_{0}=\sqrt{\hbar /(m\omega )}. "After the incident", I started to be more careful not to trip over things. The same applies to quantum tunneling. Mississippi State President's List Spring 2021, You'll get a detailed solution from a subject matter expert that helps you learn core concepts. Why does Mister Mxyzptlk need to have a weakness in the comics? Consider the square barrier shown above. % Replacing broken pins/legs on a DIP IC package. Show that for a simple harmonic oscillator in the ground state the probability for finding the particle in the classical forbidden region is approximately 16% . The classical turning points are defined by E_{n} =V(x_{n} ) or by \hbar \omega (n+\frac{1}{2} )=\frac{1}{2}m\omega ^{2} x^{2}_{n}; that is, x_{n}=\pm \sqrt{\hbar /(m \omega )} \sqrt{2n+1}. Question about interpreting probabilities in QM, Hawking Radiation from the WKB Approximation. Particle in a box: Finding <T> of an electron given a wave function. #k3 b[5Uve. hb \(0Ik8>k!9h 2K-y!wc' (Z[0ma7m#GPB0F62:b There are numerous applications of quantum tunnelling. In fact, in the case of the ground state (i.e., the lowest energy symmetric state) it is possible to demonstrate that the probability of a measurement finding the particle outside the . The answer is unfortunately no. where S (x) is the amplitude of waves at x that originated from the source S. This then is the probability amplitude of observing a particle at x given that it originated from the source S , i. by the Born interpretation Eq. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. ~ a : Since the energy of the ground state is known, this argument can be simplified. Here you can find the meaning of What is the probability of finding the particle in classically forbidden region in ground state of simple harmonic oscillatorCorrect answer is '0.18'. 1. At best is could be described as a virtual particle. 6 0 obj 23 0 obj E < V . A particle absolutely can be in the classically forbidden region. Take advantage of the WolframNotebookEmebedder for the recommended user experience. \int_{\sqrt{3} }^{\infty }y^{2}e^{-y^{2}}dy=0.0495. On the other hand, if I make a measurement of the particle's kinetic energy, I will always find it to be positive (right?) One popular quantum-mechanics textbook [3] reads: "The probability of being found in classically forbidden regions decreases quickly with increasing , and vanishes entirely as approaches innity, as we would expect from the correspondence principle.". >> Forbidden Region. 7 0 obj Wave vs. We know that for hydrogen atom En = me 4 2(4pe0)2h2n2. (iv) Provide an argument to show that for the region is classically forbidden. Quantum tunneling through a barrier V E = T . You don't need to take the integral : you are at a situation where $a=x$, $b=x+dx$. Well, let's say it's going to first move this way, then it's going to reach some point where the potential causes of bring enough force to pull the particle back towards the green part, the green dot and then its momentum is going to bring it past the green dot into the up towards the left until the force is until the restoring force drags the .