I think I have 1a. but I don't know the other parts. In class, we showed...
Equations of Simple Harmonic Motion (basic) PLEASE! show work and only answer if you know how to do it. People keeps giving me the wrong answer. Analyzing Newton's 2^nd Law for a mass spring system, we found a_x = -k/m X. Comparing this to the x-component of uniform circular motion, we found as a possible solution for the above equation: x = Acos(omega t) v_x = - omega Asin(omega t) a_x = - omega^2 Acos(omega t) with omega = square...
We study the vibrations in a diatomic molecule with the reduced mass m. Let x = R − Re, which is the bonding distance deviation from equilibrium distance. Hamiltonian operator consist of two parts: H = H(0) + H(1), where H(0) is the Hamiltonian operator to a harmonic oscillator with force constant k, and H(1) = λx3 (λ is a constant < 0). * Calculate the first order correction to the energy state v.
In class we solved the quantum harmonic oscillator problem for a diatomic molecule. As part of that solution we transformed coordinates from x, the oscillator displacement coordinate, to the unitless, y using the relationship where μ is the reduced mass of the diatomic molecule and k is the force constant. The solutions turned out to be: w(y)N,H, (y)e Where N is a normalization constant, H,(v) are the Hermit polynomials and v is the quantum number with values of v0,1,2,3,.. The...
i just nee help with #3 this is all the information that was given to mee Run your program for the sequence of steps N = 2, 4, 8, 16. Write the Matlab routine which computes the root-mean-square (RMS) error for each case as: r_N = e_N/Squareroot (u^ex_1)^2 + (u^ex_2) + middot middot middot + (u^ex_N)^2 + (u^ex_N)^2/N And the relative error as r_N = e_N/Squareroot (u^ex_1)^2 + (u^ex_2) + middot middot middot + (u^ex_N)^2 + (u^ex_N)^2/N Analyze the behavior...
problem 17. fully explain parts e and f. I have answer need explanation as to why we multiply for part e. amplitude times angular frequency to get vmax um 16. A 0.250-kg block attached to a light spring U 23. Thev At t frictionless, horizontal table. The oscillation amplitude is 0.125 m and the block moves at 3.00 m/s as it passes through equilibrium at 0. (a) Find the spring constant, k. (b) Calculate the total energy of the block-spring...
Learning Goal: To understand the application of the general harmonic equation to the kinematics of a spring oscillator. One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be x=0. The length of the relaxed spring is L. (Figure 1) The block is slowly...
help with 1-3 1) A simple harmonic oscillator consists of a 0.100 kg mass attached to a spring whose force constant is 10.0 N/m. The mass is displaced 3.00 cm and released from rest. Calculate (a) the natural frequency fo and period T (b) the total energy , and (c) the maximum speed 2) Allow the motion in problem 1 to take place in a resisting medium. After oscillating for 10 seconds, the maximum amplitude decreases to half the initial...
Problem 1: A particle of mass 5 10-8 kg is attached to a massless spring whose spring constant is 2 10-3N/m. Maximum displacement of the particle from the equilibrium position is 3 - 10-9m. The world is one- dimensional. (a) Find the (classical) amplitude of these oscillations. Yes, funny question, but I had to ask (b) Calculate the quantum of energy ho of this harmonic oscillator. he oea rthe prtic estimale the cormesponding uanitum umher. (d) If this harmonic oscillator...
There's a lot going on here and I am overwhelmed. I have no idea how to start this. -w'r, (25%) Problem 4: Any system for which the acceleration is linearly proportional to the position with a negative proportionality constant), or a = undergoes simple harmonic motion, a form of oscillatory motion. The mathematical solution to this is (t) = A coswt) where A is the amplitude and w=2nf = 2 is the angular frequency (fis the frequency in Hz and...
If the net force acting on a particle is a linear restoring force, the motion will be simple harmonic motion around the equilibrium position.The position as a function of time is x(t)=Acos(omega t + phi_0). The velocity as a function of time is v_x(t)=-omega A sin(omega t + phi_0). The maximum speedis v_{rm max} = omega A. The equations are given here in terms of x, but they can be written in terms of y, theta or some other parameter...