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# Tutor profile: Henry J.

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Henry J.
Tutor for three years, teacher for one year, studying Mechanical Engineering
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## Questions

### Subject:Calculus

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Question:

Find the maximum and minimum values for f(x,y) = X^2 + y^2, subject to the constraint function g(x,y) = 2x^2 + 3xy + 2y^2 = 7.

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Henry J.

### Subject:Physics

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Question:

An electron enters a magnetic field B = 12Ĵ Teslas with velocity v = 15î (m/s). The particle enters into circular motion from the magnetic force on the moving charge. Calculate the radius of the circular orbit of the electron.

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Henry J.

The magnetic force is causing the particle to enter circular motion, so these forces are equal to each other. F_b + F_c = 0 F_b = -F_c = qvBsin(theta) = -(mv^2)/r (notice that one of the "v" terms will cancel on both sides of the equation) q = -1.602e-19 (C) (This the charge of an electron) Abs(v) = 15 (m/s) (This is the magnitude of the velocity vector) Abs(B) = 12 (T) (This is the magnitude of the magnetic field vector) sin(theta) = 1 (because î and ĵ are orthogonal, so theta = 90 degrees and sin(90) = 1) m = 9.11e-31 (kg) (This is the mass of an electron) r is the radius of circular motion, what we're trying to find. Plugging these numbers in, we get (-1.602e-19)(12) = -((9.11e-31)(15))/r solving for r, r = 7.1083e-12 (m) = 7.1083 (pm)

### Subject:Differential Equations

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Question:

Consider a mass "m" hanging at rest on the end of a vertical spring of original length "l". The mass causes an elongation "L" of the spring in the downward (assume positive) direction. Assume that the mass is given an initial displacement x(t_0) = x_0 where x_0 > 0, and an initial velocity x'(t_0) = x'_0, where x'_0 > 0. There are four forces acting on the system: the spring force, the gravitational force, the damping force, and the driving force. Assume a massless spring. Let the damping force be proportional to the velocity (which is a relatively accurate approximation at low orders of magnitude). Find the general solution for position as a function of time for the given ordinary differential equation above. The particular solution to the system may be expressed as x_p, and constants do not need to be found.

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Henry J.

To start this problem, set the net force equal to 0, since the system is in equilibrium. Use Newton's Second Law: F_net = ma = mx'' Since the system is in equilibrium, the spring force (F_s) is equal to the gravitational force (F_g), but in the opposite direction. This is what allows the elongation, "L", to be measured. If the system is moving, F_s = k(L + x), where (L + x) is the total displacement of the system. Another force in the system is the damping force, F_d. If the damping force is proportional to the velocity of the mass, then F_d = -yv = -yx', where y is positive and is the damping coefficient. F_d is a negative expression because it is assumed that x is increasing initially (x'_0 > 0), so the damping force would act in opposition to the velocity vector, in the negative direction. The final force that needs to be accounted for is the driving force, or external force, F_e. This is the force that accounts for the movement of the system in simple harmonic motion. Now that all of the forces in the system have been accounted for, we can substitute them back into Newton's Second Law to get mx'' = F_s + F_g + F_d + F_e = -k(L + x) + mg - yx' + F_e Since F_s = -F_g, this expression will simplify to mx'' = -kx - yx' + F_e , or mx'' + yx' + kx = F_e Now we have arrived at a second order non-homogeneous ODE, and the solution to this ODE looks like x(t) = x_h + x_p, where x_h is the solution to the homogeneous version of the ODE, and x_p is the particular solution to the non-homogeneous version of the ODE. Since these two form a complementary set of solutions, their sum is needed. To find x_h, we must solve the homogeneous version of the ODE, which is expressed as mx'' + yx' + kx = 0 = x'' + (y/m)x' + (k/m)x If we assume solution of the form x = ce^(rt), differentiate twice, and substitute, we get (r^2)(ce^(rt)) + (y/m)(r)(ce^(rt)) + (k/m)(ce^(rt)) = 0 = (r^2) + (y/m)(r) + (k/m) = 0 This is a quadratic equation and can be solved with the quadratic formula. This will give roots r_1 = ((-y/m) - ((y/m)^(2) - 4(k/m))^(1/2))/2 and r_2 = ((-y/m) + ((y/m)^(2) - 4(k/m))^(1/2))/2 These roots will give us the functions for the part of the solution, x_h x_h = (C_1)e^((r_1)t) + (C_2)e^((r_2)t) x_h = (C_1)e^((((-y/m) - ((y/m)^(2) - 4(k/m))^(1/2))/2)t) + (C_2)e^(((-y/m) + ((y/m)^(2) - 4(k/m))^(1/2))/2)t) With explicitly given initial values for x(t_0) and x'(t_0), explicit values could be found for (C_1) and (C_2) respectively. Similarly, if the driving function F_e was given explicitly, x_p could be defined explicitly. As such, the position function for the mass on the spring is x(t) = x_h + x_p = (C_1)e^((((-y/m) - ((y/m)^(2) - 4(k/m))^(1/2))/2)t) + (C_2)e^(((-y/m) + ((y/m)^(2) - 4(k/m))^(1/2))/2)t) + x_p

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