Differential equation for an oscillation of a spring using Hooke’s and Newton’s second law

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Differential equation for an oscillation of a spring using Hooke&#8217;s and Newton&#8217;s second law

Prince Budha · Accepted Answer

Consider a mass less spring having spring constant 'k', one end of which is connected to a mass 'm' and other is connected to a fixed support. The mass 'm' is free to move on a friction less horizontal surface. The equilibrium position is shown in 1(a) as relaxed and no force is acting on it. If a force is applied on the mass to stretch and then released the motion of the mass is simple harmonic. Hooke's law states that the extension or compression x of an elastic body is proportional to the applied force F. F ∝ x F = -kx ............. (1) Here, -ve sign shows that F and x are in opposite direction where k is constant of proportionality called force constant or spring constant. The value of k characterizes the strength of the spring. A spring with a large k is stronger or stiffer than with a small k. By Newton's 2nd law, (F = ma = m frac{dx^2}{dt^2}) ........(2) Comparing equations (1) and (2) (Rightarrow m frac{d^{2}x}{dt^2} = -kx) (Rightarrow m frac{d^{2}x}{dt^2} + kx = 0) (Rightarrow frac{d^{2}x}{dt^2} + frac{k}{m}x = 0) i.e (frac{d^{2}x}{dt^2}) + ω2x = 0 ......... (1) This is the equation of SHM, where, ω2 = (frac{k}{m}) ω = (sqrt{frac{k}{m}}) (T = frac{2 pi}{ω} = frac{2pi sqrt{m}}{k}) The solution of equation (1) is given by, x = xmsin(ωt + Ø) Velocity We have, x = xmsin(ωt + Ø) velocity, v = (frac{dx}{dt}) = ωxmcos(ωt + Ø) => vmax = xmω (v = omega x_{m}sqrt{1-sin^{2}(omega t - Theta)}) (= omega x_{m} sqrt{1 - frac{x^2}{x_m^2}}) (= omega x_{m} sqrt{frac{x_m^2 - x^2}{x_m^2}}) (= omega sqrt{A^2 - x^2})  (where xm = A ) Acceleration We have, v = ωxmcos(ωt + Ø) Acceleration, a = (frac{dv}{dt}) = -ω2xmsin(ωt + Ø) => amax = -ω2xm ∴ a = -ω2x where -ve sign shows 'a' is directed towards mean position. Time Period The time taken by particle to complete a cycle of oscillation is called time period 'T'. This means the displacement 'x' returns to its initial value after one complete oscillation i.e in time period T and the sine function repeat in angle 2π. Therefore, x = xmsinω(t + T) = xmsin(ωt + 2π). Frequency The number of complete oscillation made by an oscillation particle in one second is called Frequency. It is denoted by, f = 1/T Here, T is time required for one complete oscillation is called time period. The angular speed ω is related to the frequency of rotation f rather simply. By definition, ω is the change in angular displacement per unit time. ω = θ/t For t = T (time period),  θ = 2π Therefore, ω = (frac{2pi}{T}) = 2πf.

Set up a differential equation for an oscillation of a spring using Hooke’s and Newton’s second law. Find the general solution of this equation and hence the expressions for the period, velocity, and acceleration of oscillation.

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Set up a differential equation for an oscillation of a spring using Hooke’s and Newton’s second law. Find the general solution of this equation and hence the expressions for the period, velocity, and acceleration of oscillation.

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