Engineering Acoustics/Simple Oscillation

The Position Equation This section shows how to form the equation describing the position of a mass on a spring. For a simple oscillator consisting of a mass m attached to one end of a spring with a spring constant s, the restoring force, f, can be expressed by the equation where x is the displacement of the mass from its rest position. Substituting the expression for f into the linear momentum equation, where a is the acceleration of the mass, we can get or, Note that the frequency of oscillation is given by To solve the equation, we can assume The force equation then becomes Giving the equation Solving for This gives the equation of x to be Note that and that C1 and C2 are constants given by the initial conditions of the system If the position of the mass at t = 0 is denoted as x0, then and if the velocity of the mass at t = 0 is denoted as u0, then Solving the two boundary condition equations gives The position is then given by This equation can also be found by assuming that x is of the form And by applying the same initial conditions, This gives rise to the same position equation Back to Main page Alternate Position Equation Forms If A1 and A2 are of the form Then the position equation can be written By applying the initial conditions (x(0)=x0, u(0)=u0) it is found that If these two equations are squared and summed, then it is found that And if the difference of the same two equations is found, the result is that The position equation can also be written as the Real part of the imaginary position equation Due to euler's rule (ejφ = cosφ + jsinφ), x(t) is of the form Example 1.1 GIVEN: Two springs of stiffness, , and two bodies of mass, FIND: The natural frequencies of the systems sketched below 100px|left|Simple Oscillator-1.2.1.a 100px|left|Simple Oscillator-1.2.1.b 100px|left|Simple Oscillator-1.2.1.c 300px|Simple Oscillator-1.2.1.c-solution 100px|left|Simple Oscillator-1.2.1.d Back to Main page