How Do You Derive Equations for Position and Velocity in Simple Harmonic Motion?

[swatikiss]

I am confused by the following problem. Any help / hints would be greatly appreciated! I understand that the velocity equation is the derivative of the position function... i just don't understand how to derive these first / last equations. THANKS!

The initial position and initial velocity of an object moving in simple harmonic motion are xi, vi, and ai; the angular frequency of oscillation is (omega).

a) Show that the position and velocity of the object for all time can be written as:

x(t) = xi cos(omega * t) + (vi / (omega) )*sin (omega * t)

v(t) = - xi (omega)sin(omega * t) + vi*cos(omega * t)

b) If the amplitude of the motion is A, show that

v^2 - ax = vi^2 - aixi = (omega)^2A^2

 

[maverick280857]

Here's a hint. Consider the substitution of initial conditions in the solutions for displacement and velocity.

 

[wais]

a) To derive the position equation, we start with the general equation for simple harmonic motion: x(t) = A*cos(omega * t + phi), where A is the amplitude and phi is the phase constant. We can rewrite this equation as x(t) = A*cos(omega * t)*cos(phi) - A*sin(omega * t)*sin(phi). Since we know that at t=0, the position is xi, we can substitute this value in for x(t) and solve for cos(phi):

x(0) = A*cos(0)*cos(phi) - A*sin(0)*sin(phi) = xi

cos(phi) = xi/A

Similarly, we can find the value of sin(phi) by using the initial velocity, vi, at t=0:

v(0) = -A*omega*sin(0)*cos(phi) + A*omega*cos(0)*sin(phi) = vi

sin(phi) = vi/(A*omega)

Substituting these values back into the general equation, we get:

x(t) = A*cos(omega * t)*xi/A - A*sin(omega * t)*vi/(A*omega)
x(t) = xi*cos(omega * t) - (vi/omega)*sin(omega * t)

For the velocity equation, we can use the fact that the derivative of cos(omega * t) is -omega*sin(omega * t) and the derivative of sin(omega * t) is omega*cos(omega * t). So we have:

v(t) = -xi*omega*sin(omega * t) - (vi/omega)*omega*cos(omega * t)
v(t) = -xi*omega*sin(omega * t) + vi*cos(omega * t)

b) We can use the position equation to find the acceleration (since a = -omega^2*x) and substitute it into the equation for kinetic energy (KE = 1/2*m*v^2). We get:

KE = 1/2*m*(-xi*omega^2*sin(omega * t) + 2*vi*omega*cos(omega * t) - vi^2)
KE = 1/2*m*(xi*omega^2*sin(omega * t) + vi^2 - vi^2)
KE = 1/2*m*