Simple harmonic motion is defined as motion in which the "acceleration is proportional to and opposite in direction to, the displacement."
$$a=-\omega^{2}x$$
or in the differential equation form,
$$\ddot{x}+\omega^{2}x=0$$
The solution for the above differential equation is $x=x_{0}sin(\omega t)$. This solution simply gives us the diplacement about the equilibrium position for any given time after the oscillation is set. Here's an example solution where the amplitude is $2m$ and angular frequency equals to 1rad $s^{-1}$.
Given the solution for the displacement we can find the other quantities, namely the velocity, acceleration and energy.
The characteristic variation for the velocity for the above displacement function is as given below,
$$\dot{x}=v_{0}cos(\omega t)=\pm\omega \sqrt{x_{0}^{2}-x^{2}} $$
Similarly, the characteristic variation for the acceleration for the above displacement function is as given below,
$$\ddot{x}=-a_{0}sin(\omega t)$$
Deduction from the above plots:
The expression for the kinetic energy is,
$$E_{k}=\frac {1}{2}m v^{2}=\frac{1}{2}m \omega^{2}(x_{0}^{2}-x^{2})$$
The expression for the potential energy is (in case of spring).
$$E_{p}=\frac{1}{2}kx^{2}$$
where, $k$ is the spring constant. It's important to realize that when we add the expression for the two energy we get a constant ($(1/2)m\omega^{2}x_{0}^{2}$)and that is in agreement with the coservation of energy.
$$a=-\omega^{2}x$$
or in the differential equation form,
$$\ddot{x}+\omega^{2}x=0$$
The solution for the above differential equation is $x=x_{0}sin(\omega t)$. This solution simply gives us the diplacement about the equilibrium position for any given time after the oscillation is set. Here's an example solution where the amplitude is $2m$ and angular frequency equals to 1rad $s^{-1}$.
The characteristic variation for the velocity for the above displacement function is as given below,
$$\dot{x}=v_{0}cos(\omega t)=\pm\omega \sqrt{x_{0}^{2}-x^{2}} $$
$$\ddot{x}=-a_{0}sin(\omega t)$$
- If you compare the displacement and acceleration plot, we can conclude that for displacement away from the equlibrium position acceleration are in opposite direction.
- When the displacement is maximum, the velocity is zero and acceleration is maximum, and when the particle is in the equilibrium position, the velocity is maximum and acceleration is zero.
The expression for the kinetic energy is,
$$E_{k}=\frac {1}{2}m v^{2}=\frac{1}{2}m \omega^{2}(x_{0}^{2}-x^{2})$$
The expression for the potential energy is (in case of spring).
$$E_{p}=\frac{1}{2}kx^{2}$$
where, $k$ is the spring constant. It's important to realize that when we add the expression for the two energy we get a constant ($(1/2)m\omega^{2}x_{0}^{2}$)and that is in agreement with the coservation of energy.
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