Simple harmonic motion is defined as motion in which the

$$a=-\omega^{2}x$$

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

Given the solution for the

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,

$$\ddot{x}=-a_{0}sin(\omega t)$$

The expression for the

$$E_{k}=\frac {1}{2}m v^{2}=\frac{1}{2}m \omega^{2}(x_{0}^{2}-x^{2})$$

The expression for the

$$E_{p}=\frac{1}{2}kx^{2}$$

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.

**"***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:- 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.

**Kinetic energy and Potential energy**for SHM: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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