Question

State the differential equation for the linear simple harmonic motion. Therefore reach the expression for acceleration, velocity, and displacement of a particle, which is undergoing a linear simple harmonic motion.

Answer 1

In any SHM, there will always be acting as a restoring force, which will try to bring the object back to the mean position. This force will result in an acceleration in the object. The double derivative of the displacement with respect to time is the acceleration of the body.

Complete step-by-step solution
In any SHM, there will always be acting as a restoring force which will try to bring the object back to the mean position. This force will result in an acceleration in the object.
Therefore
$F=-Kx$
Here $F$ be the restoring force, $x$ be the displacement of the object from the mean position, and $K$ is the force per unit displacement. The negative sign represents that the force will be opposite to the displacement. Only then can the body be brought back after the displacement.
According to the newton’s second law of motion, we can write that,
$\begin{aligned} & F=ma \\\ & \therefore a=\dfrac{F}{m} \\\ \end{aligned}$
Applying this in the equation can be written as,
$\begin{aligned} & F=ma \\\ & \therefore a=\dfrac{-K}{m}x \\\ \end{aligned}$
Let $\dfrac{K}{m}$ be the square of the angular frequency. That is,
$\begin{aligned} & F=ma \\\ & \dfrac{K}{m}\to {{\omega }^{2}} \\\ \end{aligned}$
Hence we can write that,
$\begin{aligned} & F=ma \\\ & a=-{{\omega }^{2}}x \\\ \end{aligned}$
Therefore the acceleration of the body has been found.
Therefore in calculus, the double derivative of the displacement with respect to time can be written as,
$a=\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$
Therefore the differential equation will be obtained as,
$\dfrac{{{d}^{2}}x}{d{{t}^{2}}}+{{\omega }^{2}}x=0$
We can also write the acceleration as the first derivative of the velocity as,
$a=\dfrac{dv}{dt}$
This can be written as,
$\dfrac{dv}{dt}=-{{\omega }^{2}}x$
The derivative can be written as,
$\dfrac{dv}{dt}=\dfrac{dv}{dx}\times \dfrac{dx}{dt}=v\dfrac{dv}{dx}$
That is we can write this as acceleration as,
$v\dfrac{dv}{dx}=-{{\omega }^{2}}x$
Rearranging this equation,
$vdv=-{{\omega }^{2}}xdx$
Let us integrate this equation as,
$\int{vdv}=\int{-{{\omega }^{2}}xdx}$
Performing the integration can be shown as,
$\dfrac{{{v}^{2}}}{2}=-\dfrac{{{\omega }^{2}}{{x}^{2}}}{2}+c$
In the above equation, when displacement is maximum, then,
$x=A$
And at extreme point,
$v=0$
Therefore,

& 0=-\dfrac{{{\omega }^{2}}{{A}^{2}}}{2}+c \\\ & \therefore c=\dfrac{{{\omega }^{2}}{{A}^{2}}}{2} \\\ \end{aligned}$$ Let us substitute this in the equation as, $$\begin{aligned} & \dfrac{{{v}^{2}}}{2}=\dfrac{-{{\omega }^{2}}{{A}^{2}}}{2}+c \\\ & \therefore c=\dfrac{{{\omega }^{2}}{{A}^{2}}}{2} \\\ \end{aligned}$$ Substituting this we will get, $$\begin{aligned} & \dfrac{{{v}^{2}}}{2}=\dfrac{-{{\omega }^{2}}{{x}^{2}}}{2}+\dfrac{{{\omega }^{2}}{{A}^{2}}}{2} \\\ & \therefore {{v}^{2}}={{\omega }^{2}}\left( {{A}^{2}}-{{x}^{2}} \right) \\\ \end{aligned}$$ Taking the square root on both the sides will give, $$v=\pm \omega \sqrt{\left( {{A}^{2}}-{{x}^{2}} \right)}$$ The $$\pm $$ represents that the object can move towards positive direction or negative direction. In order to get an expression for displacement, let us consider the magnitude of velocity. That is, $$\begin{aligned} & v=\omega \sqrt{\left( {{A}^{2}}-{{x}^{2}} \right)} \\\ & \Rightarrow \dfrac{dx}{dt}=\omega \sqrt{\left( {{A}^{2}}-{{x}^{2}} \right)} \\\ \end{aligned}$$ Rearranging this equation will give, $$\begin{aligned} & v=\omega \sqrt{\left( {{A}^{2}}-{{x}^{2}} \right)} \\\ & \Rightarrow \dfrac{dx}{\sqrt{\left( {{A}^{2}}-{{x}^{2}} \right)}}=\omega dt \\\ \end{aligned}$$ Let us integrate on both the sides. $$\begin{aligned} & v=\omega \sqrt{\left( {{A}^{2}}-{{x}^{2}} \right)} \\\ & \int{\dfrac{dx}{\sqrt{\left( {{A}^{2}}-{{x}^{2}} \right)}}}=\int{\omega dt} \\\ \end{aligned}$$ That is we can write that, $${{\sin }^{-1}}\left( \dfrac{x}{A} \right)=\omega t+\alpha $$ That is, $$x=A\sin \left( \omega t+\alpha \right)$$ Therefore the equation for displacement, velocity and acceleration is obtained. **Note:** Here $$\alpha $$ is the constant of integration which is the initial phase angle called as epoch. It is dependent on initial conditions. Therefore the epoch angle is the condition, which refers to the displacement of the body at the time being at zero.