Question

Obtain the differential equation of linear simple harmonic equation.

Answer 1

Simple harmonic motion is an oscillatory motion in which particle’s acceleration at any point position is directly proportional to the displacement from the mean position. All simple harmonic motions are oscillatory motions but all oscillatory motions are not simple harmonic motions.

Complete step by step answer:
Let us derive the differential equation of linear simple harmonic motion.
Consider the mass of the particle $\left( m \right)$ that executes the simple harmonic motion along a path. It has the mean position at $O$. Let ${V_0}$ be the speed of the particle at the position $P$.
There are two cases. First case:
$\Rightarrow t = 0$
When the particle at the position $P$ will be moving on the right side.
Second case:
$\Rightarrow t = t$
When the particle at the position $P$ will be moving on both the left and right side.
The given diagram is the reference of the details discussed.
With the velocity $\left( v \right)$ the restoring force is given by,
$\Rightarrow \overrightarrow F = - K\overrightarrow x$ ----- 1
Where $K$ is the constant, that is positive.
Another force can be derived as the multiplication of the mass and acceleration. That is,
$\Rightarrow \overrightarrow F = m\overrightarrow a$ ------2
Where $a$ is the acceleration and $m$ is the mass.
By equating the 1 and 2 equations we get,
$\Rightarrow m\overrightarrow a = - K\overrightarrow x$
Bring the mass term $m$ to the left-hand side’s denominator.
$\Rightarrow \overrightarrow a = - \dfrac{K}{m}\overrightarrow x$
Substitute the value for $\dfrac{K}{m}$ as ${\omega ^2}$. That is,
$\Rightarrow {\omega ^2} = \dfrac{K}{m}$
To remove the square on the right-hand side, take square root on the term of left-hand side.
$\Rightarrow \omega = \sqrt {\dfrac{K}{m}}$
But the term ${\omega ^2} = \dfrac{K}{m}$ is enough for solving the equation. substituting in the above equation we get,
$\Rightarrow \overrightarrow a = - {\omega ^2}\overrightarrow x$
Acceleration can be defined as the rate of change of velocity that changes with respect to time. It can be represented as velocity divided by time.
$\Rightarrow a = \dfrac{\overrightarrow v} {t}$
Differentiating the given equation we get,
$\Rightarrow \dfrac{{d\overrightarrow v} }{{dt}} = da$
Let us substitute the value of acceleration in the equation $\overrightarrow a = - {\omega ^2}\overrightarrow x$,
$\Rightarrow \dfrac{{d\overrightarrow v} }{{dt}} = - {\omega ^2}\overrightarrow x$
Equating the above equation to xero we get.
$\Rightarrow \dfrac{{d\overrightarrow v} }{{dt}} + {\omega ^2}\overrightarrow x = 0$
Actually $\dfrac{{d\overrightarrow v} }{{dt}}$ is the first order differential equation. The second order differential equation is $\dfrac{{{d^2}\overrightarrow v} }{{d{t^2}}}$. Writing the above equation by substituting the second order equation we get,
$\Rightarrow \dfrac{{{d^2}\overrightarrow v} }{{d{t^2}}} + {\omega ^2}\overrightarrow x = 0$
The rewritten equation is the differential equation of the linear simple harmonic motion.

Note: The minimum time at which the repetition of the particle takes place is known as the time period of the simple harmonic motion. Time period can also be defined as the shortest time taken to complete one oscillation. The time period is given by:
$\Rightarrow T = \dfrac{{2\pi }}{\omega }$.