In this section we'll derive the time period of the

$1.$ Lets consider the following

$2.$ Lets consider the Spring with spring constant $(k)$, and mass $(m)$ in the equilibrium position as follows,

Firstly, we must know the Hooke's law. i.e.,$F=-k\times x$. Where, $k$ is the spring constant for the given spring system and $x$ is the displacement from the equilibrium position.$-ve$ sign is the indication that spring's restoring force is opposite the stretch direction.

And Secondly, the Newton's second law, i.e. $F=m\times a$

We follow exactly the same process as above. We first equate the two force quantity,

\begin{align*}ma&=-kx\\ \Rightarrow \ddot{x}+\frac{k}{m}x &= 0\end{align*}

Therefore, \begin{align*}\omega=\sqrt{\frac{k}{m}}\Rightarrow f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\end{align*}

Now as the relation between the frequency and time period from here is given by,

$$f\times T=1$$

Thus, \begin{align*}T_{spring}={2\pi}\sqrt{\frac{m}{k}}\end{align*}

\begin{align*}T_{pendulum}={2\pi}\sqrt{\frac{l}{g}}\end{align*}

Here's a video example showing

**Simple Pendulum**and the**Spring**. Both of the system exhibits the Simple Harmonic Motion (SHM).$1.$ Lets consider the following

**Simple Pendulum**of length ($l$), mass ($m$) and angular displacement ($\theta$).
The restoring force ($F_{T}$) is in the direction perpendicular to the
string, $F_{T}=-mg sin(\theta)$

We also know from Newton’s second law of motion, $F=m\times
a$. As both of these equation represents the same quantity so, $-mg sin(\theta)=ma $.

We therefore have, $$a=-g sin(\theta)$$

Now, if we assume ‘$\theta$’ is very small, that is to say if
the angular displacement from the equilibrium position is small, then we have,
$sin\theta \approx \theta$. The expression for $'a'$ then simply becomes,$$a=-g\theta......(*)$$

We also know the relation from the radian measure of an angle, that $\theta =\frac{x}{l}$ where, $x$ is the arc subtended by the angle $\theta$ and $l$ is the length of the pendulum. So, $(*)$ becomes,

\begin{align*} \ddot {x}+\frac{g}{l}x&=0 \end{align*}

Now, comparing this expression with the expression in the previous notes,i.e.,

$$\ddot{x}+\omega^{2}x=0$$

Therefore, \begin{align*}\omega=\sqrt{\frac{g}{l}}\Rightarrow f=\frac{1}{2\pi}\sqrt{\frac{g}{l}}\end{align*}

$2.$ Lets consider the Spring with spring constant $(k)$, and mass $(m)$ in the equilibrium position as follows,

And Secondly, the Newton's second law, i.e. $F=m\times a$

We follow exactly the same process as above. We first equate the two force quantity,

\begin{align*}ma&=-kx\\ \Rightarrow \ddot{x}+\frac{k}{m}x &= 0\end{align*}

Again, comparing this expression with the expression in the previous notes,i.e.,

$$\ddot{x}+\omega^{2}x=0$$Therefore, \begin{align*}\omega=\sqrt{\frac{k}{m}}\Rightarrow f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\end{align*}

Now as the relation between the frequency and time period from here is given by,

$$f\times T=1$$

Thus, \begin{align*}T_{spring}={2\pi}\sqrt{\frac{m}{k}}\end{align*}

\begin{align*}T_{pendulum}={2\pi}\sqrt{\frac{l}{g}}\end{align*}

Here's a video example showing

**time period of pendulum dependence on length**, not shown in the video but we can also show similar behaviour for spring i.e.**time period of spring dependence on mass.**
## No comments:

Post a Comment