4.1 SHM (Derivation of expression for frequency)

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

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

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.

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