Simple Harmonic Motion

Using general equation of SHM :

$y=Asin\left(\omega t+{?}_0\right)=\frac{A}{2}$

$\omega t+{?}_0=\frac{\pi }{6},\frac{5\pi }{6}$

At t = 0

${?}_0=\frac{\pi }{6},\frac{5\pi }{6}$

Since it is moving in – x direction

$?=\frac{5\pi }{6}$

Resonance occurs when the frequency of an external periodic force matches the natural frequency of a system. From that, physicists know that resonance causes the amplitude of oscillations to increase significantly. This can be beneficial in devices, such as musical instruments, but dangerous in structures like bridges.

The phase in SHM tells us the position and direction of motion of the particle at a specific instant. It determines the state of oscillation and includes both displacement and time information.

The restoring force in SHM is the force that always acts towards the mean position and is directly proportional to the displacement from it. It follows F=? kx. Here, the negative sign indicates the force is in the opposite direction to the displacement.

x (t) = A sin $\omega$ $$ t + B cos $\omega t$ $$

$\Rightarrow x\left(t\right)=\sqrt[]{A^2+B^2}sin\left(\omega t+\delta \right)=\sqrt[]{A^2+B^2}cos\left(\omega t-\left(\frac{\pi }{2}-\delta \right)\right)$

At t = 0:

$x\left(0\right)=\sqrt[]{A^2+B^2}sin\left(\delta \right)$

$v\left(0\right)=\omega \sqrt[]{A^2+B^2}cos\left(\delta \right)$

$\Rightarrow {\left(x\left(0\right)\right)}^2+{\left(\frac{v\left(0\right)}{\omega }\right)}^2=A^2+B^2$

$tan?=tan\left(\frac{\pi }{2}-\delta \right)=cot\delta =\frac{v\left(0\right)}{x\left(0\right)\omega }$

$\Rightarrow ?=ta{n}^{-1}\left(\frac{v\left(0\right)}{x\left(0\right)\omega }\right)$

As we know that ${\upsilon }^2={\omega }^2\left(A^2-x^2\right)$ ${}^{}{}^{}{}^{}{}^{}$ for SHM, so

${\upsilon }_1^2={\omega }^2\left(A^2-x_1^2\right).......\left(i\right),and{\upsilon }_2^2={\omega }^2\left(A^2-x_2^2\right).........\left(ii\right)$

Subtracting equation (ii) from equation (i), we have

${\upsilon }_1^2-{\upsilon }_2^2={\omega }^2\left(x_2^2-x_1^2\right)\Rightarrow \omega =\frac{2\pi }{T}=\sqrt[]{\frac{{\upsilon }_1^2-{\upsilon }_2^2}{x_2^2-x_1^2}}\Rightarrow T=2\pi \sqrt[]{\frac{x_2^2-x_1^2}{{\upsilon }_1^2-{\upsilon }_2^2}}$

In SHM sum of kinetic and potential energy will be constant and average kinetic energy & average potential energy in one time will be remains same.

$x=Asin\omega t$   (Eq. of a particle executing SHM)

When KE = PE

$\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

$\frac{1}{2}m{\omega }^2\left(A^2-x^2\right)=\frac{1}{2}k{x}^2\left[?k=m{\omega }^2\right]$

A2 – x2 = x2

 $So,\frac{A}{\sqrt[]{2}}=Asin\omega t$

$\frac{1}{\sqrt[]{2}}=sin\omega t$

$\therefore t=\frac{T}{8}$

According to question, we can write

$\omega =\pi =\sqrt[]{\frac{g}{\mathcal{l}}}\Rightarrow \mathcal{l}=\frac{g}{{\pi }^2}=\frac{9.8}{{\left(3.14\right)}^2}=0.99395=99.4cm$