Physics Oscillations

A simple pendulum shows simple harmonic motion only for small angular displacements. Otherwise, its motion is oscillatory but not purely SHM.

Tension in the string and gravitational force are common forces. These forces are resolved into tangential and radial components.

For small displacements,  sinθ≈θ (in radians), the small-angle approximation simplifies pendulum motion equations and makes it to approximate simple harmonic motion.

$x=5sin\left(\pi t+\frac{\pi }{3}\right)m$

Amplitude  $=5 m$

$=5 m$

$\omega =\pi =\frac{2\pi }{T}$

$T=\frac{2\pi }{\pi }=2 s$

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:

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

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

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

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

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

$\frac{T}{T\text{'}}=\frac{2\pi \sqrt[]{\frac{\mathcal{l}}{g}}}{2\pi \sqrt[]{\frac{\mathcal{l}}{\frac{16}{g}}}}$

$\Rightarrow \frac{T}{T\text{'}}=\frac{1}{\left(\frac{1}{4}\right)}=4$

$\Rightarrow T\text{'}=\frac{T}{4}$

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

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