Physics Oscillations

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

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

Range R = $\frac{u^{2}sin2\theta }{g}$

For 42° and 48° Range will be same

$H_{max}\alpha$  maximum q

So maximum height will be for 48°

Maximum energy = 10 J

$\frac{1}{2}K{x}^2=10$              

K = 5

Given T_pendulum = T_spring

$2\pi \sqrt[]{\frac{\mathcal{l}}{g}}=2\pi \sqrt[]{\frac{m}{K}}$             

$\sqrt[]{\frac{4}{g}}=\sqrt[]{\frac{5}{5}}$           

g = 4m/s²

Time period T = 2π $\sqrt[]{\frac{l}{g_{eff}}}$

$\Rightarrow T_i=T=2\pi \sqrt[]{\frac{l}{g}}$    

$\Rightarrow T_f=T\text{'}=2\pi \sqrt[]{\frac{4l/3}{g\left(1-\frac{\rho }{\sigma }\right)}}=2\pi \sqrt[]{\frac{16l}{9g}}$

$\Rightarrow T\text{'}=\frac{4}{3}T$

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

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

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

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

Velocity of bob after collision,

v_{1 $=\sqrt[]{5gR}=\sqrt[]{5*10*0.5}=5m/s$}

From Conservation of Momentum,

$\frac{10}{1000}*v=\frac{-10}{1000}*100+1*5$  

->v = 400 m/s