Oscillations Test - 61

The equilibrium position is $$\displaystyle \frac{du}{dx}=0$$
$$\displaystyle \frac{-2a}{x_0^3}+\frac{b}{x_0^2}=0$$
or $$\displaystyle x_0=\frac{2a}{b}$$
$$\displaystyle u(x)=\frac{a}{x_0^2}-\frac{b}{x_0}+\left| (x-x_0)\frac{dU}{dx}\right |_{x=x_0}+\left |\frac{1}{2}(x-x_0)^2\frac{d^2u}{dx^2}\right|_{x=x_0}$$
$$\displaystyle \left|\frac{d^2u}{dx^2}\right|_{x=x_0}=\frac{6a}{x_0^4}-\frac{2b}{x_0^3}=\frac{6a}{\left(\displaystyle \frac{2a}{b}\right)^4}-\frac{2b}{\left(\displaystyle \frac{2a}{b}\right)^3}=\frac{b^4}{8a^3}$$
Thus $$\displaystyle u(x)=U(x_0)+\frac{1}{2}\left (\frac{b^4}{8a^3} \right) y^2$$
comparing with $$\displaystyle \frac{1}{2}m\omega^2 y^2=\frac{1}{2}\frac{b^4}{8a^3} y^2$$
$$\displaystyle \omega = \sqrt{\frac{b^4}{8a^3m}}$$
or $$\displaystyle T=2\pi\sqrt{\frac{8a^3m}{b^4}}$$

$$\omega ^{ 2 }=\dfrac { k }{ m } \\ \Rightarrow { \left( 2\pi f \right)  }^{ 2 }=\dfrac { k }{ m } \\ \Rightarrow k=4\pi ^{ 2 }f^{ 2 }m$$
and
$$F=kx_{ 0 }=4\pi ^{ 2 }f^{ 2 }mx_{ 0 }\\ \Rightarrow F=4\times \pi ^{ 2 }\times 10^{ 28 }\times 1.68\times 10^{ -27 }\times 10^{ -11 }N\\ \Rightarrow F=6.63\times 10^{ -9 }N$$

$$F=-amr\\ \Rightarrow F_{ x }\hat { i } +F_{ y }\hat { j } =-am\left( x\hat { i } +y\hat { j }  \right) \\ \Rightarrow m\dfrac { d^{ 2 }x }{ dt^{ 2 } } \hat { i } +m\dfrac { d^{ 2 }y }{ dt^{ 2 } } \hat { j } =-am\left( x\hat { i } +y\hat { j }  \right)$$

$$ \dfrac { d^{ 2 }x }{ dt^{ 2 } } =-ax$$ and $$\dfrac { d^{ 2 }y }{ dt^{ 2 } } =-ay\quad $$ ...............................................................$$(i)$$

Thus $$x=x_o\sin{(\omega t+\theta)}$$ and $$\omega^2=a$$ .........................................................................................$$(ii)$$
$$\dfrac{dx}{dt}=x_o\omega\cos{(\omega t+\theta)}$$
since $$v_x=0$$ when $$t=0$$
$$\Rightarrow x_o\omega\cos{(\theta)}=0$$
$$\Rightarrow \theta=\dfrac{\pi}{2}$$
$$\Rightarrow x=x_o \sin\left( \omega t+\dfrac { \pi  }{ 2 }  \right) $$
$$x=x_o \cos\left( \omega t \right) $$
$$\Rightarrow x=r_o \cos\left( \omega t \right)$$ ...............................................$$(iii)$$ $$\because x(t=0)=r_o $$

Now from $$(i)$$, we have, 
$$y=y_0\sin{(\omega t+\beta)}$$
$$y=y_0\sin{(\omega t)}                        \because y(t=0)=0$$
$$\dfrac{dy}{dt}=y_0\omega\cos {(\omega t)}$$
since $$v_y=v_o$$ when $$t=0$$
$$\Rightarrow v_o=y_0\omega$$
$$\Rightarrow y=\dfrac{v_o}{\omega}\sin{(\omega t)}$$ .....$$(iv)$$

from $$(iii)$$ and $$(iv)$$ we have,
$$\left(\dfrac{x}{r_0}\right)^2+\left(\dfrac{y}{v_0/\omega}\right)^2=1$$
$$\Rightarrow \left(\dfrac{x}{r_0}\right)^2+\omega^2\left(\frac{y}{v_0}\right)^2=1$$
$$\Rightarrow \left(\dfrac{x}{r_0}\right)^2+a\left(\dfrac{y}{v_0}\right)^2=1            \because \omega^2=a$$ from $$(ii)$$

the spring is compressed by y therefore the force on masses is $$F = Ky$$

maximum velocity $$\displaystyle v_{max}=\omega A\left ( \sqrt{\frac{K}{m}} \right )a$$
Change in momentum $$\Delta P=2mv_{max}=2m\left ( \sqrt{\frac{K}{m}} \right )a$$
Force exerted by the mass$$m$$ on wall $$W_2$$ is    $$\displaystyle F=\frac{\Delta P}{\Delta t}=\frac{\Delta P}{T/2}=\frac{2ma\sqrt{k/m}}{\pi \sqrt{m/k}}=\frac{2aK}{\pi }$$

$$\quad V(x) = k|x|^3$$
Since, $$\quad F = -\displaystyle\frac{dV(x)}{dx} = -3k|x|^2 \quad                             ...(i)$$
$$\quad x = a\sin(\omega t)$$
This equation always fits to the differential equation
$$\quad \displaystyle\frac{d^2x}{dt^2} = -\omega^2x$$ or $$m\displaystyle\frac{d^2x}{dt} = -m\omega^2x$$
$$\quad \Rightarrow F = - m\omega^2x\quad                         ...(ii)$$
Equations $$(i)$$ and $$(ii)$$ give
$$\quad -3k|x|^2 = -m\omega^2x$$
$$\quad \Rightarrow \omega = \sqrt{\displaystyle\frac{3kx}{m}} = \sqrt{\displaystyle\frac{3ka}{m}}[\sin(\omega t)]^{1/2}$$
$$\quad\Rightarrow \omega \propto \sqrt{a}$$
$$\quad \Rightarrow T \propto \displaystyle\frac{1}{\sqrt{a}}$$

$$\displaystyle \omega =\frac{2\pi }{T}=\frac{2\pi }{16}=\frac{\pi }{8}$$ rad/s
At t=0, particle crosses the mean position. Hence its velocity is maximum. So, velocity as a function of time can be written as
$$v=v_{max}\cos \omega t$$
or $$v=\omega A\cos \omega t$$
$$\therefore $$   $$\displaystyle 1=\left ( \frac{\pi }{8} \right )A\cos \left ( \frac{\pi }{8} \right )\left ( 2 \right )=\left ( \frac{\pi }{8\sqrt{2}} \right )A$$
$$\therefore $$   $$\displaystyle A=\frac{8\sqrt{2}}{\pi }$$ m

Maximum velocity $$V_{max}=\omega A$$

Comparing with equation $$\displaystyle x=A\sin (\omega t+\delta )$$ we see that the amplitude = 5 m
and time period = $$\displaystyle \frac{2\pi }{\omega }=\frac{2\pi }{\pi s^{-1}}=2s$$
The maximum speed = $$\displaystyle A\omega =5m\times \pi s^{-1}=5\pi \: m/s$$
The velocity at time $$\displaystyle t=\frac{dx}{dt}=A\omega \cos (\omega t+\delta )$$
At              t = 1 s
$$\displaystyle v=(5m)(\pi s^{-1})\cos \left ( \pi +\frac{\pi }{3} \right )=-\frac{5\pi }{2}\: m/s$$

From equation $$x=10+3sin(10t)$$