Question

#dont dare to give silly answers☬

Of course u silly answers will be reported..㋡㋡

#my frnd need it..​

Answer 1

Newton's Second Law gives,

$\longrightarrow\sf{m\,\dfrac{d^2x}{dt^2}=-\alpha\,x}$

$\longrightarrow\sf{\dfrac{d^2x}{dt^2}=-\dfrac{\alpha}{m}\,x\quad\quad\dots(1)}$

$\longrightarrow\sf{\dfrac{\alpha}{m}\,x+\dfrac{d^2x}{dt^2}=0\quad\quad\dots(2)}$

This differential equation shows that $\sf{x}$ multiplied with some constant is added to its second derivative wrt $\sf{t,}$ giving zero.

Hence let $\sf{x}$ be in the form,

$\longrightarrow\sf{x=A\cos\theta+B\sin\theta\quad\quad\dots(3)}$

for some constants $\sf{A}$ and $\sf{B.}$

Differentiating it wrt $\sf{t,}$

$\longrightarrow\sf{\dfrac{dx}{dt}=\dfrac{d}{dt}\,(A\cos\theta)+\dfrac{d}{dt}\,(B\sin\theta)}$

$\longrightarrow\sf{\dfrac{dx}{dt}=-A\sin\theta\cdot\dfrac{d\theta}{dt}+B\cos\theta\cdot\dfrac{d\theta}{dt}}$

Again differentiating wrt $\sf{t,}$

$\longrightarrow\sf{\dfrac{d^2x}{dt^2}=-\dfrac{d}{dt}\,\left(A\sin\theta\cdot\dfrac{d\theta}{dt}\right)+\dfrac{d}{dt}\,\left(B\cos\theta\cdot\dfrac{d\theta}{dt}\right)}$

$\begin{aligned}\longrightarrow\sf{\dfrac{d^2x}{dt^2}}=&-\sf{A\left(\cos\theta\left(\dfrac{d\theta}{dt}\right)^2+\sin\theta\cdot\dfrac{d^2\theta}{dt^2}\right)}\\&+\sf{B\left(-\sin\theta\left(\dfrac{d\theta}{dt}\right)^2+\cos\theta\cdot\dfrac{d^2\theta}{dt^2}\right)}\end{aligned}$

$\begin{aligned}\longrightarrow\sf{\dfrac{d^2x}{dt^2}}=&-\sf{\left[A\cos\theta\left(\dfrac{d\theta}{dt}\right)^2+A\sin\theta\cdot\dfrac{d^2\theta}{dt^2}\right.}\\&+\sf{\left.B\sin\theta\left(\dfrac{d\theta}{dt}\right)^2-B\cos\theta\cdot\dfrac{d^2\theta}{dt^2}\right]}\end{aligned}$

$\longrightarrow\sf{\dfrac{d^2x}{dt^2}}=-\sf{\left[\big(A\cos\theta+B\sin\theta\big)\left(\dfrac{d\theta}{dt}\right)^2+\big(A\sin\theta-B\cos\theta\big)\,\dfrac{d^2\theta}{dt^2}\right]}$

From (3),

$\longrightarrow\sf{\dfrac{d^2x}{dt^2}}=-\sf{x\left(\dfrac{d\theta}{dt}\right)^2-\big(A\sin\theta-B\cos\theta\big)\,\dfrac{d^2\theta}{dt^2}}$

From (1),

$\longrightarrow\sf{\dfrac{\alpha}{m}\,x=\left(\dfrac{d\theta}{dt}\right)^2x+\big(A\sin\theta-B\cos\theta\big)\,\dfrac{d^2\theta}{dt^2}}$

Equating corresponding coefficients, we get,

$\longrightarrow\sf{\dfrac{\alpha}{m}=\left(\dfrac{d\theta}{dt}\right)^2}$

Since $\sf{\omega^2=\dfrac{\alpha}{m},}$

$\longrightarrow\sf{\omega^2=\left(\dfrac{d\theta}{dt}\right)^2}$

$\longrightarrow\sf{\dfrac{d\theta}{dt}=\omega}$

$\longrightarrow\sf{d\theta=\omega\,dt}$

$\longrightarrow\sf{\theta=\omega\,t}$

Hence (3) becomes,

$\longrightarrow\underline{\underline{\sf{x(t)=A\cos(\omega\,t)+B\sin(\omega\,t)}}}$

From (2),

$\longrightarrow\sf{\alpha xv+mv\,\dfrac{d^2x}{dt^2}=0}$

$\longrightarrow\sf{\dfrac{1}{2}\alpha\cdot2xv+\dfrac{1}{2}\,m\cdot2v\cdot\dfrac{d^2x}{dt^2}=0}$

$\longrightarrow\sf{\dfrac{1}{2}\alpha\cdot\dfrac{d(x^2)}{dx}\cdot\dfrac{dx}{dt}+\dfrac{1}{2}\,m\cdot\dfrac{d\left[\left(\dfrac{dx}{dt}\right)^2\right]}{d\left(\dfrac{dx}{dt}\right)}\cdot\dfrac{d\left(\dfrac{dx}{dt}\right)}{dt}=0}$

$\longrightarrow\sf{\dfrac{1}{2}\alpha\cdot\dfrac{d(x^2)}{dt}+\dfrac{1}{2}\,m\cdot\dfrac{d\left[\left(\dfrac{dx}{dt}\right)^2\right]}{dt}=0}$

$\longrightarrow\sf{\dfrac{1}{2}\,m\cdot d\left[\left(\dfrac{dx}{dt}\right)^2\right]+\dfrac{1}{2}\alpha\cdot d(x^2)=0}$

On integrating, we get,

$\longrightarrow\underline{\underline{\sf{\dfrac{1}{2}\,m\left(\dfrac{dx}{dt}\right)^2+\dfrac{1}{2}\alpha\,x^2=E}}}$

for some constant $\sf{E.}$

The quantity $\sf{\dfrac{1}{2}\,\alpha\,x^2}$ represents the potential energy due to elasticity of the spring at the extension $\sf{x.}$