Spring and SHM - 1

Method 1:

As we see, $\ddot x = g-\omega^2x$ where $\ddot x = a$ , the acceleration of the block.

So we get a second degree non-homogenous differential equation, $\ddot x + \omega^2x = g$

Taking auxiliary equation, $\ddot x+\omega^2x=0$ with general solution $x=e^{rt}$ , $\ddot x = r^2x$ . So, $(r^2+\omega^2)\cdot e^{rt} = 0\\\therefore r = i\omega\\\begin{aligned}x_c &= \Re \left\{e^{i\omega t}\right\}\\x_c &=\cos \omega t\end{aligned}$

Now taking particular solution, the D.Eq is of the form $A\ddot x + Bx = C$

C is of the form $c\cdot t^0$ . So, particular solution $x_p = c\cdot t^0$ . So, $\ddot x_p = 0$ .

Substituting in our D.Eq, $0+\omega^2x_p =\omega^2\cdot c = g\\\Rightarrow c = \dfrac g{\omega^2}\\x_p =\dfrac g{\omega^2}$

Thus, the solution for $x$ is $\begin{aligned}x &= C_1x_c+x_p\\&=C_1\cos\omega t +\dfrac g{\omega^2}\end{aligned}$ where $\omega = \sqrt{\dfrac k m}$ .

Taking initials conditions to get $C_1$ , $x_{t=0} = 0\\x = C_1cos\omega t+\dfrac g{\omega^2}\\\Rightarrow 0 = C_1 + \dfrac g{\omega^2}\\\Rightarrow C_1 = -\dfrac g{\omega^2}$

Hence, the solution is $x = \dfrac g{\omega^2}\left( 1-\cos\omega t\right)$

Thus $\displaystyle\int_0^3 x~dt = \int_0^3 \dfrac g{\omega^2}\left( 1-\cos\omega t\right)~dt = 9.714246925$

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Method 2:

We know that the spring oscillates harmonically.Also $\ddot x = g-\omega^2x$

At equilibrium position, $\ddot x = 0\\\Rightarrow x_{eq} = \dfrac g{\omega^2}$

So, writing equations of harmonic oscillations, $x_{eq}-x = a\cos(\omega t)$

Substituting initial conditions, at $t=0$ , $x=0$ . So, $a=x_{eq}=\dfrac g{\omega^2}$

So, the equation of motion becomes, $x = x_{eq}-x_{eq}\cos(\omega t)\\\boxed{x=\dfrac{mg}k\left(1-\cos \omega t\right)}$

Thus $\displaystyle\int_0^3 x~dt = \int_0^3 \dfrac g{\omega^2}\left( 1-\cos\omega t\right)~dt = 9.714246925$

May be I was lucky to get the answer , plz do correct me if I were wrong.

Using law of conservation of energy

$E=0.5 kx^2-mgx-0.5 mv^2$

$\large{\frac{dE}{dt}=0}$

We get $\large{kxv-mgv-mv \frac{dv}{dt}=0}$

$\large{\Rightarrow \frac{dv}{dt}=\frac{kx-mg}{m}}$

$\large{\Rightarrow \displaystyle \int^{3}_{0} dv=\frac{k}{m} \displaystyle \int^{3}_{0} x.dt- g \displaystyle \int^{3}_{0} dt}$