A mass of $2\,kg$ is attached to the spring of spring constant $50 \,Nm^{-1}$. The block is pulled to a distance of $5 \,cm $ from its equilibrium position at $x= 0$ on a horizontal frictionless surface from rest at $t = 0$. Write the expression for its displacement at anytime $t$.
A mass of $2\,kg$ is attached to the spring of spring constant $50 \,Nm^{-1}$. The block is pulled to a distance of $5 \,cm $ from its equilibrium position at $x= 0$ on a horizontal frictionless surface from rest at $t = 0$. Write the expression for its displacement at anytime $t$.
Spring-block system is shown as in figure and it executes SHM with amplitude of $5 \mathrm{~cm}$ from mean position.
Here, spring constant $k=50 \mathrm{~N} / \mathrm{m}$, amplitude $\mathrm{A}=5 \mathrm{~cm}$
mass attached $m=2 \mathrm{~kg}$
$\therefore$ Angular frequency $\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{2}}=5 \mathrm{rad} / \mathrm{s}$
Displacement of block at time $t$
$y(t)=$ Asin $(\omega t+\phi)$, where $\phi=$ initial phase
at time $t=0$
$y(0)=$ Asin $(\phi)$
A $=$ Asin $\phi$
$\therefore 1=\sin \phi$$\quad[\because$ at time $t=0, y=+$ A $]$
$\therefore \phi=\frac{\pi}{2}$ rad
$\therefore$ required equation,
$y(t)=\text { Asin }[\omega t+\phi]$
$\therefore y(t)=5 \sin \left(\omega t+\frac{\pi}{2}\right)$
$y(t)=5 \cos 5 t$ is a required equation where $t$ is in second and $y$ is in cm.
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