A particle is moving on $$x-axis$$ has potential energy $$U=2$$−$$20x+5x2$$ Joules along $$x-axis$$. The particle is released at x $$=$$ $$3$$. The maximum value of ' x.' will be: [x is in meters and U is in joules]

Question

A particle is moving on $$x-axis$$ has potential energy $$U=2$$−$$20x+5x2$$ Joules along $$x-axis$$. The particle is released at x $$=$$ $$3$$. The maximum value of ' x.' will be: [x is in meters and U is in joules]

Infinity Learn · Accepted Answer

$$U=2$$−$$20x+5x2F=$$−$$dUdx=20-10xAt$$ equilibrium position; F $$=$$ $$020$$ $$-$$ $$10x$$ $$=$$ $$0$$⇒x $$=2It$$ will oscillate about x $$=$$ $$2$$ m.Now m mx $$=$$ F $$=-10$$ (X-2).. ⇒x..$$+10(x-2)$$ $$=0$$ . The above equation represents an SHM whose equation is given by               $$x-2$$ $$=$$ A $$cos$$ $$($$ω$$t+$$φ$$)$$ At $$t=0$$ , $$x=3m$$ and at $$t=0$$ , $$v=$$ $$dxdt=0$$ Solving $$A=$$ $$1m$$ and φ$$=0$$ . Therefore $$x-2=1$$ $$cos$$ ωt .  Since maximum value of $$cos$$ ωt is $$+1$$,  we get $$xmax=(2+1$$ $$)$$ $$m=3m$$...

A particle is moving on x-axis has potential energy $U = 2 - 20 x + 5 x^{2}$ Joules along x-axis. The particle is released at x = 3. The maximum value of ' x.' will be: [x is in meters and U is in joules]

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

A particle is moving on x-axis has potential energy U=2−20x+5x2 Joules along x-axis. The particle is released at x = 3. The maximum value of ' x.' will be: [x is in meters and U is in joules]

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

A particle is moving on x-axis has potential energy $U = 2 - 20 x + 5 x^{2}$ Joules along x-axis. The particle is released at x = 3. The maximum value of ' x.' will be: [x is in meters and U is in joules]