Question

The potential energy of a particle of mass 100g moving along x axis is given by U=5x(x-4) where x is in metres .The period of oscillation is

Answer 1

SOLUTION.

Function

U = 5x² - 20x

AS we know when a particle performing SHM the maximum value Potential energy is at Its maximum displacement i.e at its amplitude.

Differentiate the equation.

du = (10x - 20) dx

du/dx = 10x - 20

For U to be max du/dx has to be 0

0 = 10x - 20

AMPLITUDE = 2m

Value of potential energy at X= 2m

U = 5×2(2-4)

U = -10J

CHANGE IN POTENTIAL ENERGY FROM X=2 to X=2 is 10J

At X= 0 potential energy is zero hence all the potential energy at X= 2 get converted into kinetic energy

1/2mv² = ∆U

(1/2)(0.1)v² = 10

v² = 200

MAX VELOCITY = 10√2m/sec

And we know that in SHM max velocity is A×w

A×w= 10√2

2×(2π/T) = 10√2

T = (π√2)/5

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