Question

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

Answer 1

Answer:

Given:

Potential energy function of a particle of mass 100 g along x axis is :

U = 5x(x - 4)

To find:

Period of oscillation of the particle

Concept:

Since Potential energy is function is provided , the force function can be found out by following differential :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \red{ \bold{F = - \dfrac{dU}{dx}}}}}$

Calculation:

∴ U = 5x(x - 4)

=> U = 5x² - 20x

Now , force is :

$\sf{F = - \dfrac{dU}{dx}}$

$\sf{F = - \dfrac{d \{5 {x}^{2} - 20x \}}{dx}}$

$\sf{ \implies \: F = - \: (10x - 20)}$

$\sf{ \implies \: F = 20 - 10x}$

Considering only distance related functions , we can say that this motion is an SHM .

So , time period :

$\sf{t = 2\pi \sqrt{ \dfrac{m}{k} }}$

$\sf{ \implies \: t = 2\pi \sqrt{ \dfrac{0.1}{10} } }$

$\sf{ \implies \: t = 2\pi \sqrt{ {10}^{ - 2} } }$

$\sf{\implies \: t = 2\pi(0.1)}$

$\sf{ \implies \: t = 0.2\pi \: sec}$

Hence , final answer is:

$\boxed{ \huge{ \red{ \sf{\: t = 0.2\pi \: sec}}}}$

Answer 2

$\underline{ \boxed{ \bold{ \mathfrak{ \huge{ \purple{Answer}}}}}}$

Given :

The potential energy of a particle of mass 100 g moving alpng x-axis in function of distance given by...

U = 5x (x - 4)

To Find :

Time period of oscillation

Formula :

Relation between potential energy and force is given by....

$\: \: \: \: \: \: \: \: \: \: \: \dag \: \: \underline{ \boxed{ \bold{ \rm{ \pink{F = - \frac{dU}{dx} }}}}} \: \: \dag$

Time period in SHM is given by...

$\: \: \: \: \: \: \: \: \: \: \: \dag \: \: \underline{ \boxed{ \bold{ \rm{ \purple{T = 2\pi \sqrt{ \frac{m}{k}}}}}} } \: \: \dagger$

Calculation :

$\mapsto \rm \: U = 5x(x - 4) = 5 {x}^{2} - 20x \\ \\ \mapsto \rm \: F = - \frac{dU}{dx} = - \frac{d(5 {x}^{2} - 20x )}{dx} \\ \\ \therefore \rm \: F = - (10x - 20) = 20 - 10x \\ \\ \mapsto \rm \: it \: is \: clear \: that \: force \: depends \\ \rm \: on \: distance \: therefore \: this \: is \: SHM \\ \\ \mapsto \rm \: force \: in \: shm \: is \: given \: by \: \blue{F = - kx} \\ \\ \mapsto \rm \: so \: \green{k = 10} \\ \\ \mapsto \rm \: T = 2\pi \sqrt{ \frac{m}{k} } = 2\pi \sqrt{ \frac{0.1}{10} } \\ \\ \therefore \rm \: \underline{ \boxed{ \bold{ \orange{ \rm{T = 0.2\pi = 0.628 \: s}}}}} \: \red{ \star}$