Question

The displacement of a particle of mass 100g from its equilibrium position is guven by

Answer 1

The potential energy of a particle of mass 100g moving along x axis is U=5x(x-4), where x is in metres. the period of oscillation is as follows:  

F=-dU/dx, given U=5x(x-4)

F= - 10(x-2)

F= - 10X

We know F= -Kx;

spring constant, K= 10.

Time period, T= 2*pi*(m/k)^1/2

Putting all the values , finally, T = pi/5 seconds.

Answer 2

The time period of oscillation is 0.628 sec.

Explanation:

Given that,

Mass of particle = 100 g

Suppose the potential energy is $U_{x}=5x(x-4)$.

Determine the time period of oscillation.

We need to calculate the force

Using formula of potential energy

$F=-\dfrac{dU}{dX}$

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

$F=-10x-20$....(I)

We know the restoring force,

$F=-kx$...(II)

The value of spring constant

On comparing equation (I) and (II)

$k=10$

We need to calculate the time period of oscillation

Using formula of time period

$T=2\pi\sqrt{\dfrac{m}{k}}$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{100\times10^{-3}}{10}}$

$T=0.628\ sec$

Hence, The time period of oscillation is 0.628 sec.

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Topic : Time period

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