Physics, asked by manju2478, 11 months ago

A particle is free to move on x-axis in which of the following case the particle will execute oscillation about X=1
(1)F= (x-1)
(2)F= -(x-1)^2
(3)F= -(x-1)^3
(4)F= (x-1)^3​

Answers

Answered by Anonymous
23

Answer: 3rd option. See the force us restoring in 3rd case. When x=2, force is negative i.e. towards x=1, when x=0, force is positive i.e. towards x= 1. So particle will oscilate about x=1

Explanation:

Answered by wwwsaiganesharisetty
0

Answer:

The potential is not quadratic so the motion is not sinusoidal or simple harmonic. The motion is periodic since the particle is bounded by the potential. The energy is constant and so the particle will move back and forth between a maximum negative position and the position x=x

0

.

The potential energy is continuous and hence K.E.=TE−U is continuous. Therefore, speed is continuous.

The derivative of the potential energy is not continuous at x = 0. Since, force is equal to the negative derivative of the potential energy, the force is not continuous at x=0. Therefore, acceleration (

m

F

) is not continuous at x=0.

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