Physics, asked by nafeesahmad7714, 9 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?​
A. F=(x-1)
B. F=-(x-1)^2
C. F=-(x-1)^3
D. F=F=(x-1)^3

Answers

Answered by SmritiSami
7

(A) We have , F = (x-1)

•) Now, we have to find the motion of the particle on which this force is acting.

Now , for F = (x-1)

•) If x = 1 . F = 0

•) If x > 1 , F will be positive

•) If x < 1 , F will be negative

Now also equation of Simple Harmonic Motion is F = -kx

•) Hence , motion of the particle is Simple Hamonic Motion

(B) We have , F = -(x-1)^2

•) Now, we have to find the motion of the particle on which this force is acting.

Now , for F = -(x-1)^2

•) If x = 1 , F = 0

•) If x > 1 , F will be negative

•) If x < 1 , F will be negative

•) Hence , motion of the particle will be rectilinear

(C) We have , F = - (x-1)^3

•) Now, we have to find the motion of the particle on which this force is acting .

Now , for F = -(x-1)^3

•) If x = 1 , F = 0

•) If x > 1 , F will be negative

•) If x < 1 , F will be positive

•) Hence , motion of the particle will be Oscillatory around the point x = 1, but particle will be not be in Simple Harmonic Oscillation .

(D) We have , F = (x-1)^3

•) Now, we have to find the motion of the particle on which this force is acting .

Now , for F = (x-1)^3

•) If x = 1 , F = 0

•) If x > 1 , F will be positive

•) If x < 1 , F will be negative

•) Hence , motion of the particle will be Oscillatory around the point x = 1, but particle will be not be in Simple Harmonic Oscillation .

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