Question

A particle of mass ‘m’ performs oscillatory motion of amplitude A on the x –axis, about the origin
the restoring force on the particle is given by F = cx³ . Where C is constant and x is the
instantaneous displacement. The period of oscillation is

Answer 1

Answer:it is inversely proportional to A

Explanation:it is what i want to know

Answer 2

Hence the period of oscillation is T∝  1 / √ a

Explanation:

V(x)=K∣x∣^3

[K] = [V]  /  [x]^3  =  [ML^ 2  T^ −2  ] / [L^3 ]  

​ [K] = [ML^−1  T^ −2  ]

Now time period may depends on  

T ∝ (mass)^x  (amplitude)^y  (K)^ z

[M ^0  L^0 T]=[M]^x  [L]^y  [ML^−1 T  −2  ]^z

Equating powers we get,

−2z = 1, z =   −1 / 2

y − z = 0, y =   − 1  / 2

as

T ∝ (amplitude)^y

T ∝ (a)^ -1/2

T∝  1 / √ a

Hence the period of oscillation is T∝  1 / √ a

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