Method 1: Dimensional Analysis

This method is suggested by the problem statement itself and is the quickest way to find the proportionality.

1. Identify the quantities and their dimensions:

   • Time period, T: [T]
   • Mass, m: [M]
   • Amplitude, A: [L]
   • The constant α from the potential energy V(x) = αx⁴.

2. Determine the dimensions of α: The potential energy V(x) has dimensions of energy, which is [M L² T⁻²]. The position x has dimensions of length, [L]. $[V] = [α] [x]^4$ $[M L^2 T^{-2}] = [α] [L]^4$ Solving for the dimensions of α: $[α] = {{[M L^2 T^{-2}]} \over {[L]^4}} = [M L^{-2} T^{-2}]$

3. Set up the proportionality relationship: Let the time period T be proportional to some combination of m, α, and A raised to powers a, b, and c respectively: $T \propto m^a α^b A^c$

4. Equate the dimensions on both sides: $[T^1] = [M]^a [M L^{-2} T^{-2}]^b [L]^c$ $[M^0 L^0 T^1] = [M]^a [M^b L^{-2b} T^{-2b}] [L]^c$ $[M^0 L^0 T^1] = [M^{a+b} L^{-2b+c} T^{-2b}]$

5. Solve the system of equations for the exponents: By comparing the powers of M, L, and T on both sides, we get three equations:

   • For M: a + b = 0 ...(i)
   • For L: -2b + c = 0 ...(ii)
   • For T: -2b = 1 ...(iii)

   From equation (iii): $b = -{1 \over 2}$ Substitute b into equation (i): $a + (-{1 \over 2}) = 0 \implies a = {1 \over 2}$ Substitute b into equation (ii): $-2(-{1 \over 2}) + c = 0 \implies 1 + c = 0 \implies c = -1$

6. Form the final proportionality: Substitute the values of a, b, and c back into the proportionality relation: $T \propto m^{1/2} α^{-1/2} A^{-1}$ $T \propto \sqrt{m} {1 \over \sqrt{\alpha}} {1 \over A}$ $T \propto {1 \over A}\sqrt{m \over \alpha}$

Method 2: Using Energy Conservation (Verification)

1. Total Energy: The total energy E is conserved. $E = K + V = (1/2)mv^2 + αx^4$.
2. At amplitude A: The velocity v=0, so the total energy is $E = αA^4$.
3. Velocity as a function of x: $(1/2)mv^2 + αx^4 = αA^4$. This gives $v = dx/dt = \sqrt{{2α \over m}(A^4 - x^4)}$.
4. Time Period Integral: The time period is four times the time taken to travel from x=0 to x=A. $T = 4 \int_0^A {dx \over v} = 4 \int_0^A {dx \over \sqrt{{2α \over m}(A^4 - x^4)}} = 4\sqrt{m \over 2α} \int_0^A {dx \over \sqrt{A^4 - x^4}}$
5. Change of Variables: Let x = Au, so dx = A du. The limits of integration change from 0 to 1. $T = 4\sqrt{m \over 2α} \int_0^1 {A du \over \sqrt{A^4 - (Au)^4}} = 4\sqrt{m \over 2α} \int_0^1 {A du \over A^2\sqrt{1 - u^4}} = \left(4\sqrt{m \over 2α} \int_0^1 {du \over \sqrt{1 - u^4}}\right) {1 \over A}$
6. The term in the parenthesis is a constant (the integral is a definite integral evaluating to a dimensionless number). Therefore, the time period T is proportional to (1/A)√(m/α).

Conclusion

Both methods show that the time period T is proportional to (1/A)√(m/α). This corresponds to option B.