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A) What is a homogeneous equation? B) Determine the values of alpha and beta for the relation T = Ck^alpha m^beta where T is period, k is spring constant, and m is mass.

Physics

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a) An equation is homogeneous if the dimensions of all terms on both sides of the equation are identical. This means that the units on the left side of the equation must match the units on the right side, ensuring the equation is physically consistent.

b) Determine the values of α\alpha and β\beta.

  1. Write down the dimensions of each physical quantity in the given relation T=CkαmβT = Ck^\alpha m^\beta.

    • The dimension of period TT is time: [T][T].
    • The constant CC is dimensionless: [1][1].
    • The dimension of mass mm is: [M][M].
    • The dimension of spring constant kk: From Hooke's Law, F=kxF = kx, so k=Fxk = \frac{F}{x}. The dimension of force FF is [M][L][T]−2[M][L][T]^{-2}. The dimension of length xx is [L][L]. Therefore, the dimension of kk is [M][L][T]−2[L]=[M][T]−2\frac{[M][L][T]^{-2}}{[L]} = [M][T]^{-2}.
  2. Substitute the dimensions into the given equation.

    [T]=[1]⋅([M][T]−2)α⋅([M])β[T] = [1] \cdot ([M][T]^{-2})^\alpha \cdot ([M])^\beta [T]1=[M]α[T]−2α[M]β[T]^1 = [M]^\alpha [T]^{-2\alpha} [M]^\beta

    Combine the powers of the same fundamental dimensions:

    [M]0[L]0[T]1=[M]α+β[L]0[T]−2α[M]^0 [L]^0 [T]^1 = [M]^{\alpha+\beta} [L]^0 [T]^{-2\alpha}
  3. Equate the powers of the fundamental dimensions on both sides of the equation.

    For mass [M][M]:

    0=α+β(1)0 = \alpha + \beta \quad (1)

    For time [T][T]:

    1=−2α(2)1 = -2\alpha \quad (2)
  4. Solve the system of equations for α\alpha and β\beta.

    From equation (2):

    α=−12\alpha = -\frac{1}{2}

    Substitute the value of α\alpha into equation (1):

    0=−12+β0 = -\frac{1}{2} + \beta β=12\beta = \frac{1}{2}

    The values are α=−12\alpha = -\frac{1}{2} and β=12\beta = \frac{1}{2}.
    The relation is T=Ck−1/2m1/2=CmkT = Ck^{-1/2}m^{1/2} = C\sqrt{\frac{m}{k}}.

    The values of α\alpha and β\beta are α=−12,β=12\boxed{\alpha = -\frac{1}{2}, \beta = \frac{1}{2}}.

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