The frequency of vibration f of a mass m suspended from a spring of spring constant K is given by a relation o

Question

The frequency of vibration f of a mass m suspended from a spring of spring constant K is given by a relation of this type f = C m^x K^y; where C is a dimensionless quantity. The value of x and y are:

Accepted Answer

We use the method of dimensional analysis to equate the dimensions of both sides of the equation $f = C m^x K^y$.

1.  **Identify Dimensions** :
    *   Frequency ($f$): Dimensions are $[T^{-1}]$.
    *   Mass ($m$): Dimensions are $[M]$.
    *   Spring Constant ($K$): Defined as Force per unit length. Dimensions are $\frac{[MLT^{-2}]}{[L]} = [MT^{-2}]$.
    *   Constant ($C$): Dimensionless $[M^0L^0T^0]$.

2.  **Apply Principle of Homogeneity** :
    Equating the dimensions on both sides:
    $$ [f] = [m]^x [K]^y $$
    $$ [M^0 L^0 T^{-1}] = [M]^x [MT^{-2}]^y $$
    $$ [M^0 L^0 T^{-1}] = [M]^{x+y} [T]^{-2y} $$

3.  **Solve for x and y:**
    *   Comparing dimensions of Time ($T$): $-1 = -2y \Rightarrow y = 1/2$.
    *   Comparing dimensions of Mass ($M$): $0 = x + y \Rightarrow x = -y \Rightarrow x = -1/2$.

Thus, the correct values are $x = -1/2$ and $y = 1/2$. This corresponds to the physical formula $f = \frac{1}{2\pi}\sqrt{\frac{K}{m}}$.

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