The frequency of vibration f of a mass m suspended from a spring of spring constant k is given by a relation f = amˣkʸ, where a is a dimensionless constant. The values of x and y are

The correct option is x = -1/2 , y = 1/2.

To find the values of $x$ and $y$, we can use the method of dimensional analysis.

The given relation is:
$f=a{m}^x{k}^y$

Let us write down the dimensions of each quantity involved in the relation:
1. Frequency ($f$) is the number of oscillations per unit time, so its dimension is:
$\left[f\right]=\left[T^{-1}\right]=\left[M^0{L}^0{T}^{-1}\right]$
2. Mass ($m$) has the dimension:
$\left[m\right]=\left[M\right]=\left[M^1{L}^0{T}^0\right]$
3. Spring constant ($k$) is defined as force per unit length ($k=\frac{F}{d}$). The dimension of force is $\left[ML{T}^{-2}\right]$ and length is $\left[L\right]$. Thus, the dimension of $k$ is:
$\left[k\right]=\frac{\left[ML{T}^{-2}\right]}{\left[L\right]}=\left[M{T}^{-2}\right]=\left[M^1{L}^0{T}^{-2}\right]$
4. The constant $a$ is dimensionless, so it does not affect the dimensional equation.

Substituting these dimensions into the given relation:
$\left[M^0{L}^0{T}^{-1}\right]={\left[M\right]}^x{\left[M{T}^{-2}\right]}^y$
Simplifying the right-hand side:
$\left[M^0{L}^0{T}^{-1}\right]=\left[M^{x+y}{T}^{-2}\right]$

By equating the powers of corresponding quantities on both sides, we get:
For mass ($M$):
$x+y=0$ --- (Equation 1)
For time ($T$):
$-2y=-1$ --- (Equation 2)

From Equation 2, we find $y$:
$y=\frac{1}{2}$

Substituting the value of $y$ in Equation 1:
$x+\frac{1}{2}=0$
$x=-\frac{1}{2}$

Thus, the values are $x=-1/2$ and $y=1/2$.