The frequency `f` of vibrations of a mass `m` suspended from a spring of spring constant `k` is given by `f = Cm^(x) k^(y)` , where `C` is a dimensionnless constant. The values of `x and y` are, respectively,

Options

A. `(1)/(2) , (1)/(2)`
B. ` -(1)/(2) , -(1)/(2)`
C. `(1)/(2) , - (1)/(2)`
D. `- (1)/(2) , (1)/(2)`

Explanation

To solve the problem, we need to determine the values of \( x \) and \( y \) in the equation \( f = C m^x k^y \), where \( f \) is the frequency of vibrations, \( m \) is the mass, and \( k \) is the spring constant. We will use dimensional analysis to find these values.

### Step-by-Step Solution:

1. **Identify the Dimensions**:
- The dimension of frequency \( f \) is \( [f] = M^0 L^0 T^{-1} \).
- The dimension of mass \( m \) is \( [m] = M^1 L^0 T^0 \).
- The spring constant \( k \) has the dimension \( [k] = M^1 L^0 T^{-2} \).

2. **Write the Equation with Dimensions**:
- The equation given is \( f = C m^x k^y \).
- Substituting the dimensions, we have:
\[ [f] = [C] [m]^x [k]^y \]
- Since \( C \) is dimensionless, we can ignore its dimensions:
\[ [f] = [m]^x [k]^y \]

3. **Substituting the Dimensions**:
- Substitute the dimensions of \( m \) and \( k \):
\[ M^0 L^0 T^{-1} = (M^1 L^0 T^0)^x (M^1 L^0 T^{-2})^y \]
- This simplifies to:
\[ M^0 L^0 T^{-1} = M^{x+y} L^{0} T^{-2y} \]

4. **Equating the Powers**:
- Now, we equate the powers of \( M \), \( L \), and \( T \) from both sides:
- For \( M \): \( 0 = x + y \) (Equation 1)
- For \( L \): \( 0 = 0 \) (This is trivially satisfied)
- For \( T \): \( -1 = -2y \) (Equation 2)

5. **Solving the Equations**:
- From Equation 2:
\[ -1 = -2y \implies y = \frac{1}{2} \]
- Substitute \( y \) into Equation 1:
\[ 0 = x + \frac{1}{2} \implies x = -\frac{1}{2} \]

6. **Final Values**:
- Therefore, the values of \( x \) and \( y \) are:
\[ x = -\frac{1}{2}, \quad y = \frac{1}{2} \]

### Conclusion:
The values of \( x \) and \( y \) are:
- \( x = -\frac{1}{2} \)
- \( y = \frac{1}{2} \)