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. `x = (1)/(2),y = (1)/(2)`
B. `x =- (1)/(2), y =- (1)/(2)`
C. `x = (1)/(2), y =- (1)/(2)`
D. `x =- (1)/(2), y =(1)/(2)`

Explanation

To solve the problem of finding the values of \( x \) and \( y \) in the equation \( f = C m^{x} k^{y} \), we can follow these steps:

### Step 1: Understand the relationship between frequency, mass, and spring constant
The frequency \( f \) of vibrations of a mass \( m \) suspended from a spring with spring constant \( k \) can be expressed in terms of angular frequency \( \omega \):
\[ \omega = \sqrt{\frac{k}{m}} \]
And since \( \omega = 2\pi f \), we can relate frequency to the spring constant and mass.

### Step 2: Express frequency in terms of mass and spring constant
From the angular frequency formula, we can derive:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
This can be rewritten as:
\[ f = \frac{1}{2\pi} k^{1/2} m^{-1/2} \]

### Step 3: Identify the powers of \( m \) and \( k \)
From the expression \( f = C m^{x} k^{y} \), we can compare it with our derived formula:
\[ f = C m^{-1/2} k^{1/2} \]
This gives us:
- The exponent of \( m \) is \( x = -\frac{1}{2} \)
- The exponent of \( k \) is \( y = \frac{1}{2} \)

### Step 4: Write down the final values of \( x \) and \( y \)
Thus, the values of \( x \) and \( y \) are:
\[ x = -\frac{1}{2}, \quad y = \frac{1}{2} \]

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