Question

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

• A. $x=\frac{1}{2}, y=\frac{1}{2}$
• B. $x=-\frac{1}{2}, y=-\frac{1}{2}$
• C. $x=\frac{1}{2}, y=-\frac{1}{2}$
• D. $x=-\frac{1}{2}, y=\frac{1}{2}$

Answer 1

The Correct Option is D

To solve the problem, we need to find the dimensions of the mass $m$, the spring constant $k$, and the frequency $f$. The relation given is:

f = am^x k^y

We are tasked with finding the values of x and y.

Step 1: Dimensional Analysis

Let's start by analyzing the dimensions for each parameter:

• Frequency, [f] = T^{-1} (where T is time).
• Mass, [m] = M.
• Spring constant, [k] = M T^{-2}.

Step 2: Express Dimensions

According to the given relation, we have:

[f] = [m^x] \cdot [k^y]

This means:

T^{-1} = (M)^x \cdot (M T^{-2})^y

Expanding and simplifying the right-hand side gives:

T^{-1} = M^x \cdot M^y \cdot T^{-2y}

Which further simplifies to:

T^{-1} = M^{x+y} \cdot T^{-2y}

Step 3: Equate the Dimensions

Comparing dimensions on both sides:

• For mass (M): x + y = 0.
• For time (T): -2y = -1, hence y = \frac{1}{2}.

Step 4: Solve Equations

From x + y = 0, if y = \frac{1}{2}, then:

x = -\frac{1}{2}

Conclusion

Substituting these values back, we find that the values of x and y should be:

Option: x = -\frac{1}{2}, y = \frac{1}{2} is correct.

This matches with the correct answer given in the options.