ntal frequency of a stretched string is given by:

$$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$

where:

$$f$$ is the frequency of the standing wave

$$L$$ is the length of the string

$$T$$ is the tension in the string

$$\mu$$ is the linear mass density of the string

The tension $$T$$ in the wire due to a suspended mass $$m$$ is given by:

$$T = mg$$

where $$g$$ is the acceleration due to gravity.

In the first case, the wire resonates with five antinodes. For a wire with $$n$$ antinodes, the length of the wire $$L$$ is an integer multiple of half the wavelength $$\lambda / 2$$:

$$L = \frac{n \lambda}{2}$$

Thus, if there are five antinodes (n=5):

$$L = \frac{5\lambda}{2}$$

Similarly, for three antinodes (n=3):

$$L = \frac{3\lambda}{2}$$

Since the frequency is the same in both cases, we equate the frequencies using the formula for the fundamental frequency:

$$f_1 = \frac{5}{2L} \sqrt{\frac{T_1}{\mu}}$$

$$f_2 = \frac{3}{2L} \sqrt{\frac{T_2}{\mu}}$$

But since $$f$$ remains constant and the length $$L$$ of the wire remains constant between the two cases, we have:

$$\frac{5}{2L} \sqrt{\frac{T_1}{\mu}} = \frac{3}{2L} \sqrt{\frac{T_2}{\mu}}$$

Simplifying this, we get:

$$5 \sqrt{T_1} = 3 \sqrt{T_2}$$

Square both sides:

$$25 T_1 = 9 T_2$$

Recall that the tension $$T$$ is given by $$T = mg$$:

$$25 \cdot 9g = 9 \cdot Mg$$

Cancel out $$g$$:

$$225 = 9M$$

Solve for $$M$$:

$$M = 25 \mathrm{~kg}$$

Therefore, the value of '$$M$$' is $$\boxed{25 \ \mathrm{kg}}$$.

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