91Ó°ÊÓ

The length of the wire is L = 10.0 m

The mass of the wire isM = 100 g = 0.100 kg

The tension in the wire is$\mathrm{Ï„}=250\mathrm{N}$

By simplifying the equation for frequencies using the formulas of the possible wavelengths, their corresponding frequencies, the wave speed, and the linear mass density, we can find them, and the lowest frequencies for the standing wave.

Formula:

The resonant frequency of a wave,$f=\frac{v}{\mathrm{λ}}.....\left(1\right)$

The speed of the wave $v=\sqrt{\frac{\mathrm{τ}}{\mathrm{μ}}}......\left(2\right)$

The nth frequency of a wave, $f_n=\frac{nv}{2L}.......\left(3\right)$

The possible wavelengths are given by:

$\mathrm{λ}=\frac{2L}{n}$

where, L is the length of the wire andis an integer

The corresponding frequencies are given by

Using the linear density formula in equation (2), we get

$v=\sqrt{\frac{\mathrm{Ï„}L}{M}}(âˆ\mu \mathrm{μ}=\frac{M}{L})$

Thus, substituting the above value in equation (3) and the given values, we get

$$\begin{gathered} f_n=\frac{n}{2}\sqrt{\frac{\mathrm{τ}}{LM}} \\ {f}_n=\frac{n}{2}\sqrt{\frac{250}{10.0×0.100}} \\ =7.91\mathrm{n}\mathrm{Hz}..........\left(4\right) \end{gathered}$$

The lowest frequency is given by:

$f_1=7.91\mathrm{Hz}$

Hence, the lowest value of frequency is 7.91 Hz

The second lowest frequency using equation (4) is given by:

$$\begin{gathered} f_2=7.91×2 \\ =15.8\mathrm{Hz} \end{gathered}$$

Hence, the value of second lowest frequency is 15.8 Hz

Similarly, the third lowest frequency using equation (4) is given by:

$$\begin{gathered} f_3=7.91×3 \\ =23.7\mathrm{Hz} \end{gathered}$$

Hence, the value of third lowest frequency is 23.7 Hz