Question

The heaviest and lightest strings on a certain violin have linear densities of3.0and0.29 g/m.What is the ratio of the diameter of the heaviest string to that of the lightest string, assuming that the strings are of the same material?

Short answer

The ratio of the diameter of the heaviest to that of the lightest string is 3.22

Step-by-step solution

The given data

i) The linear density of the heaviest string, ${\mathrm{\mu }}_1=3.0\mathrm{g}/\mathrm{m}$

ii) The linear density of the lightest string, ${\mathrm{\mu }}_2=0.29\mathrm{g}/\mathrm{m}$

Understanding the concept of wave equation

The mass of the unit length of the wire is known as linear density. By writing the mass as the product of density and volume, in the formula for linear density. Finally, by taking the ratio of the linear densities, we can find the ratio of the diameters.

The volume of a cylinder of heightI ,

$V={\mathrm{\pi d}}^2\frac{\mathrm{I}}{4}$ (i)

The mass of a body in terms of density,

role="math" $m=\rho V$ (ii)

Here, d is the diameter, I is the length of the cylinder, p is the density of the wire and V is the volume.

Calculation for the ratio of the diameters

Let, the strings are long narrow cylinders, one of diameterand other of diameter$d_2$.

So that, the mass per unit length that is the linear density of a body is given as:

role="math" localid="1660978010280" $$\begin{gathered} \mathrm{\mu }=\frac{\mathrm{m}}{\mathrm{I}} \\ =\frac{{\displaystyle \frac{{\mathrm{p\pi d}}^2\mathrm{I}}{4}}}{\mathrm{I}}\left(\mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{m}\mathrm{using}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\right) \\ =\frac{{\mathrm{\rho \pi d}}^2}{4}.......................\left(\mathrm{a}\right) \end{gathered}$$

Now, using equation (a), their ratio of linear densities is given by:

$$\begin{gathered} \frac{{\mu }_1}{{\mu }_2}=\frac{{\displaystyle \frac{p{\mathrm{\pi d}}_1^2}{4}}}{\frac{p{\mathrm{\pi d}}_2^2}{4}} \\ \frac{{\mu }_1}{{\mu }_2}=\frac{{\displaystyle d_1^2}}{d_2^2} \end{gathered}$$

Therefore, the ratio of the diameters is given as:

$$\begin{gathered} \frac{d_1}{d_2}=\sqrt{\frac{{\mu }_1}{{\mu }_2}} \\ =\sqrt{\frac{3.0}{0.29}} \\ =3.22 \end{gathered}$$

Hence, the value of the ratio is 3.22