In a laboratory experiment, one end of a horizontal string is tied

to a support while the other end passes over a frictionless pulley

and is tied to a 1.5 kg sphere. Students determine the frequencies

of standing waves on the horizontal segment of the string, then

they raise a beaker of water until the hanging 1.5 kg sphere is

completely submerged. The frequency of the fifth harmonic with

the sphere submerged exactly matches the frequency of the third

harmonic before the sphere was submerged. What is the diameter

of the sphere?

the sphere is a geometrical shape with a set of three-dimensional points with space lying the same distance

Here, tension and force are equal to the weight that the ball carries. for the second case, weight is subtracted from the buoyant force. the buoyant force has a sphere diameter which is D and the liquid with density is p.

Thus, $F_B=pgV=pg\frac{4}{3}\pi R^3=pg\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3=\frac{1}{6}{\mathrm{\pi pgD}}^3$

The frequency and the string are stretched by tension T

The linear density $\mu$and the anti mode number is m which is provided as

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

after the substitution, the denoted mass or m will be

$\frac{3\sqrt{Mg}}{2L\sqrt{\mu }}=\frac{5\sqrt{Mg-{\displaystyle \frac{1}{6}}{\mathrm{\pi pgD}}^3}}{2L\sqrt{\mu }}$

The simplification of the equation is

The square on both sides are

$9M=25(M-\frac{1}{6}{\mathrm{\pi pD}}^3)\Rightarrow \frac{1}{6}{\mathrm{\pi pD}}^3=\frac{16}{25}M$

The diameter can be expressed as,

role="math" localid="1649066728962" $D=\sqrt[3]{\frac{16·6M6}{25\mathrm{\pi p}}}M\Rightarrow \boxed{D=\sqrt[3]{\frac{16·6M6M}{25\mathrm{\pi p}}}}$

The step we have is

$D=\sqrt[3]{\frac{16·6·1.5}{25\pi ·1000}}=\boxed{3.93cm}$