A solid copper object hangs at the bottom of a

Question

steel wire of negligible mass. The top end of the wire is
fixed. When the wire is struck, it emits sound with a fundamental
frequency of 300 Hz. The copper object is then submerged
in water so that half its volume is below the water
line. Determine the new fundamental frequency.
is this the question?

Answer 1

It is the question if you say it is.

Partially submerging the copper object will reduce the tension in the steel wire. There will be a buoyancy force equal to half the volume of the copper object multiplied by the density of water. Let V be the volume of the copper and g be the acceleration of gravity. The specific gravity of copper is 8.92, and that of water is 1.00.

Copper object weight = 8.92 V*g
Buoyancy force = 1.00*(V/2)*g = 0.50 V*g
Effective weight = 8.42 V*g
The tension T gets multiplied by a factor 8.42/8.92 = 0.9439.

The speed of waves in the wire AND the fundamental frequency get multiplied by the square root of this factor, or 0.9716.

The new fundamental frequency becomes
300*0.9716 = 291.5 Hz
That is about half of a half-tone on the musical scale.