Question

A solid copper object hangs at the bottom of a steel wire of negligible mass. The top end of the wire is fixed. When the wire is struck, it emits a sound with a fundamental frequency of 135 Hz. The copper object is then submerged in water so that half its volume is below the water line. Determine the new fundamental frequency.
Hz

Answer 1

Density of copper= 8960 kg/m^3
" " "" water = 1000 kg/m^3
v= √(T/ρ)= λf
λ= 1/f1 √(T1/ρ)= 1/f2 √(T2/ρ)
1 is before submersion and 2 is after
f2= f1 √(T2/T1)

T2= T1-B
B is the buoyant force
B= ρ(w)V(w)g
(w) denotes the density and volume of water, otherwise I'll be referring to the hanging mass
V(w)= V/2
Since half of the hanging object is submerged
V= m/ρ

Put it all together

f2= f1 √(T2/T1)= f1 √((T1-B)/T1)
= f1 √(1-(B/T1))
= f1 √(1-((ρ(w)(V/2)g)/mg))
= f1 √(1-((ρ(w)(m/ρ)g)/2mg))
Cancel the m and g
= f1 √(1-(p(w)/2ρ))
= 135√(1-(1000/2(8960)))

= 131.2 Hz

Hope that helps

Answer 2

Nice try