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.

Question

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