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A steel wire is used to suspend an object with a specific gravity of ρ, and the fundamental frequency of transverse standing waves in the wire is 300 Hz. When the object is partially submerged in water such that half of its volume is underwater, what will the new fundamental frequency (in Hz) become?
- 300√((2ρ−1)/2ρ)
- 300√(2ρ−1)
- 300(2ρ−1)/2ρ
- 300(2ρ−1)/2
Correct answer: 300√((2ρ−1)/2ρ)
Solution
Fundamental frequency f is proportional to sqrt(T). In air T = V*rho*g; with half the volume submerged T' = V*rho*g - (V/2)*1*g = Vg(rho - 1/2). Thus f' = 300*sqrt((rho-1/2)/rho) = 300*sqrt((2rho-1)/(2rho)), which is option 0, not the stored option 2.
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