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A body of specific gravity ρ is suspended from a light steel wire. The wire supports transverse standing waves with a fundamental frequency of 300 Hz. The body is then placed in water such that exactly half of its volume is below the surface. What is the new fundamental frequency of the wire, in Hz?

1. 300((2ρ−1)/(2ρ))¹/2
2. 300((2ρ)/(2ρ−1))¹/2
3. 300((2ρ)/(2ρ−1))
4. 300((2ρ−1)/(2ρ))

Correct answer: 300((2ρ−1)/(2ρ))¹/2

Solution

f ~ sqrt(T). Initial tension = rho*rho_w*V*g; in water (half submerged) tension = rho_w*V*g*(rho - 1/2). So f_new = 300*sqrt((rho-1/2)/rho) = 300*((2rho-1)/(2rho))^(1/2).

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