A large rock that weighs 164.0 N is suspended from the lower end of a thin wire

Question

A large rock that weighs 164.0 N is suspended from the lower end of a thin wire that is 3.00 m long. The density of the rock is 3200 kg/m^3. The mass of the wire is small enough that its effect on the tension in the wire can be neglected. The upper end of the wire is held fixed. When the rock is in air, the fundamental frequency for transverse standing waves on the wire is 43.0 Hz. When the rock is totally submerged in a liquid, with the top of the rock just below the surface, the fundamental frequency for the wire is 22.0 Hz. What is the density of the liquid?

Explanation / Answer

Tension in the string in air = 164 N
we have
Frequency of vibration = K sqrt(T1/u) where K and u are constants
Now when in is immersed in a liquid
Tension in the string
mg - Vg = 164 - (164/(9.8*3200))9.8

again we have frequency of vibration as K sqrt(T2/u)

F1/F2 = sqrt(T1/T2)

43/22 = sqrt(164/(164 - (.051)))

3.818 = 164/(164 - (.051)))

164(2.818) = .194

on solving we get = 2382.21 kg/m^3

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