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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...Asked by Hazel
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/m3. 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 30.0 Hz . What is the density of the liquid?
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Answered by
Ryan
I solved it! It is:
((m_rock*g)-(m_rock/(g*density_rock))*density_liquid*g)/(lower frequency squared) = (m_rock*g)/(higher frequency squared)
Then just solve for density_liquid :)
((m_rock*g)-(m_rock/(g*density_rock))*density_liquid*g)/(lower frequency squared) = (m_rock*g)/(higher frequency squared)
Then just solve for density_liquid :)
Answered by
Steve
Close but it's actually,
((m_rock*g)-((m_rock*g)/(density_rock*density_liquid)))/(lower frequency squared) = (m_rock*g)/(higher frequency squared)
((m_rock*g)-((m_rock*g)/(density_rock*density_liquid)))/(lower frequency squared) = (m_rock*g)/(higher frequency squared)
Answered by
Simon
For anyone finding this in the future, Ryan has the correct answer.
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