{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

4_prelab_solutions

# 4_prelab_solutions - Standing Waves Prelab 1 1.1 Guitar...

This preview shows pages 1–3. Sign up to view the full content.

Standing Waves Prelab 1 Guitar String 1.1 Linear Mass Density For a steel wire of diameter D = 0 . 010 inches and density ρ = 7 . 80 g/cm 3 , what is the linear mass density μ of the string? The definition of volume mass density ρ is mass per unit volume, ρ = m V The mass m of the guitar string is therefore m = ρ V In order to calculate the volume of the steel wire, we will model the wire as a cylinder of length L and diameter D , V = A L = ( π r 2 ) L = π D 2 2 L = π D 2 L 4 We can now rewrite the expression for the mass of the wire, m = ρ V = ρ π D 2 L 4 The definition of linear mass density μ is mass per unit length, μ = m L Using the expression for the mass, μ = m L = ρ π D 2 4 = π ρ D 2 4 The length of the wire cancels out: the linear mass density of a very long wire is equal to that of a short piece of wire. We can use the known values for ρ and D in order to calculate μ , μ = π ρ D 2 4 = π ( 7 . 80 g cm 3 ) (0 . 010 in ) 2 4 Remembering that 1 in = 2 . 54 cm , μ = π 7 . 80 10 - 3 kg (10 - 2 m ) 3 (0 . 010 · 2 . 54 cm ) 2 4 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
μ = π 7 . 80 10 - 3 kg 10 - 6 m 3 (0 . 010
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

4_prelab_solutions - Standing Waves Prelab 1 1.1 Guitar...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online