Please answer all parts! A 2.00 times 102 gram mass hangs from the end of an ela

Question

Please answer all parts!

A 2.00 times 102 gram mass hangs from the end of an elastic string wrapped over a pulley at a fixed distance of 1.25 m from a wall, as shown. If a frequency of 31.7 Hz produces 3 anti-nodes, what is the linear mass density of the string? What frequency would be needed to produce 5 anti-nodes? If the frequency on the string is 25.0 Hz, what different hanging mass would be needed to produce 4 anti-nodes? Sadly, this lighter mass makes the string stretch back a distance 0.250 cm causing it to have a larger mass density. Knowing this, what should be the mass used to achieve the same results as part c?

Explanation / Answer

Part A)
v = sqrt(T/u)

f(wavelength) = sqrt(T/u)

(31.7)(.833) = sqrt[(.2)(9.8)/u]

u = .0028 kg/m

Part B)
f(.5) = sqrt[(.2)(9.8)/(.0028)]

f = 52.9 Hz

Part C)

(25)(.625) = sqrt[(m)(9.8)/(.0028)]

m = .0698 kg which is 69.8 grams

Part D)

u = .0028 kg/m times 1.25m = .0035 kg

.0035/1.2475 = .002805611 kg/m

(25)(.625) = sqrt[(m)(9.8)/(.002805611)]

m = .00447 kg

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