Question

An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large object suspended from it. Assume that the wire has a mass of 4.00 g and a length of 1.60 m and that a 3.00 kg object is suspended from it. A pulse requires 43.8 ms to traverse the length of the wire. Calculate the local g from these data. (You may neglect the mass of the wire when calculating the tension in it.)

Answer 1

0.195 m/s²

Step-by-step explanation:

Since a standing wave is set up in the wire, its frequency f = n/2l√(T/μ). For the fundamental frequency, n = 1. So f = 1/2l√(T/μ)

where l = length of wire = 1.60 m, T₀ = tension in wire = weight of object = mg (neglecting wires mass), m = mass of object = 3.00 kg, g = acceleration due to gravity on the small planet, μ = linear density of wire = m₀/l , m₀= mass of wire = 4.00 g = 0.004 kg and f = 1/T where T = period of pulse = 43.8 ms = 0.0438 s

f = 1/2l√(T₀/μ) = 1/T ⇒ T₀ = 4l²μ/T²

mg = 4l²μ/T²

g = 4l²μ/mT² = 4l²m₀/l/mT² = 4lm₀/mT²

g = 4lm₀/mT² = 4 × 1.60 × 0.004/(3.00 × 0.0438²) = 0.195 m/s²