Question

A 300 cm rope under a tension of 120 N is set into oscillation. The mass density of the rope is 120 g/cm. What is the frequency of the first harmonic mode (m

Answer 1

Answer:

Explanation:

f = $\sqrt{T/(m/L)} / 2L$

T = 120 N

L = 3.00 m

(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m

                                                  (wow that's massive for a "rope")

f = $\sqrt{120/12} /(2(3))$)

f = $\sqrt{10\\}$/6 = 0.527 Hz

This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.

A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N