A 500.0 cm rope under a tension of 70.0 N.is set into oscillation.The mass density of the rope is 120.0 g/cm.What is the frequency of the first harmonic mode (m = 2)?
A)0.48 Hz
B)2.92 Hz
C)0.0048 Hz
D)0.153 Hz

Question

Quizplus · Accepted Answer

The frequency of the first harmonic mode (m = 2) is given by:f = (1/2L) * sqrt(T/μ) * mwhere L is the length of the rope, T is the tension, μ is the mass density, and m is the mode number.Plugging in the given values, we get:f = (1/2 * 500.0 cm) * sqrt(70.0 N / 120.0 g/cm) * 2= 0.48 HzTherefore, the correct answer is A.

A 5000 Cm Rope Under a Tension of 70