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 is given by:$$f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$where:* f_1 is the frequency of the first harmonic mode* L is the length of the rope* T is the tension in the rope* $\mu$ is the mass density of the ropeSubstituting the given values into the equation, we get:$$f_1 = \frac{1}{2(500.0 cm)}\sqrt{\frac{70.0 N}{120.0 g/cm}} = 0.48 Hz$$Therefore, the frequency of the first harmonic mode is 0.48 Hz, which corresponds to choice A.

A 5000 Cm Rope Under a Tension of 70