A thin 2.00-m string of mass 50.0 g is fixed at both ends and under a tension of 70.0 N.If it is set into small-amplitude oscillation,what is the frequency of the first harmonic mode?
A) 6.61 Hz
B) 13.2 Hz
C) 26.5 Hz
D) 52.9 Hz

Question

Quizplus · Accepted Answer

The frequency of the first harmonic (fundamental frequency) for a string fixed at both ends is given by $ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $, where $ L $ is the length of the string, $ T $ is the tension, and $ \mu $ is the linear mass density. Here, $ L = 2.00 $ m, $ T = 70.0 $ N, and $ \mu = \frac{0.050 \text{ kg}}{2.00 \text{ m}} = 0.025 \text{ kg/m} $. Plugging in these values, $ f_1 = \frac{1}{2 \times 2.00} \sqrt{\frac{70.0}{0.025}} = 13.2 \text{ Hz} $.

A Thin 200-M String of Mass 50