A string with a mass/length of 0.00200 kg/m is stretched between two points with a tension of 120 N. If the fundamental frequency is 440 Hz, then what is the distance between the two points?
A) 17.5 cm
B) 20.3 cm
C) 27.8 cm
D) 30.0 cm
E) 35.0 cm

Question

Quizplus · Accepted Answer

The fundamental frequency (f) of a string can be calculated using the formula $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length of the string, $T$ is the tension in the string, and $\mu$ is the mass per unit length of the string. Rearranging this formula to solve for $L$, we get $L = \frac{1}{2f} \sqrt{\frac{T}{\mu}}$. Substituting the given values ($f = 440 \, \text{Hz}$, $T = 120 \, \text{N}$, $\mu = 0.00200 \, \text{kg/m}$), we find $L = \frac{1}{2 \times 440} \sqrt{\frac{120}{0.00200}} = \frac{1}{880} \sqrt{60000} = \frac{1}{880} \times 244.949 = 0.278 \, \text{m}$ or 27.8 cm.

A String with a Mass/length of 0