A string fixed at both ends is 50.0 cm long and has a tension that causes the frequency of its fundamental to be 262 Hz.If the tension is increased by 4%,what does the fundamental frequency become?
A)252 Hz
B)257 Hz
C)264 Hz
D)267 Hz
E)272 Hz

Question

Quizplus · Accepted Answer

The frequency of the fundamental mode of a string fixed at both ends is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, 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. When the tension is increased by 4%, the new tension $T'$ is $1.04T$. The new frequency $f'$ can be found by substituting $T'$ into the formula, resulting in $f' = \frac{1}{2L} \sqrt{\frac{1.04T}{\mu}} = f \sqrt{1.04}$. Since the original frequency $f$ is 262 Hz, $f' = 262 \sqrt{1.04}$, which calculates to approximately 267 Hz.

A String Fixed at Both Ends Is 50