An 80 cm long string,fixed at both ends vibrates at a fundamental frequency of 450 Hz and has a mass per unit length of 5.51 * 10^-4 kg/m.What is the tension in the string? A)200 N B)191 N C)320 N D)231 N E)286 N

Question

Quizplus · Accepted Answer

The fundamental frequency (f) 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. Rearranging this formula to solve for $T$, we get $T = \mu (2Lf)^2$. Substituting the given values: $L = 0.8$ m, $f = 450$ Hz, and $\mu = 5.51 \times 10^{-4}$ kg/m, we find $T = 5.51 \times 10^{-4} \times (2 \times 0.8 \times 450)^2 = 285.76896$ N, which rounds to 286 N.

An 80 Cm Long String,fixed at Both Ends Vibrates at a Fundamental