Wire A is the same mass per unit length as wire B.However wire A is twice as long as wire B and has three times as much tension on it.Calculate the fundamental frequency of wire A divided by wire B.
A)0.87
B)0.66
C)0.43
D)0.75
E)1.50

Question

Quizplus · Accepted Answer

The fundamental frequency of a wire is given by:$$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$where:* f is the fundamental frequency* L is the length of the wire* T is the tension on the wire* $\mu$ is the mass per unit length of the wireSince wire A has the same mass per unit length as wire B, $\mu$ is the same for both wires. Since wire A is twice as long as wire B, $L_A = 2L_B$. Since wire A has three times as much tension on it, $T_A = 3T_B$. Substituting these values into the equation for the fundamental frequency, we get:$$f_A = \frac{1}{2(2L_B)}\sqrt{\frac{3T_B}{\mu}} = \frac{1}{4L_B}\sqrt{\frac{3T_B}{\mu}}$$$$f_B = \frac{1}{2L_B}\sqrt{\frac{T_B}{\mu}}$$Dividing $f_A$ by $f_B$, we get:$$\frac{f_A}{f_B} = \frac{\frac{1}{4L_B}\sqrt{\frac{3T_B}{\mu}}}{\frac{1}{2L_B}\sqrt{\frac{T_B}{\mu}}} = \frac{1}{2}\sqrt{\frac{3}{1}} = \frac{1}{2}\sqrt{3} = 0.866$$Therefore, the fundamental frequency of wire A divided by wire B is 0.866, which is closest to choice A, 0.87.

Wire a Is the Same Mass Per Unit Length as Wire