A piano wire has a radius of 0.500 mm. One end is fixed and the other is wrapped around a tuning peg that has a diameter of 3.00 mm and is 70.0 cm away. After the wire just becomes taut, the peg is given one and a half turns. What is the fundamental frequency of oscillation of the wire? The density of steel is 7860 kg/m³ and Young's modulus for steel is 20 × 10¹⁰ N/m². A) 1020 Hz B) 512 Hz C) 2050 Hz D) 342 Hz E) 256 Hz

Question

Quizplus · Accepted Answer

The length of the wire can be calculated by first finding the length of wire that is wrapped around the peg. This can be found using the equation:

$L_{peg} = 	heta r = (1.5 \cdot 2\pi)\cdot\frac{3.00	ext{ mm}}{2}=14.14$ mm

The length of wire not wrapped around the peg can be found using Pythagoras' theorem:

$L_{wire} = \sqrt{(70.0	ext{ cm})^2 + (14.14	ext{ mm})^2}=70.1$ cm

The mass of the wire can be found using the density of steel and the volume of the wire:

$m = ho V = ho \pi r^2 L = (7860	ext{ kg/m}^3)\pi (0.500	ext{ mm})^2(70.1	ext{ cm})=0.8643$ g

The tension in the wire can be found using Young's modulus and the strain (which is equal to the change in length divided by the original length):

$\Delta L = L_{wire}-L_0 = 70.1	ext{ cm}-2r=69.1$ cm
$\epsilon = \frac{\Delta L}{L_0}=\frac{69.1	ext{ cm}}{70.0	ext{ cm}}-1=-0.013$
$T=\frac{F}{A}=\frac{Y\epsilon A}{L_0}=(20	imes 10^{10}	ext{ N/m}^2)(-0.013)\pi(0.500	ext{ mm})^2=5.19$ N

Finally, the fundamental frequency of the wire can be found using the equation:

$f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}=\frac{1}{2L}\sqrt{\frac{T}{ho A}}$

$f_1=\frac{1}{2(70.1	ext{ cm})}\sqrt{\frac{5.19	ext{ N}}{(7860	ext{ kg/m}^3)\pi (0.500	ext{ mm})^2}}=512$ Hz

Therefore, the answer is B) 512 Hz.

A Piano Wire Has a Radius of 0