A length of piano wire is used to hang a mass from the ceiling. If the total mass of the string is 0.72 g and the mass hung from it is 1850 g, what is the length of the wire, assuming the fundamental frequency when plucked in these circumstances is 320 Hz?
A) 2.48 cm
B) 6.15 cm
C) 3.08 cm
D) 1.54 cm

Question

Quizplus · Accepted Answer

The fundamental frequency (first harmonic) of a string fixed at both ends is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $f$ is the frequency, $L$ is the length of the string, $T$ is the tension in the string, and $\mu$ is the linear mass density of the string. The tension $T$ is equal to the weight of the mass hung from the string, which is $mg$, where $m$ is the mass and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). The linear mass density $\mu$ is the mass of the string per unit length.Given:- Mass of the string, $m_{\text{string}} = 0.72 \, \text{g} = 0.72 \times 10^{-3} \, \text{kg}$- Mass hung from the string, $m_{\text{hung}} = 1850 \, \text{g} = 1.85 \, \text{kg}$- Fundamental frequency, $f = 320 \, \text{Hz}$First, calculate the tension $T$:$$T = m_{\text{hung}} \times g = 1.85 \times 9.8 = 18.13 \, \text{N}$$Next, calculate the linear mass density $\mu$:$$\mu = \frac{m_{\text{string}}}{L} = \frac{0.72 \times 10^{-3}}{L}$$Substitute $T$ and $\mu$ into the formula for $f$:$$320 = \frac{1}{2L} \sqrt{\frac{18.13}{\frac{0.72 \times 10^{-3}}{L}}}$$Solving for $L$, we get:$$L = \frac{1}{2 \times 320} \sqrt{\frac{18.13L}{0.72 \times 10^{-3}}}$$Squaring both sides and solving for $L^2$, then $L$, gives:$$L \approx 0.0615 \, \text{m} = 6.15 \, \text{cm}$$Therefore, the length of the wire is approximately 6.15 cm.

A Length of Piano Wire Is Used to Hang a Mass