Answer to Question #288778 in Mechanics | Relativity for NICKO

Question

A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m long and has a mass of 3.00 g. What is the frequency of its fundamental mode of vibration? What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to 10,000 Hz?

Answer 1

Explanations & Calculations

• The harmonics possible for this setup is given by the formula $f_n=n\frac{V}{2L}$ .
• Then the fundamental frequency corresponds to $\small n=1$ .
• The speed of waves on the string is $\small V=\sqrt{\frac{T}{\mu}}$
• Here $\small \mu$: the mass per unit length is $\mu= \frac{0.003\,kg}{0.400\,m}=\small 0.0075\,kgm^{-1}$
• Therefore,

$\qquad\qquad \begin{aligned} \small f_1&=\small 1\times\frac{\sqrt{\frac{800\,N}{0.0075\,kgm^{-1}}}}{2\times0.400\,m}\\ &=\small 408.25\,s^{-1}\,\text{or}\,\,408.25\,Hz \end{aligned}$

• This string setup can produce all the consecutive harmonics for $\small n=1,2,3...$
• Therefore, the highest harmonic would be at some $n$ value that would near or equal the maximum hearing frequency.
• Then,

$\qquad\qquad \begin{aligned} \small n\times408.25\,Hz&=\small10000\,Hz\\ \small n&=\small 24.49\\ \small n & \approx \small 24 \end{aligned}$

• It's the $\small 24^{th}$ harmonic.