Question

A piano tuner stretches a steel piano wire with a tension of 1070 N . The wire is 0.400 m long and has a mass of 4.00 g . A. What is the frequency of its fundamental mode of vibration?

B. What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to 1.00×10^4 Hz ?

Answer 1

The fundamental frequency of the piano wire is approximately 409 Hz, based on the given mass, length, and tension. The highest harmonic that could be heard by a person capable of hearing frequencies up to 10,000 Hz would be the 24th harmonic.

Step-by-step explanation:

To find the fundamental frequency (f1) of the piano wire, we can use the formula for the fundamental frequency of a stretched string, which is f1 = (1/2L) * sqrt(T/μ), where L is the length of the string, T is the tension, and μ is the mass per unit length (linear mass density).

First, we need to calculate the linear mass density of the piano wire:

• Mass (m) = 4.00 g = 0.004 kg
• Length (L) = 0.400 m
• Linear mass density (μ) = mass / length = 0.004 kg / 0.400 m = 0.010 kg/m

Now, we plug in the values into the formula:

• Tension (T) = 1070 N
• Length (L) = 0.400 m
• μ = 0.010 kg/m

f1 = (1 / (2 * 0.400 m)) * sqrt(1070 N / 0.010 kg/m) = (1 / 0.8 m) * sqrt(107000 N/m) = (1 / 0.8) * 327.16 Hz ≈ 409 Hz.

For part B, we need to determine the highest harmonic that can be heard if the upper limit of human hearing is 10,000 Hz. Since the fundamental frequency is 409 Hz, the harmonics will be integer multiples of this value. The highest harmonic number (n) will be the largest integer such that:

n * f1 ≤ 10,000 Hz

Doing the division gives us n ≤ 10,000 / 409, and we round down to the nearest whole number because we cannot have a fraction of a harmonic. So, the highest harmonic heard would be:

n = floor(10,000/409) = 24 (since 24 * 409 ≈ 9816 Hz which is below the 10,000 Hz threshold).

Answer 2

A. 409 Hz

The fundamental frequency of a string is given by:

`f_1=(1)/(2L)\sqrt{(T)/(m/L)}`

where

L is the length of the wire

T is the tension in the wire

m is the mass of the wire

For the piano wire in this problem,

L = 0.400 m

T = 1070 N

m = 4.00 g = 0.004 kg

So the fundamental frequency is

`f_1=(1)/(2(0.400))\sqrt{(1070)/((0.004)/(0.400))}=409 Hz`

B. 24

For this part, we need to analyze the different harmonics of the piano wire. The nth-harmonic of a string is given by

`f_n = nf_1`

where
`f_1` is the fundamental frequency.

Here in this case

`f_1 = 409 Hz`

A person is capable to hear frequencies up to

`f = 1.00 \cdot 10^4 Hz`

So the highest harmonics that can be heard by a human can be found as follows:

`f=nf_1\\n= (f)/(f_1)=(1.00\cdot 10^4)/(409)=24.5 \sim 24`