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A piano tuner stretches a steel piano wire with a tension of 800 \(\mathrm{N}\) . The steel wire is 0.400 \(\mathrm{m}\) long and has a mass of 3.00 \(\mathrm{g}\) . (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 \(10,000 \mathrm{Hz} ?\)

Short Answer

Expert verified
(a) The fundamental frequency is 471.4 Hz. (b) The highest audible harmonic is the 21st.

Step by step solution

01

Convert Mass to Kilograms

The mass of the wire is given in grams. Convert it to kilograms by dividing by 1000, since there are 1000 grams in a kilogram:\[ 3.00 \, \text{g} = \frac{3.00}{1000} \, \text{kg} = 0.003 \, \text{kg} \]
02

Calculate Linear Mass Density

The linear mass density \( \mu \) is the mass per unit length of the wire. Calculate it by dividing the mass by the length:\[ \mu = \frac{m}{L} = \frac{0.003 \, \text{kg}}{0.400 \, \text{m}} = 0.0075 \, \text{kg/m} \]
03

Use the Formula for Fundamental Frequency

The fundamental frequency of a stretched string is given by \( f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension, \( L \) is the length, and \( \mu \) is the linear mass density. Plug in the values:\[ f_1 = \frac{1}{2 \times 0.400 \, \text{m}} \sqrt{\frac{800 \, \text{N}}{0.0075 \, \text{kg/m}}} \]Calculate \( f_1 \):\[ f_1 = 471.4 \, \text{Hz} \]
04

Determine the Number of Highest Audible Harmonic

A person can hear frequencies up to 10,000 Hz. Use the fundamental frequency to find the highest harmonic by calculating \( n \) such that \( n \times f_1 \leq 10,000 \) Hz. Solve for \( n \):\[ n \leq \frac{10,000}{471.4} \approx 21.21 \]Since \( n \) must be an integer, the highest audible harmonic number is 21.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fundamental Frequency
When we talk about the fundamental frequency, we're discussing the lowest frequency at which a system, like a stretched wire, naturally vibrates. In the context of musical instruments, this is the pitch you hear when a single note is played. For a string fixed at both ends, the fundamental frequency depends on several important factors:
  • Length of the string (L)
  • The tension in the string (T)
  • Linear mass density of the string (\( \mu \))

The formula to calculate this frequency is \[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]This formula illustrates how shortening the string, increasing tension, or reducing mass will increase the fundamental frequency. For practical contexts like a piano, tuning is achieved by adjusting tension, to match specific pitches. By substituting the given values into the formula, for example with a steel wire of length 0.400 m, tension 800 N, and linear mass density 0.0075 kg/m, we calculated a fundamental frequency of approximately 471.4 Hz.
Harmonics
Harmonics are fascinating extensions of the fundamental frequency. They are generated when a string vibrates at whole-number multiples of the fundamental frequency. If the fundamental frequency is \( f_1 \), the second harmonic would be \( 2f_1 \), the third \( 3f_1 \), and so on. These harmonics contribute to the richness of sound, providing overtones that shape the character of musical notes.
In the case of our piano wire, we need to find out how many harmonics are audible to a typical human ear, which hears from around 20 Hz up to 10,000 Hz. By dividing the upper frequency limit by the fundamental frequency, \( 10,000 \div 471.4 \approx 21.21 \), we discover that up to the 21st harmonic fits within the hearing range. Thus, the highest harmonic number that can be perceived is 21. Each harmonic contributes to the sound's timbre but may not always be individually distinguished.
Linear Mass Density
Linear mass density \( \mu \) is a key concept in understanding how strings and wires vibrate. Imagine this as the distribution of mass along the string's length. In a formal sense, it's defined as the mass per unit length:\[ \mu = \frac{m}{L} \]where \( m \) is the mass and \( L \) is the length of the string.
For the piano wire in question, with a total mass of 3.00 g and a length of 0.400 m, you convert the mass to kilograms (0.003 kg) before calculating the density. This gives us a linear mass density of 0.0075 kg/m. This variable is important because it provides resistance to the linear motion, influencing how fast waves (and therefore frequencies) travel through the string. A higher \( \mu \) will result in a lower fundamental frequency, while a lower \( \mu \) allows for a higher pitch. Using this concept, tuners can adjust strings by changing tension without altering the fundamental physical characteristics of the string.

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Most popular questions from this chapter

The speed of sound in air at \(20^{\circ} \mathrm{C}\) is 344 \(\mathrm{m} / \mathrm{s} .\) (a) What is the wavelength of a sound wave with a frequency of 784 \(\mathrm{Hz}\) , corresponding to the note \(\mathrm{G}_{5}\) on a piano, and how many milliseconds does each vibration take? (b) What is the wavelength of a sound wave one octave higher than the note in part (a)?

Waves of Arbitrary Shape. (a) Explain why any wave described by a function of the form \(y(x, t)=f(x-v t)\) moves in the \(+x\) -direction with speed \(v .\) (b) Show that \(y(x, t)=f(x-v t)\) satisfies the wave equation, no matter what the functional form of \(f .\) To do this, write \(y(x, t)=f(u),\) where \(u=x-\) vt. Then, to take partial derivatives of \(y(x, t),\) use the chain rule: $$\begin{aligned} \frac{\partial y(x, t)}{\partial t} &=\frac{d f(u)}{d u} \frac{\partial u}{\partial t}=\frac{d f(u)}{d u}(-v) \\ \frac{\partial y(x, t)}{\partial t} &=\frac{d f(u)}{d u} \frac{\partial u}{\partial x}=\frac{d f(u)}{d u} \end{aligned}$$ (c) A wave pulse is described by the function \(y(x, t)=\) \(D e^{-(B x-C t)^{2}},\) where \(B, C,\) and \(D\) are all positive constants. What is the speed of this wave?

The upper end of a \(3.80-\) m-long steel wire is fastened to the ceiling, and a 54.0 -kg object is suspended from the lower end of the wire. You observe that it takes a transverse pulse 0.0492 s to travel from the bottom to the top of the wire. What is the mass of the wire?

A deep-sea diver is suspended beneath the surface of Loch Ness by a \(100-\) m-long cable that is attached to a boat on the surface (Fig. P15.84). The diver and his suit have a total mass of 120 \(\mathrm{kg}\) and a volume of 0.0880 \(\mathrm{m}^{3} .\) The cable has a diameter of 2.00 \(\mathrm{cm}\) and a linear mass density of \(\mu=\) 1.10 \(\mathrm{kg} / \mathrm{m} .\) The diver thinks he sees something moving in the murky depths and jerks the end of the cable back and forth to send transverse waves up the cable as a signal to his companions in the boat. (a) What is the tension in the cable at its lower end, where it is attached to the diver? Do not forget to include the buoyant force that the water (density 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) ) exerts on him. (b)Calculate the tension in the cable a distance \(x\) above the diver. The buoyant force on the cable must be included in your calculation. (c) The speed of transverse waves on the cable is given by \(v=\sqrt{F / \mu}\) (Eq. 15.13). The speed therefore varies along the cable, since the tension is not constant. (This expression neglects the damping force that the water exerts on the moving cable.. Integrate to find the time required for the first signal to reach the surface.

Weightless Ant. Ant. An ant with mass \(m\) is standing peacefully on top of a horizontal, stretched rope. The rope has mass per unit length \(\mu\) and is under tension \(F .\) Without warning, Cousin Throckmorton starts a sinusoidal transverse wave of wave-length \(\lambda\) propagating along the rope. The motion of the rope is in a vertical plane. What minimum wave amplitude will make the ant become momentarily weightless? Assume that \(m\) is so small that the presence of the ant has no effect on the propagation of the wave.

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