/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 The lowest note on a grand piano... [FREE SOLUTION] | 91影视

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Chapter 21: Problem 18

The lowest note on a grand piano has a frequency of \(27.5 \mathrm{Hz}\) The entire string is \(2.00 \mathrm{m}\) long and has a mass of \(400 \mathrm{g}\). The vibrating section of the string is \(1.90 \mathrm{m}\) long. What tension is needed to tune this string properly?

Short Answer

Expert verified
The tension needed to properly tune the string is approximately \(2173.05 N\).

Step by step solution

01

Calculate the wave speed

First, we calculate the wave speed using the formula \(v = f \lambda\), where \(f = 27.5 Hz\) is the frequency and \(\lambda = 2 \times 1.90 m = 3.80 m\) is the wavelength. When you substitute these values into the formula, you get \(v = 27.5 Hz \times 3.80 m = 104.5 m/s\).
02

Determine the mass per unit length

The mass per unit length \(\mu\) can be found by dividing the total mass of the string \(m = 400 g\) by the total length of the string \(l = 2.00 m\). Bear in mind that the mass should be converted to kg to match the SI unit, therefore, \(m = 400 g = 0.4 kg\). So, we have \(\mu = \frac{0.4 kg}{2.00 m} = 0.2 kg/m\).
03

Calculate the tension

To calculate the tension, we substitute the wave speed and mass per unit length into the wave speed formula \(v = \sqrt{\frac{T}{\mu}}\) and solve for the tension \(T\). This gives us \(T = \mu v^2\). Substituting the previously calculated values, we find \(T = 0.2 kg/m \times (104.5 m/s)^2 = 2173.05 N\).

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Key Concepts

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

Wave Speed Calculation
Understanding wave speed is crucial in analyzing how vibrations in strings produce sound. In our case, the wave speed is calculated using the formula:
  • \( v = f \lambda \)
where:
  • \( f \) is the frequency of the vibration,
  • \( \lambda \) is the wavelength.
For the piano string, we know the frequency \( f = 27.5 \text{ Hz} \) and the effective vibrating length of the string is \( 1.90 \text{ m} \), which results in a wavelength \( \lambda = 2 \times 1.90 \text{ m} = 3.80 \text{ m} \).
This is because it is vibrating in its fundamental mode, forming a half-wavelength. When you multiply these values together, you calculate the wave speed as:
  • \( v = 27.5 \text{ Hz} \times 3.80 \text{ m} = 104.5 \text{ m/s} \)
Wave speed tells us how quickly a wave propagates along the string, playing a vital role in determining the pitch produced.
Mass per Unit Length
The mass per unit length (\( \mu \)) of a string is a fundamental property that affects its vibration. It's essentially how heavy the string is for every meter of its length. You can determine it by using:
  • \( \mu = \frac{m}{l} \)
where:
  • \( m \) is the total mass of the string,
  • \( l \) is the total length of the string.
For our piano string, \( m = 400 \text{ g} = 0.4 \text{ kg} \) and \( l = 2.00 \text{ m} \).
Converting the mass into kilograms matches the SI units, ensuring our final calculation is accurate:
  • \( \mu = \frac{0.4 \text{ kg}}{2.00 \text{ m}} = 0.2 \text{ kg/m} \)
This value is essential as a heftier string (higher \( \mu \)) reduces the wave speed, altering the note produced.
Tension in Strings
Tension is the force applied to a string to make it taut enough to play a specific note.
To find the tension \( T \) that is required, we use the wave speed formula connected to tension:
  • \( v = \sqrt{\frac{T}{\mu}} \)
Rearranging it gives us the formula for tension:
  • \( T = \mu v^2 \)
Using the previously calculated values:
  • \( \mu = 0.2 \text{ kg/m} \)
  • \( v = 104.5 \text{ m/s} \)
we substitute to find:
  • \( T = 0.2 \text{ kg/m} \times (104.5 \text{ m/s})^2 = 2173.05 \text{ N} \)
This means a tension of approximately \( 2173 \text{ N} \) is required to tune the string to the desired frequency, enabling it to produce the correct tone when struck. Proper tension ensures that the string vibrates at the right speed, thus creating the intended musical note.

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Most popular questions from this chapter

Il A \(40-\mathrm{cm}-\) long tube has a \(40-\mathrm{cm}-\) long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As the insert is slowly pulled out, the sound from the tuning fork creates standing waves in the tube when the total length \(L\) is \(42.5 \mathrm{cm}, 56.7 \mathrm{cm},\) and \(70.9 \mathrm{cm} .\) What is the frequency of the tuning fork? Assume \(v_{\text {sound }}=343 \mathrm{m} / \mathrm{s}\).

Two sinusoidal waves with equal wavelengths travel along a string in opposite directions at \(3.0 \mathrm{m} / \mathrm{s}\). The time between two successive instants when the antinodes are at maximum height is \(0.25 \mathrm{s} .\) What is the wavelength?

As the captain of the scientific team sent to Planet. Physics, one of your tasks is to measure \(g\). You have a long, thin wire labeled \(1.00 \mathrm{g} / \mathrm{m}\) and a \(1.25 \mathrm{kg}\) weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the \(1.25 \mathrm{kg}\) weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is \(100 \mathrm{Hz}\). Next, with the \(1.25 \mathrm{kg}\) weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 314 s to complete 100 oscillations. Pulling out your trusty calculator, you get to work. What value of \(g\) will you report back to headquarters?

Two loudspeakers face each other from opposite walls of a room. Both are playing exactly the same frequency, thus setting up a standing wave with distance \(\lambda / 2\) between antinodes. Assume that \(\lambda\) is much less than the room width, so there are many antinodes. a. Yvette starts at one speaker and runs toward the other at speed \(v_{Y} .\) As the does so, she hears a loud-soft-loud modulation of the sound intensity. From your perspective, as you sit at rest in the room, Yvette is running through the nodes and antinodes of the standing wave. Find an expression for the number of sound maxima she hears per second. b. From Yvette's perspective, the two sound waves are Doppler shifted. They're not the same frequency, so they don't create a standing wave. Instead, she hears a loud-soft-loud modulation of the sound intensity because of beats. Find an expression for the beat frequency that Yvette hears. c. Are your answers to parts a and b the same or different? Should they be the same or different?

A vertical tube, open at both ends, is lowered into a tank of water until it is partially filled. The top portion of the tube, above the water, is filled with a gas that, because it is denser than air, remains in the tube. A 50.0 -cm-long, 1.00 g horizontal wire is stretched just above the top of the tube with \(440 \mathrm{N}\) of tension. Bowing the wire at its center causes the wire to vibrate at its fundamental frequency. The water level in the tube is adjusted until the sound from the vibrating wire sets up a standing sound wave in the gas. The water is then lowered another \(30.5 \mathrm{cm}\) until the next standing sound wave is detected. Use this information to determine the speed of sound in the gas.
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