/*! 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 25 A string 180 -cm-long resonates ... [FREE SOLUTION] | 91Ó°ÊÓ

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A string 180 -cm-long resonates in a standing wave that has three segments when driven by a 270 -Hz vibrator. What is the speed of the waves on the string?

Short Answer

Expert verified
The wave speed is 324 m/s.

Step by step solution

01

Understanding the Problem

We are given that a string, which is 180 cm long, forms a standing wave with three segments or loops. This means the string length is 1.5 wavelengths long (since each segment is half a wavelength). The frequency of the wave is given as 270 Hz. Our task is to find the speed of the wave on the string.
02

Calculating the Wavelength

The standing wave has three segments, which corresponds to 1.5 wavelengths fitting into the length of the string (180 cm). Therefore, we find the wavelength by dividing the string length by 1.5. The length of the string is 180 cm, so the wavelength \( \lambda \) is: \[ \lambda = \frac{180 \text{ cm}}{1.5} \] Convert to meters: \( \lambda = \frac{180}{1.5} = 120 \text{ cm} = 1.2 \text{ m} \).
03

Using the Wave Speed Formula

The wave speed \( v \) on the string can be calculated using the formula: \[ v = f \times \lambda \] Where \( f \) is the frequency (270 Hz) and \( \lambda \) is the wavelength (1.2 m). Substituting the given values:\[ v = 270 \times 1.2 \]
04

Calculating the Wave Speed

Perform the multiplication: \[ v = 270 \times 1.2 = 324 \text{ m/s} \]. Therefore, the speed of the wave on the string is 324 meters per second.

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Key Concepts

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

Understanding Standing Waves
Standing waves occur when two waves of the same frequency and amplitude travel in opposite directions and interfere with each other. This results in a wave pattern that appears to stand still. Let's think about a string tied at both ends. When it vibrates, it forms loops, or segments, clearly visible as it resonates. Each loop corresponds to a half of a wavelength (also known as an antinode) while the points that remain stationary are called nodes. For example, in the exercise mentioned, the string forms three such loops, which means the length of the string is equivalent to 1.5 wavelengths.
This oscillation and formation of nodes and antinodes are how musical instruments, such as guitars and violins, produce sound. The length and tension of the string determine the frequency of the standing waves that can be formed. By understanding this, we can link the number of segments to the wavelength and frequency, helping to solve further problems like the one in the exercise.
Calculating Wavelength in Standing Waves
Calculating the wavelength in a standing wave involves understanding the relationship between the string segment numbers and wavelengths. If a string resonates with three segments forming, it means there are three half-wavelengths fitting along its length. Thus, the length of the string corresponds to one and a half times the wavelength.
  • Standards for calculating such a wavelength involve dividing the total length of the string by the number of 0.5 wavelengths.
  • In our exercise, the string is 180 cm long, which is \(180 \text{ cm} \div 1.5\) to find one wavelength.
  • After solving, we find that the wavelength comes out to be 1.2 meters when converted from centimeters.
Breaking it down into such clear steps ensures an undeniable understanding of the inner workings of wave behavior and their calculations.
Exploring Wave Frequency and Its Role
Wave frequency refers to how often the wave oscillates in one second and is measured in Hertz (Hz). Higher frequencies mean more oscillations occur per second. In the given exercise, the string vibrates at a frequency of 270 Hz.
  • By knowing the frequency, combined with the wavelength calculated previously, we can determine the wave speed using the formula \[ v = f \times \lambda \].
  • It's crucial to grasp that frequency plays a pivotal role in determining the characteristics of the wave speed as it links directly to the physical properties of the wave.
  • When applying this to the string, and through the calculation \[ v = 270 \times 1.2 = 324 \text{ m/s}\], we arrive at the wave speed, which clearly tells us how quickly wave oscillations are traveling along the string.
Understanding this relationship between frequency, wavelength, and wave speed is fundamental because it governs how different waves perform under various conditions and mediums.

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Most popular questions from this chapter

A rod \(200 \mathrm{~cm}\) long is clamped \(50 \mathrm{~cm}\) from one end, as shown in Fig. 22-6. It is set into longitudinal vibration by an electrical driving mechanism at one end. As the frequency of the driver is slowly increased from a very low value, the rod is first found to resonate at \(3 \mathrm{kHz}\). What is the speed of sound (compression waves) in the rod?The clamped point remains stationary, and so a node exists there. Since the ends of the rod are free, antinodes exist there. The lowest-frequency resonance occurs when the rod is vibrating in its longest possible segments. In Fig. 22-6 we show the mode of vibration that corresponds to this condition. Since a segment is the length from one node to the next, then the length from \(A\) to \(N\) in the figure is one-half segment. Therefore, the rod is two segments long. This resonance form satisfies our restrictions about positions of nodes and antinodes, as well as the condition that the bar vibrate in the longest segments possible. Since one segment is \(\lambda / 2\) long, $$ L=2(\lambda / 2) \quad \text { or } \quad \lambda=L=200 \mathrm{~cm} $$ Then, from \(\lambda=v T=v / f\), $$ v=\lambda f=(2.00 \mathrm{~m})\left(3 \times 10^{3} \mathrm{~s}^{-1}\right)=6 \mathrm{~km} / \mathrm{s} $$

A metal bar \(6.0 \mathrm{~m}\) long, clamped at its center and vibrating longitudinally in such a manner that it gives its first overtone, vibrates in unison with a tuning fork marked 1200 vibration/s. Compute the speed of sound in the metal.

A flexible cable, \(30 \mathrm{~m}\) long and weighing \(70 \mathrm{~N}\), is stretched by a force of \(2.0 \mathrm{kN}\). If the cable is struck sideways at one end, how long will it take the transverse wave to travel to the other end and return?

Radar waves with \(3.4 \mathrm{~cm}\) wavelength are sent out from a transmitter. Their speed is \(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\). What is their frequency?

A long, narrow pipe closed at one end does not resonate to a tuning fork having a frequency of \(300 \mathrm{~Hz}\) until the length of the air column reaches \(28 \mathrm{~cm} .(a)\) What is the speed of sound in air at the existing room temperature? (b) What is the next length of column that will resonate to the fork?

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