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For a string stretched between two supports, two successive standing-wave frequencies are 525 Hz and 630 Hz. There are other standing-wave frequencies lower than 525 Hz and higher than 630 Hz. If the speed of transverse waves on the string is 384 m/s, what is the length of the string? Assume that the mass of the wire is small enough for its effect on the tension in the wire to be ignored.

Short Answer

Expert verified
The length of the string is 1.83 meters.

Step by step solution

01

Understand the Problem

We have two successive standing-wave frequencies of a stretched string: 525 Hz and 630 Hz. We need to find the length of the string, knowing the speed of wave propagation along the string is 384 m/s.
02

Identify Successive Harmonics

For standing waves, the difference between two successive harmonics \(f_{n+1} - f_n\) gives the fundamental frequency \(f_1\). Here, \(f_{n+1} = 630\) Hz and \(f_n = 525\) Hz. The fundamental frequency \(f_1\) is the difference: \(f_1 = 630 - 525 = 105\) Hz.
03

Apply the Wave Speed Formula

The wave speed \(v\) on a string is related to its frequency \(f\) and wavelength \(\lambda\) by the equation \(v = f \cdot \lambda\). For the fundamental frequency, where \(\lambda_1 = 2L\), we can substitute the known wave speed and frequency: \(384 = 105 \cdot 2L\).
04

Solve for the Length of the String

Rearrange the equation from Step 3 to solve for \(L\). Thus, \(L = \frac{384}{2 \times 105}\). After simplifying, \(L = \frac{384}{210} = 1.83\) meters.

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Key Concepts

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

Standing Waves
Standing waves are a fascinating phenomenon that occurs when two waves of the same frequency and amplitude travel in opposite directions along a medium, such as a string, and interfere with each other. This interference results in a wave pattern that appears to be stationary, hence the name "standing wave." The points where the waves always cancel out are called nodes, and the points of maximum amplitude are called antinodes. These stationary patterns are formed at specific frequencies, called resonance frequencies or harmonics.

In our example, we have a string fixed at both ends, which naturally supports the formation of standing waves. The length of the string determines the possible wavelengths and consequently the frequencies of the standing waves observed. The string can vibrate at multiple harmonics, each corresponding to a different standing wave pattern.
Harmonics
Harmonics are integral multiples of the fundamental frequency, the lowest frequency at which a standing wave can form. The fundamental frequency, often called the first harmonic, is the simplest standing wave pattern, with nodes at both ends and an antinode in the middle. Its wavelength is twice the length of the string, given by \( \lambda_1 = 2L \).

Higher harmonics, like the second, third, and so on, have additional nodes and antinodes. The second harmonic (or first overtone) has one additional node and antinode compared to the first harmonic, and its frequency is twice that of the fundamental frequency. This pattern continues, with each higher harmonic having an increased frequency and a more complex standing wave pattern.
  • The fundamental frequency is important because it can be used to determine other harmonics.
  • The difference between successive harmonics gives us the fundamental frequency.
In the given problem, the difference between the two successive harmonics 630 Hz and 525 Hz provided the fundamental frequency of 105 Hz. This helped in finding the length of the string once the speed was known.
Wave Speed Formula
The wave speed formula is crucial in linking the physical properties of waves to their behavior on a particular medium. It is given by the equation \( v = f \cdot \lambda \), where \( v \) is the wave speed, \( f \) is the frequency of the wave, and \( \lambda \) is the wavelength.

This formula tells us that for a constant wave speed, if the frequency increases, the wavelength must decrease, and vice versa. In the context of a stretched string, the wave speed depends on the tension in the string and its mass per unit length, but in our problem, these effects are negligible.
  • The wave speed of 384 m/s is used in conjunction with the fundamental frequency to find the string's length.
  • Applying the wave speed formula, along with the fundamental frequency and understanding that the fundamental wavelength is twice the string length, allows us to derive the equation for the length.
By rearranging \( v = f_1 \cdot 2L \) to solve for \( L \), we can determine that \( L = \frac{v}{2f_1} = \frac{384}{2 \times 105} = 1.83 \) meters, reflecting the length of the string which supports these standing waves with given frequencies.

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Most popular questions from this chapter

With what tension must a rope with length 2.50 m and mass 0.120 kg be stretched for transverse waves of frequency 40.0 Hz to have a wavelength of 0.750 m?

A 0.800-m-long string with linear mass density \(\mu = 7.50\) g/m is stretched between two supports. The string has tension \(F\) and a standing-wave pattern (not the fundamental) of frequency 624 Hz. With the same tension, the next higher standing-wave frequency is 780 Hz. (a) What are the frequency and wavelength of the fundamental standing wave for this string? (b) What is the value of \(F\)?

In your physics lab, an oscillator is attached to one end of a horizontal string. The other end of the string passes over a frictionless pulley. You suspend a mass \(M\) from the free end of the string, producing tension \(Mg\) in the string. The oscillator produces transverse waves of frequency \(f\) on the string. You don't vary this frequency during the experiment, but you try strings with three different linear mass densities \(\mu\). You also keep a fixed distance between the end of the string where the oscillator is attached and the point where the string is in contact with the pulley's rim. To produce standing waves on the string, you vary \(M\); then you measure the node-to-node distance \(d\) for each standingwave pattern and obtain the following data: (a) Explain why you obtain only certain values of \(d\). (b) Graph \(\mu d^2\) (in kg \(\cdot\) m) versus \(M\) (in kg). Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straight-line fit to the data to determine the frequency \(f\) of the waves produced on the string by the oscillator. Take \(g = 9.80 \, \mathrm{m/s}^2\). (d) For string A (\(\mu = 0.0260\) g/cm), what value of \(M\) (in grams) would be required to produce a standing wave with a node-to-node distance of 24.0 cm? Use the value of \(f\) that you calculated in part (c).

A horizontal wire is stretched with a tension of 94.0 N, and the speed of transverse waves for the wire is 406 m/s. What must the amplitude of a traveling wave of frequency 69.0 Hz be for the average power carried by the wave to be 0.365 W?

A thin string 2.50 m in length is stretched with a tension of 90.0 N between two supports. When the string vibrates in its first overtone, a point at an antinode of the standing wave on the string has an amplitude of 3.50 cm and a maximum transverse speed of 28.0 m/s. (a) What is the string's mass? (b) What is the magnitude of the maximum transverse acceleration of this point on the string?

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