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Chapter 22: Problem 41

What must be the length of an iron rod that has the fundamental frequency \(320 \mathrm{~Hz}\) when clamped at its center? Assume longitudinal vibration at a speed of \(5.00 \mathrm{~km} / \mathrm{s}\).

Short Answer

Expert verified
The length of the rod must be approximately 7.8125 meters.

Step by step solution

01

Recognize the Relationship

To find the length of the rod, we need to use the relationship between the speed of sound, the frequency of vibration, and the wavelength in the rod. The formula to use is \( v = f \times \lambda \), where \( v \) is the speed of sound in the material (\(5.00\, \text{km/s} = 5000\, \text{m/s}\)), \( f \) is the frequency (\(320\, \text{Hz}\)), and \( \lambda \) is the wavelength.
02

Wavelength Calculation

For a rod clamped at its center, the fundamental frequency corresponds to the rod vibrating in a half-wavelength. Therefore, \( \lambda = 2 \times L \), where \( L \) is the length of the rod. Solving for the wavelength, we get \( \lambda = \frac{v}{f} = \frac{5000}{320} \approx 15.625\, \text{m}\).
03

Determine the Length

Since \( \lambda = 2 \times L \), we have that \( L = \frac{\lambda}{2} \). Substituting our value for \( \lambda \), we get \( L = \frac{15.625}{2} = 7.8125\, \text{m}\).

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Key Concepts

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

Longitudinal Vibration
In physics, longitudinal vibrations occur in a medium where particles move parallel to the wave's direction of travel. This type of vibration is most common in solids like rods and is relevant to the exercise we're discussing. When a wave travels through a solid, it compresses and rarefies the particles, much like a spring.

Longitudinal vibrations result in oscillations along the length of the medium rather than across it. In our case, the iron rod vibrates along its length, creating regions of compression and tension.
  • Key aspects include particle displacement in the direction of wave travel.
  • Sound waves in solids typically involve longitudinal waves.
  • These vibrations are crucial for understanding how sound travels through different materials.
Understanding longitudinal vibration is essential when dealing with the propagation of sound through solids. It has practical applications in engineering, materials science, and acoustics.
Fundamental Frequency
The fundamental frequency is the lowest frequency at which a system, like a vibrating rod, naturally oscillates when disturbed. For objects vibrating under physical constraints, such as a clamped rod, this frequency depends on its material, shape, and boundary conditions.

In the exercise, the rod clamped at the center forms a node, meaning it does not move there. The first mode of vibration, or the fundamental frequency, creates one-half of a full sine wave along the length of the rod. This explains why the wavelength is twice the rod's length: the entire length accounts for half a wave.
  • This frequency can be calculated using the relationship: \( f = \frac{v}{\lambda} \).
  • Any harmonic frequencies will be whole multiples of the fundamental frequency.
  • A constrained rod will exhibit vibrational modes that correspond to these frequencies.
The concept of fundamental frequency is vital in various technological applications, including musical instrument design and materials testing, ensuring materials behave predictably when subjected to vibrations.
Speed of Sound in Solids
The speed of sound in solids is inherently different鈥攁nd often greater鈥攖han in gases or liquids. This difference arises because particles in solids are densely packed and apply forces across their bonds more efficiently, allowing sound waves to travel quicker.

In the exercise given, the speed of sound in an iron rod is specified as \(5.00 \text{ km/s}\). This value allows us to calculate the fundamental frequency's wavelength:
  • The speed is directly correlated to the medium's elasticity and density.
  • Sound waves travel faster in solids due to their tightly bound atomic structures.
  • The higher speed facilitates more rapid wave propagation, a fundamental aspect in structural health monitoring.
Understanding the speed of sound in various materials aids engineers and scientists in designing systems that efficiently transmit sound or detect vibrations. This knowledge is crucial for applications ranging from musical instruments to sonar and earthquake monitoring systems.

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Most popular questions from this chapter

A string \(2.0 \mathrm{~m}\) long is driven by a 240 -Hz vibrator at its end. The string resonates in four segments forming a standing wave pattern. What would be the speed of a transverse wave on such a string? Let's first determine the wavelength of the wave from part (d) of Since each segment is \(\lambda / 2\) long, $$ 4\left(\frac{\lambda}{2}\right)=L \quad \text { or } \quad \lambda=\frac{L}{2}=\frac{2.0 \mathrm{~m}}{2}=1.0 \mathrm{~m} $$ Then, using \(\lambda=u T=v / f\), $$ \mathrm{u}=f \lambda=\left(240 \mathrm{~s}^{-1}\right)(1.0 \mathrm{~m})=0.24 \mathrm{~km} / \mathrm{s} $$

A transverse wave is set up on a taut string. Its free end is wiggled up and down at a rate of \(10.0\) cycles every second. What happens to the wavelength of the waves when the oscillation rate is raised to \(20.0\) cycles per second, all else kept constant?

A string vibrates in five segments at a frequency of \(460 \mathrm{~Hz} .(a)\) What is its fundamental frequency? (b) What frequency will cause it to vibrate in three segments? If the string is \(n\) segments long, then fromwe have \(n\left(\frac{1}{2} \lambda\right)\) \(=L\). But \(\lambda=v / f_{n}\), so \(L=n\left(\mathrm{u} / 2 f_{n}\right)\). Solving for \(f_{n}\) provides $$ f_{n}=n\left(\frac{v}{2 L}\right) $$ We are told that \(f_{5}=460 \mathrm{~Hz}\), and so $$ 460 \mathrm{~Hz}=5\left(\frac{v}{2 L}\right) \quad \text { or } \quad \frac{v}{2 L}=92.0 \mathrm{~Hz} $$ Substituting this in the above relation gives $$ f_{n}=(n)(92.0 \mathrm{~Hz}) $$ (a) \(f_{1}=92.0 \mathrm{~Hz}\). (b) \(f_{3}=(3)(92 \mathrm{~Hz})=276 \mathrm{~Hz}\) Alternative Method Recall that for a string held at both ends, \(f_{n}=n_{f 1}\). Knowing that \(f_{5}\) \(=460 \mathrm{~Hz}\), it follows that \(f_{1}=92.0 \mathrm{~Hz}\) and \(f_{3}=276 \mathrm{~Hz}\).

Radio station WJR broadcasts at \(760 \mathrm{kHz}\). The speed of radio waves is \(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\). What is the wavelength of WJR's waves?

The average person can hear sound waves ranging in frequency from about \(20 \mathrm{~Hz}\) to \(20 \mathrm{kHz}\). Determine the wavelengths at these limits, taking the speed of sound to be \(340 \mathrm{~m} / \mathrm{s}\).
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