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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?

Short Answer

Expert verified
Mass of the string is approximately calculated using wave dynamics, and maximum transverse acceleration is found using angular frequency.

Step by step solution

01

Understanding Harmonics and Overtones

The first overtone of a vibrating string is equivalent to the second harmonic. In a standing wave, the string vibrates in segments. For the second harmonic, the string vibrates with one full wave over its length, meaning two loops are formed, and there are nodes at both ends and one node in the middle.
02

Determine Wave Speed

For the second harmonic, the wavelength \( \lambda \) is equal to the length of the string, \( L = 2.50 \) m. The wave speed \( v \) can be calculated using the relation \( v = \frac{2L}{T} \), where \( T \) is the period of the wave. Since we don't directly know the period, we will later use the relationship with tension and mass per unit length to find the wave speed.
03

Relate Parameters and Calculate Mass

Using the formula for wave speed on a string, \( v = \sqrt{\frac{T}{\mu}} \), where \( T = 90.0 \) N is the tension and \( \mu \) is the mass per unit length. Re-arranging for \( \mu \), \( \mu = \frac{T}{v^2} \). From the information given for the first overtone's transverse speed, \( v_{max} = \omega A = 28.0 \) m/s where amplitude \( A = 0.0350 \) m, we can determine \( \omega \), and therefore, the wave speed \( v \).
04

Calculate Angular Frequency and Wave Speed

For the given maximum transverse speed, \( \omega = \frac{28.0}{0.0350} \) rad/s. Angular frequency \( \omega \) is related to wave speed and wavelength by \( \omega = \frac{2 \pi v}{\lambda} \). Solve for \( v \) to obtain the wave speed.
05

Find the String's Mass

With the wave speed \( v \) found, substitute into the expression for \( \mu = \frac{T}{v^2} \) to calculate mass per unit length. Subsequently, determine the total mass of the string as \( m = \mu L \).
06

Calculate Maximum Transverse Acceleration

The maximum transverse acceleration is given by \( a_{max} = \omega^2 A \). Use the \( \omega \) determined earlier and the amplitude \( A = 0.0350 \) m to find \( a_{max} \).
07

Compute and Conclude

Use all the known values and relationships to compute the answers to parts (a) and (b). Ensure all calculations are done with proper SI units for consistency and clear results.

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Key Concepts

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

Harmonic Frequencies
In the world of wave mechanics, harmonic frequencies are crucial when analyzing the vibration of strings. For a string secured at both ends, different modes of vibration, or harmonics, can form. The simplest form of vibration is termed the fundamental frequency or first harmonic, where the string vibrates as a single segment.
When we refer to the first overtone, we're discussing the second harmonic. With the second harmonic, the string vibrates in two segments, creating a pattern with two peaks. This pattern results in nodes, points that remain stationary, and antinodes, points that move with the maximum amplitude.
Understanding harmonic frequencies helps us identify the wave patterns and the number of segments or loops formed as the string vibrates between fixed ends. This concept is foundational to the problem, which involves calculating related properties like wave speed and transverse acceleration.
Standing Waves
Standing waves are formed when two identical waves traveling in opposite directions superimpose. This creates points of no movement (nodes) and points of maximum movement (antinodes) along the medium, such as a string.
The appearance of these nodes and antinodes depends on the harmonic being observed. As mentioned, a string vibrating in its second harmonic will have two loops, marked by nodes at both ends and an additional node in the middle. This rigid structure of standing waves allows for easy calculation and prediction of properties like amplitude and wave speed.
Understanding standing waves and their formation provides insight into how energy is transmitted through the medium and allows for precise calculations in wave mechanics, including understanding harmonic frequencies and wave speed.
Wave Speed Calculation
Wave speed is a critical parameter in wave mechanics. For a stretched string, wave speed is determined using the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) represents the tension in the string and \( \mu \) is the mass per unit length of the string.
In our exercise, the string's length is known, and tension is provided as 90.0 N. To compute the wave speed, we first need to determine the relationship between other variables, such as the angular frequency \( \omega \) and amplitude \( A \). With the maximum transverse speed given, \( v_{max} = \omega A \). Solving for \( \omega \), we can find the wave speed using \( \omega = \frac{2\pi v}{\lambda} \), where \( \lambda = 2.50 \) m for the second harmonic. By understanding these relationships, you can accurately compute the speed at which waves travel along the string, a key element in solving problems involving vibrating strings.
Mass per Unit Length
Mass per unit length, denoted as \( \mu \), plays a significant role in determining the dynamics of a vibrating string. It measures how much mass exists along each meter of the string and can be calculated once the wave speed is known.
The formula \( \mu = \frac{T}{v^2} \) is used, where \( T \) is the tension applied to the string, and \( v \) is the calculated wave speed. This gives us the mass per unit length in kilograms per meter (kg/m).
Once \( \mu \) is determined, you can calculate the total mass of the string by multiplying \( \mu \) with the string's length \( L \). Both parameters are interconnected, helping us solve complex problems involving string vibration mechanics efficiently.
Transverse Acceleration
Transverse acceleration is another vital aspect of wave mechanics, particularly when analyzing the dynamics of points along a vibrating string. It refers to the acceleration at which a point on the string moves perpendicular to the string's initial tension direction.
This can be expressed mathematically as \( a_{max} = \omega^2 A \), where \( \omega \) represents the angular frequency, and \( A \) is the amplitude. In the exercise, amplitude is given, and \( \omega \) can be derived from other known quantities. By computing \( a_{max} \), we gain a better understanding of the forces acting on the antinodes of the standing wave.
This concept is essential when determining dynamic behaviors of waves on strings and must be understood to accurately analyze motion and forces involved in wave mechanics.

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Most popular questions from this chapter

A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m long and has a mass of 3.00 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 Hz?

Adjacent antinodes of a standing wave on a string are 15.0 cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 cm and period 0.0750 s. The string lies along the \(+x\)-axis and is fixed at \(x = 0\). (a) How far apart are the adjacent nodes? (b) What are the wavelength, amplitude, and speed of the two traveling waves that form this pattern? (c) Find the maximum and minimum transverse speeds of a point at an antinode. (d) What is the shortest distance along the string between a node and an antinode?

The speed of sound in air at 20\(^\circ\)C is 344 m/s. (a) What is the wavelength of a sound wave with a frequency of 784 Hz, corresponding to the note 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 (twice the frequency) than the note in part (a)?

A piano wire with mass 3.00 g and length 80.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplitude 1.6 mm travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

(a) A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed \(v\), frequency \(f\), amplitude \(A\), and wavelength \(\lambda\). Calculate the maximum transverse velocity and maximum transverse acceleration of points located at (i) \(x = \lambda/2\), (ii) \(x = \lambda\)/4, and (iii) \(x = \lambda\)8, from the left-hand end of the string. (b) At each of the points in part (a), what is the amplitude of the motion? (c) At each of the points in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement?

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