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Three pieces of string, each of length \(L\), are joined together end to end, to make a combined string of length 3\(L\). The first piece of string has mass per unit length \(\mu_1\), the second piece has mass per unit length \(\mu2 = 4\mu1\), and the third piece has mass per unit length \(\mu_3 = \mu_1/4\). (a) If the combined string is under tension F, how much time does it take a transverse wave to travel the entire length 3L? Give your answer in terms of \(L, F\), and \(\mu_1\). (b) Does your answer to part (a) depend on the order in which the three pieces are joined together? Explain.

Short Answer

Expert verified
(a) \( T = \frac{3.5L \sqrt{\mu_1}}{\sqrt{F}} \). (b) No, the order doesn't affect the time.

Step by step solution

01

Find the Wave Speed in Each Segment

The wave speed in a string segment can be calculated using the formula: \( v = \sqrt{\frac{F}{\mu}} \), where \( F \) is the tension and \( \mu \) is the mass per unit length. - For the first piece: \( v_1 = \sqrt{\frac{F}{\mu_1}} \).- For the second piece: \( v_2 = \sqrt{\frac{F}{4\mu_1}} \).- For the third piece: \( v_3 = \sqrt{\frac{F}{\mu_1/4}} = 2\sqrt{\frac{F}{\mu_1}} \).
02

Calculate Time Taken for Wave to Travel Each Segment

The time taken for a wave to travel through each segment is given by \( t = \frac{L}{v} \). - For the first piece: \( t_1 = \frac{L}{v_1} = \frac{L}{\sqrt{\frac{F}{\mu_1}}} = \frac{L \sqrt{\mu_1}}{\sqrt{F}} \).- For the second piece: \( t_2 = \frac{L}{v_2} = \frac{L}{\sqrt{\frac{F}{4\mu_1}}} = \frac{2L \sqrt{\mu_1}}{\sqrt{F}} \).- For the third piece: \( t_3 = \frac{L}{v_3} = \frac{L}{2\sqrt{\frac{F}{\mu_1}}} = \frac{L \sqrt{\mu_1}}{2\sqrt{F}} \).
03

Sum Up the Times for the Entire String

The total time \( T \) for the wave to travel the entire length of the string is the sum of the times for each segment:\[ T = t_1 + t_2 + t_3 = \frac{L \sqrt{\mu_1}}{\sqrt{F}} + \frac{2L \sqrt{\mu_1}}{\sqrt{F}} + \frac{L \sqrt{\mu_1}}{2\sqrt{F}} \].This simplifies to:\[ T = \frac{3.5L \sqrt{\mu_1}}{\sqrt{F}} \].
04

Analyze Dependence on Order of Joining

The time calculation depends only on the sum of the individual times, each determined by tension \( F \) and the specific mass per unit length \( \mu \). As such, the order in which the different pieces are joined does not affect the total time. Each piece contributes a fixed amount to the total time regardless of its position.

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Key Concepts

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

Transverse Waves
Transverse waves are a type of wave where the medium's displacement is perpendicular to the direction of wave propagation. Imagine them like waves on a string, where the string moves up and down while the wave travels horizontally. This concept is vital for understanding the behavior of waves in strings or wires, which are common in physics and engineering problems.

In the exercise, a transverse wave travels through three different segments of string that are connected end to end. Each segment has different mass characteristics, which influences how the wave travels through each part. This is because the wave speed in a medium is influenced by the tension in the string and the string's mass per unit length.
  • Wave speed is determined by the tension and mass per unit length: the formula is \( v = \sqrt{\frac{F}{\mu}} \).
  • This means that in sections of the string with less mass, waves can travel faster.
  • It is crucial to understand that transverse waves are key to the behavior of this particular exercise's problem involving waves on a stretched string.
Mass per Unit Length
Mass per unit length is a concept used to describe how a material's mass is distributed along its length. It is typically denoted as \( \mu \) and plays a critical role in determining the speed at which waves can travel through the medium. In this exercise, understanding mass per unit length is essential.

The waves' velocity in a string is impacted by its mass per unit length, with lighter strings allowing faster wave travel under the same tension. This exercise features three segments of string with different mass densities:
  • The first string segment has a mass per unit length of \( \mu_1 \).
  • The second segment is heavier, with \( \mu_2 = 4\mu_1 \).
  • The third is lighter with \( \mu_3 = \frac{\mu_1}{4} \).
The varying mass per unit length across these segments affects the wave speeds as follows:
  • Heavier segments result in slower wave speeds.
  • Lighter segments allow waves to travel more quickly.
The combined effects of these differences are important for calculating the total time a wave takes to traverse the entire length of the string.
Wave Travel Time
Wave travel time is the duration it takes for a wave to travel through a specific distance. In the context of strings, this time is affected by the wave speed, which in turn depends on tension and mass per unit length. Calculating wave travel time involves understanding both these elements and how they interact over the given length of the material.

For this exercise:
  • The total travel time \( T \) is derived by summing up the times for each segment of the string:
  • Each segment's time \( t \) can be calculated using the formula \( t = \frac{L}{v} \), where \( v \) is the wave speed and \( L \) is the segment length.
  • As segments have different mass per unit lengths, they each contribute different times to the final result.
The wave travel time is calculated as:
\[ T = \frac{3.5L \sqrt{\mu_1}}{\sqrt{F}} \]
This expression shows that the order of segments does not affect the travel time, meaning each segment's length and mass are the primary time determinants. By understanding how to apply these principles, students can solve similar problems involving complex wave and string dynamics.

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Most popular questions from this chapter

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192 m/s and a frequency of 240 Hz. The amplitude of the standing wave at an antinode is 0.400 cm. (a) Calculate the amplitude at points on the string a distance of (i) 40.0 cm; (ii) 20.0 cm; and (iii) 10.0 cm from the left end of the string. (b) At each point in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement? (c) Calculate the maximum transverse velocity and the maximum transverse acceleration of the string at each of the points 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 fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is \(y(x, t) = 2.30 \, \mathrm{mm} \, \mathrm{cos} [(6.98 \, \mathrm{rad/m})x \space + 1742 \, \mathrm{rad/s})t]\). Being more practical, you measure the rope to have a length of 1.35 m and a mass of 0.00338 kg. You are then asked to determine the following: (a) amplitude; (b) frequency; (c) wavelength; (d) wave speed; (e) direction the wave is traveling; (f) tension in the rope; (g) average power transmitted by the wave.

A continuous succession of sinusoidal wave pulses are produced at one end of a very long string and travel along the length of the string. The wave has frequency 70.0 Hz, amplitude 5.00 mm, and wavelength 0.600 m. (a) How long does it take the wave to travel a distance of 8.00 m along the length of the string? (b) How long does it take a point on the string to travel a distance of 8.00 m, once the wave train has reached the point and set it into motion? (c) In parts (a) and (b), how does the time change if the amplitude is doubled?

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?

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