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Two strings are attached to poles, however the first string is twice the linear mass density mu of the second. If both strings have the same tension, what is the ratio of the speed of the pulse of the wave from the first string to the second string?

Short Answer

Expert verified
The ratio of the speed of the pulse of the wave from the first string to the second string is \(\frac{1}{\sqrt{2}}\).

Step by step solution

01

Write down the given information and the wave speed formula

The linear mass density (mu) of the first string is twice that of the second string (mu1 = 2 * mu2). Both strings have the same tension (T). The wave speed formula is given by: \(v = \sqrt{\frac{T}{\mu}}\), where \(v\) is the speed of the wave, \(T\) is the tension, and \(\mu\) is the linear mass density.
02

Write down the wave speed formula for both strings

For the first string, the speed of the wave can be written as: \(v_1 = \sqrt{\frac{T}{2*\mu_2}}\). For the second string, we can write: \(v_2 = \sqrt{\frac{T}{\mu_2}}\).
03

Compute the ratio of the wave speeds

To find the ratio of the speed of the pulse of the wave from the first string to the second string, we can divide the wave speed of the first string by the wave speed of the second string: \(Ratio = \frac{v_1}{v_2}\). Substitute the values of \(v_1\) and \(v_2\) from Step 2: \(Ratio = \frac{\sqrt{\frac{T}{2*\mu_2}}}{\sqrt{\frac{T}{\mu_2}}}\)
04

Simplify the ratio

To simplify the ratio, we can rewrite the division of the two square roots as a single square root of a fraction: \(Ratio = \sqrt{\frac{\frac{T}{2*\mu_2}}{\frac{T}{\mu_2}}}\). Now, let's simplify the fraction inside the square root: \(\frac{\frac{T}{2*\mu_2}}{\frac{T}{\mu_2}} = \frac{1}{2}\) So, the ratio becomes: \(Ratio = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}\).
05

Write the final answer

The ratio of the speed of the pulse of the wave from the first string to the second string is \(\frac{1}{\sqrt{2}}\).

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Key Concepts

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

Linear Mass Density
Linear mass density, represented by the symbol \( \mu \), is a measure of how much mass exists along a unit length of a string. It is expressed in units of mass per length, such as kg/m. Understanding linear mass density is crucial because it affects how a wave travels along a string.
For example, in the context of our exercise, the first string has a linear mass density that is twice that of the second string. This means that for the same length of string, the first string is heavier than the second, which influences the speed at which waves can travel through each string.
  • A higher linear mass density means more mass under the same tension, resulting in slower wave propagation.
  • Conversely, a lower linear mass density allows for faster wave movement.
Therefore, knowing the linear mass density helps you understand and predict wave behavior on strings, which is essential in calculating wave speed ratios as seen in this exercise.
Tension in Strings
Tension in strings, denoted by \( T \), is the force exerted along the string, pulling it tight. Tension plays a vital role in determining the speed of waves along the string. In our scenario, both strings have equal tension, which is a constant force pulling both strings with the same intensity.
Tension affects wave speed because it counteracts the mass per unit length (linear mass density) of the string. The relationship between tension and wave speed is captured by the formula: \( v = \sqrt{\frac{T}{\mu}} \).
  • If the tension is increased while keeping the linear mass density constant, the wave speed increases.
  • Conversely, if the tension decreases, the wave speed will reduce.
This relationship shows why the same tension in strings with different linear mass densities results in different wave speeds, allowing us to calculate ratios of wave speeds as demonstrated in this exercise.
Wave Pulse
A wave pulse is a single disturbance that moves through a medium, such as a string. Understanding wave pulses is important in comprehending how waves transfer energy along the string.
In the problem, we are focused on how the wave pulses travel at different speeds in the two strings due to varying linear mass densities. The speed of a wave pulse depends on the condition of the string, specifically the tension and the linear mass density.
  • The greater the tension relative to the linear mass density, the faster the wave pulse travels along the string.
  • If two strings have different mass densities but the same tension, their wave speeds will vary, as seen when calculating the wave speed ratio \( \frac{v_1}{v_2} \).
This understanding helps in visualizing how wave pulses move and interact in various mediums, particularly in the exercise where the speed ratio derived gives insights into the relative movement of wave pulses on the two strings.

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Most popular questions from this chapter

A wave traveling on a Slinky() that is stretched to \(4 \mathrm{m}\) takes \(2.4 \mathrm{s}\) to travel the length of the slinky and bach again. (a) What is the speed of the wave? (b) Using the same Slinky stretched to the same length, a standing wave is created which consists of three antinodes and four nodes At what frequency must the Slinky be oscillating?

Two sinusoidal waves are moving through a medium in the same direction, both having amplitudes of \(3.00 \mathrm{cm}, \mathrm{a}\) wavelength of \(5.20 \mathrm{m},\) and a period of \(6.52 \mathrm{s}\), but one has a phase shift of an angle \(\phi\). What is the phase shift if the resultant wave has an amplitude of \(5.00 \mathrm{cm} ?[\)Hint: Use the trig identity \(\sin u+\sin v=2 \sin \left(\frac{u+v}{2}\right) \cos \left(\frac{u-v}{2}\right)\).

Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where low frequency sound is originating from. (a) Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 18 \(\mathrm{cm}\) apart, and the speed of sound generated is \(340 \mathrm{m} / \mathrm{s}\). How long is the interval between when the sound arrives at the right ear and the sound arrives at the left ear? (b) Assume the same person was scuba diving and a low- frequency sound source was to the right of the scuba diver. How long is the interval between when the sound arrives at the right ear and the sound arrives at the left ear, if the speed of sound in water is \(1500 \mathrm{m} / \mathrm{s}\) ? (c) What is significant about the time interval of the two situations?

A sinusoidal transverse wave has a wavelength of 2.80 m. It takes 0.10 s for a portion of the string at a position \(x\) to move from a maximum position of \(y=0.03 \mathrm{m}\) to the equilibrium position \(y=0 .\) What are the period, frequency, and wave speed of the wave?

The note \(E_{4}\) is played on a piano and has a frequency of \(f=393.88 .\) If the linear mass density of this string of the piano is \(\mu=0.012 \mathrm{kg} / \mathrm{m}\) and the string is under a tension of \(1000.00 \mathrm{N},\) what is the speed of the wave on the string and the wavelength of the wave?

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