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A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation y(x,t)=(8.50mm)cos(172rad/m x4830rad/s t) Assume that the tension of the string is constant and equal to W. (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight W? (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling down the string?

Short Answer

Expert verified
(a) 0.0534 s, (b) 6.565 N, (c) 41 wavelengths, (d) y(x,t)=(8.50mm)cos(172rad/mx+4830rad/st).

Step by step solution

01

Identify Given Information

We know the length of the string, L=1.50 m, and the weight of the string is 0.0125 N. The wave equation for the string is given by y(x,t)=(8.50 mm)cos(172 rad/m x4830 rad/s t). From the wave equation, we identify the wave number k=172 rad/m and the angular frequency ω=4830 rad/s.
02

Calculate Wave Speed (a)

The wave speed v can be derived from the wave equation using v=ωk. Given ω=4830 rad/s and k=172 rad/m, we calculate:v=4830172=28.081 m/s
03

Time for a Pulse to Travel Full Length (a)

The time T it takes for a pulse to travel the full length of the string is given by T=Lv. Using L=1.50 m and v=28.081 m/s:T=1.5028.0810.0534 s
04

Calculate Tension and Weight (b)

The tension in the string is equal to the weight W. The wave speed can also be given by v=Tμ, where μ is the linear mass density of the string. To find μ, use the weight of the string and its length: μ=0.01251.50=0.00833 kg/m.Solve for T:v2=TμT=v2μ=(28.081)2×0.008336.565 NHence, W=6.565 N.
05

Wavelength on the String (c)

The wavelength λ is given by λ=2πk. Plug in k=172 rad/m:λ=2π1720.0365 mTo find how many wavelengths are on the string, use L=1.50 m:n=Lλ=1.500.036541.1Therefore, approximately 41 full wavelengths fit on the string.
06

Equation for Waves Traveling Down the String (d)

For waves traveling down the string, the wave number remains the same, but the sign of the angular frequency changes:y(x,t)=(8.50 mm)cos(172 rad/m x+4830 rad/s t)

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Key Concepts

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

Wave Speed
Wave speed is a fundamental concept in wave mechanics that describes how quickly a wave can travel through a medium. In simple terms, it tells us how fast the energy or information carried by the wave moves from one point to another.

To calculate the wave speed (v), we can use information from the wave equation. The formula to find wave speed is:
  • v=ωk
where ω is the angular frequency, and k is the wave number.

In the original exercise, the angular frequency is 4830 rad/s and the wave number is 172 rad/m. By applying our formula, the wave speed becomes:
  • v=4830172=28.081 m/s
This calculation shows us how fast the wave pulse moves along the string.
Wavelength
A wavelength is an essential concept in understanding waves. It's the distance over which the wave's shape repeats. You can visualize it as the length from crest to crest or trough to trough of the wave.

We can find the wavelength (λ) using the wave number k. The relationship between them is given by the formula:
  • λ=2πk
Given k=172 rad/m:
  • λ=2π1720.0365 m
This means that each wavelength is approximately 0.0365 meters long.

The original exercise also finds how many wavelengths fit on the string. For a string of length 1.50 meters:
  • n=1.500.036541.1
So, about 41 waves fit perfectly along the length of the string at any moment.
Tension in String
The tension in a string is a force that stretches it, playing a crucial role in determining how waves travel along it. In physical terms, the tension keeps the string tight and affects the speed of wave propagation in the string.

To connect tension (T), wave speed (v), and linear mass density (μ) of the string, we use the equation:
  • v=Tμ
The linear mass density, μ, is the mass per unit length and in this problem is 0.00833 kg/m.

Rearranging to find tension T gives:
  • T=v2μ
Plug in the values from the original exercise (v=28.081 m/s,μ=0.00833 kg/m):
  • T=(28.081)2×0.008336.565 N
Hence, the tension is approximately 6.565 N, equivalent to the supporting weight W, ensuring the string operates under constant tension.
Wave Equation
The wave equation is a mathematical representation that describes the characteristics of wave behavior along a medium.

For our string scenario, the provided wave equation is:
  • y(x,t)=(8.50 mm)cos(172 rad/m x4830 rad/s t)
This equation tells us the wave's movement along the string over time.

Different elements in the wave equation include:
  • The amplitude (8.50 mm), representing the maximum displacement of wave particles from their rest position.
  • The wave number (172 rad/m), which is linked to wavelength as we discussed.
  • The angular frequency (4830 rad/s), showing how rapidly the wave oscillates.
For waves traveling downwards, the formula is adjusted slightly:
  • y(x,t)=(8.50 mm)cos(172 rad/m x+4830 rad/s t)
Here, the sign of the angular frequency changes, indicating a different direction for the wave.

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Most popular questions from this chapter

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?

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?

A wave on a string is described by y(x,t)=Acos(kxωt). (a) Graph y,vy, and ay as functions of x for time t=0. (b) Consider the following points on the string: (i) x= 0; (ii) x=π/4k; (iii) x=π/2k; (iv) x=3π/4k; (v) x=πk; (vi) x=5π/4k; (vii) x=3π/2k; (viii) x=7π/4k. For a particle at each of these points at t=0, describe in words whether the particle is moving and in what direction, and whether the particle is speeding up, slowing down, or instantaneously not accelerating.

One string of a certain musical instrument is 75.0 cm long and has a mass of 8.75 g. It is being played in a room where the speed of sound is 344 m/s. (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength 0.765 m? (Assume that the breaking stress of the wire is very large and isn't exceeded.) (b) What frequency sound does this string produce in its fundamental mode of vibration?

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)=2.30mmcos[(6.98rad/m)x +1742rad/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.

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