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Chapter 7: Problem 51

A \(2 \mathrm{~m}\) -long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is \(200 \mathrm{~m} \mathrm{~s}^{-1}\) and the amplitude is \(0 \cdot 5 \mathrm{~cm}\). (a) Find the wavelength and the frequency. (b) Write the equation giving the displacement of different points as a function of time. Choose the \(X\) -axis along the string with the origin at one end and \(t=0\) at the instant when the point \(x=50 \mathrm{~cm}\) has reached its maximum displacement.

Short Answer

Expert verified
(a) The wavelength is 2 m and the frequency is 100 Hz. (b) The displacement equation is \( y(x, t) = 0.005 \sin(\pi x) \cos(200\pi t) \).

Step by step solution

01

Understand the Problem

We need to find the wavelength and the frequency of the first overtone of a string that is fixed at both ends, which is vibrating. We also need to write the equation of motion of the string. The given data includes the length of the string, the wave speed, and the amplitude.
02

Determine the Wavelength

For a string fixed at both ends, the first overtone is the second harmonic, which means the string has a node at both ends and an antinode in the middle. The length of the string is equal to one wavelength of the first overtone.Given that the string length is \(2 \text{ m}\), the wavelength \(\lambda\) is \(2 \text{ m}\).
03

Calculate the Frequency

The wave speed \(v\) is given as \(200 \text{ m/s}\). To find the frequency \(f\), we use the wave equation:\[ v = f \times \lambda \]From the above, \(200 = f \times 2\), solve for \(f\):\[ f = \frac{200}{2} = 100 \text{ Hz} \]
04

Write the Wave Equation

The general form of the wave equation for displacement \(y(x,t)\) is:\[ y(x, t) = A \sin(kx) \cos(\omega t + \phi) \]where \(A\) is the amplitude, \(k\) is the wave number, and \(\omega\) is the angular frequency. For this example, the amplitude \(A\) is \(0.5 \text{ cm} = 0.005 \text{ m}\), wave number \(k = \frac{2\pi}{\lambda}\), angular frequency \(\omega = 2\pi f\).Thus, \(k = \frac{2\pi}{2} = \pi\) and \(\omega = 2\pi \times 100 = 200\pi\).The initial condition \(t = 0\) when \(x = 50 \text{ cm} = 0.5 \text{ m}\) reaches maximum displacement gives us \(\phi = 0\). Therefore, the displacement equation is:\[ y(x, t) = 0.005 \sin(\pi x) \cos(200\pi t) \]

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Key Concepts

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

Wavelength Calculation
In wave motion, understanding the wavelength is crucial. The wavelength (\( \lambda \)) is the distance between two consecutive points that are in phase, such as two adjacent crests of a wave. For a string fixed at both ends, the first overtone is the second harmonic. This means that there will be one antinode and two nodes along the string鈥檚 length.
In this exercise, the string is\( 2 \mathrm{~m} \)in length. With the first overtone being the second harmonic, the length of the string is one full wavelength. Thus,\( \lambda = 2 \mathrm{~m} \). This connection between the string's length and wavelength is a common characteristic of standing waves on fixed strings.
Remember:
  • In the second harmonic, the string vibrates in such a way that it forms one complete wave.
  • The wavelength can be visually observed as the distance from one antinode to the next.
Frequency Determination
Frequency is the number of cycles a wave completes in one second, and it is measured in Hertz (Hz). To calculate the frequency (\( f \)) of the wave on the string, the wave equation\( v = f \times \lambda \)can be used, where\( v \)is the wave speed.
For the given problem:
  • The wave speed\( v \)is\( 200 \mathrm{~m/s} \)
  • The wavelength\( \lambda \)is\( 2 \mathrm{~m} \)
Solving the equation\( v = f \times 2 \), we find:\( f = \frac{200}{2} = 100 \text{ Hz} \)
It's crucial to note that:
  • The frequency determines how fast the wave oscillates.
  • Higher frequencies result in waves that repeat more quickly.
  • In this scenario, the frequency of\( 100 \text{ Hz} \)means the wave repeats 100 times every second.
Wave Equation
Wave equations allow us to describe the motion of waves mathematically. For a string fixed at both ends, the displacement of any point on the string can be modeled using the wave equation. This equation gives a detailed picture of how waves move over time and space.
The standard form of the wave equation is:\[ y(x, t) = A \sin(kx) \cos(\omega t + \phi) \]Here's how each component works:
  • \( A \)is the amplitude, indicating the maximum displacement of the wave, given as\( 0.5 \mathrm{~cm} = 0.005 \mathrm{~m} \).
  • \( k \)is the wave number, calculated as\( k = \frac{2\pi}{\lambda} \).For this problem,\( \lambda \)is\( 2 \mathrm{~m} \), so\( k = \pi \).
  • \( \omega \)is the angular frequency, found using\( \omega = 2\pi f \).Given\( f = 100 \mathrm{~Hz} \), we find\( \omega = 200\pi \).
    • The result:\[ y(x, t) = 0.005 \sin(\pi x) \cos(200\pi t) \]This equation tells us the displacement at any point\( x \) at any time\( t \). Note that the wave has a maximum displacement at\( x = 0.5 \mathrm{~m} \),when\( t = 0 \), reflecting the initial condition given.

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Most popular questions from this chapter

Two long strings \(A\) and \(B\), each having linear mass density \(1 \cdot 2 \times 10^{-2} \mathrm{~kg} \mathrm{~m}^{-1}\), are stretched by different tensions \(4 \cdot 8 \mathrm{~N}\) and \(7 \cdot 5 \mathrm{~N}\) respectively and are kept parallel to each other with their left ends at \(x=0 .\) Wave pulses are produced on the strings at the left ends at \(t=0\) on string \(A\) and at \(t=20 \mathrm{~ms}\) on string \(B .\) When and where will the pulse on \(B\) overtake that on \(A\) ?

Two wires of different densities but same area of cross section are soldered together at one end and are stretched to a tension \(T\). The velocity of a transverse wave in the first wire is double of that in the second wire. Find the ratio of the density of the first wire to that of the second wire.

A \(2 \cdot 00 \mathrm{~m}\) -long rope, having a mass of \(80 \mathrm{~g}\), is fixed at one end and is tied to a light string at the other end. The tension in the string is \(256 \mathrm{~N}\). (a) Find the frequencies of the fundamental and the first two overtones. (b) Find the wavelength in the fundamental and the first two overtones.

A heavy ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at a speed of \(60 \mathrm{~cm} \mathrm{~s}^{-1}\) on the string when the car is at rest and \(62 \mathrm{~cm} \mathrm{~s}^{-1}\) when the car accelerates on a horizontal road. Find the acceleration of the car. Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).

A circular loop of string rotates about its axis on a frictionless horizontal plane at a uniform rate so that the tangential speed of any particle of the string is \(v .\) If a small transverse disturbance is produced at a point of the loop, with what speed (relative to the string) will this disturbance travel on the string?
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