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Chapter 9: Problem 75

A wave is represented by the equation: \(y=0.1 \sin (100 \pi t-k x)\). If wave velocity is \(100 \mathrm{~m} / \mathrm{s}\), its wave number is equal to (A) \(1 \mathrm{~m}^{-1}\) (B) \(2 \mathrm{~m}^{-1}\) (C) \(\pi \mathrm{m}^{-1}\) (D) \(2 \pi \mathrm{m}^{-1}\)

Short Answer

Expert verified
The wave number (k) can be calculated by using the given wave velocity (v = 100 m/s) and the frequency (f) extracted from the wave equation. First, find the wavelength (\(\lambda\)) using the equation \(v = f\lambda\), then calculate the wave number using the equation \(k = \dfrac{2\pi}{\lambda}\). The wave number is equal to \(\pi m^{-1}\) (option C).

Step by step solution

01

Identifying the given information from the wave equation

The wave equation is given by: \(y=0.1 \sin (100 \pi t-kx)\) From the equation: Amplitude (A) = 0.1 Frequency term (inside the sine function) = \(100 \pi t\) Wave number term (inside the sine function) = \(kx\) Wave velocity (v) = \(100 m/s\) We need to find the value of the wave number (k).
02

Write the equation relating wave velocity, frequency, and wavelength:

We know that the wave velocity (v) is related to the frequency (f) and wavelength (\(\lambda\)) by: \(v = f\lambda\)
03

Define the relationship between wave number, frequency, and wavelength:

We know that wave number, frequency, and wavelength have the following relationships: \(k = \dfrac{2\pi}{\lambda}\) \(f = \dfrac{2\pi}{T}\) where T is the time period and k is the wave number.
04

Extract frequency from the wave equation:

We look at the frequency term inside the sine function (\(100 \pi t\)) and note that the frequency is given as follows: \(f= \dfrac{100\pi}{2\pi} = 50Hz\)
05

Calculate the wavelength:

From the equation \(v = f\lambda\), we can find the wavelength (\(\lambda\)): \(\lambda = \dfrac{v}{f} = \dfrac{100}{50} = 2m\)
06

Calculate the wave number:

Now, using the equation \(k = \dfrac{2\pi}{\lambda}\), we can find the wave number (k): \(k = \dfrac{2\pi}{2} = \pi m^{-1}\) So, the wave number is equal to \(\pi m^{-1}\), which is option (C).

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Key Concepts

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

Wave Equation
The wave equation is a mathematical description of a moving wave. It represents the displacement of the wave at a given position and time. When it comes to periodic waves鈥攕uch as sound waves, light waves, and water waves鈥攖he wave equation often takes the form of a sinusoidal function. In our exercise, the wave equation is given as:
\(y=0.1 \sin (100 \pi t-k x)\)

This equation tells us that the wave displacement y varies with time t and position x. Here, the amplitude is 0.1, indicating the maximum displacement of the wave from its rest position. The term within the sine function includes the angular frequency \(100 \pi\) times the time variable t, and the wave number k times the position variable x. The angular frequency is related to the rate at which the wave oscillates, while the wave number relates to the spatial frequency of the wave.
Wave Velocity
Wave velocity, or phase velocity, is the speed at which the crests and troughs of a wave propagate through space. It is a fundamental property determining how fast the wave travels. The wave equation can be used to learn about this velocity when combined with the frequency and wavelength of the wave.

Using the relationship \(v = f\lambda\), we calculate the wave velocity by multiplying the frequency (f) with the wavelength (\(\lambda\)). In the context of our exercise, the wave velocity is given as 100 m/s, providing a crucial piece of information to find other wave properties such as wavelength and wave number.
Wavelength and Frequency Relationship
The relationship between wavelength and frequency is critical to understanding wave properties and behaviors. They are inversely related to each other鈥攁s the frequency of a wave increases, its wavelength decreases, and vice versa. Mathematically, the wavelength is given by \(\lambda = \dfrac{v}{f}\), where v is the wave velocity and f is the frequency.

In the given exercise, we first calculated the frequency by examining the wave equation's angular frequency term. To find the wavelength, we rearranged the relationship and divided the wave velocity by the frequency. This calculation provided the necessary values to compute the wave number by using the relationship \(k = \dfrac{2\pi}{\lambda}\). Knowing these key concepts helps in understanding how different parameters of a wave are interconnected and how changing one affects the others.

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Most popular questions from this chapter

A particle moves on the \(x\)-axis as per the equation \(x=x_{0} \sin ^{2} \omega t .\) The motion is simple harmonic (A) With amplitude \(x_{0}\) (B) With amplitude \(2 x_{0}\) (C) With time period \(\frac{2 \pi}{\omega}\) (D) With time period \(\frac{\pi}{\omega}\)

A particle of mass \(m\) moves in a straight line. If \(v\) is the velocity at a distance \(x\) from a fixed point on the line and \(v^{2}=a-b x^{2}\), where \(a\) and \(b\) are constant, then (A) The motion continues along the positive \(x\)-direction only. (B) The motion is simple harmonic. (C) The particle oscillates with a frequency equal to \(\frac{\sqrt{b}}{2 \pi}\) (D) The total energy of the particle is \(m a\).

The equation of a particle executing SHM is given by \(x=3 \cos \left(\frac{\pi}{2}\right) t \mathrm{~cm}\), where \(t\) is in second. The distance travelled by the particle in the first \(8.5 \mathrm{~s}\) is \(\left(24+\frac{n}{\sqrt{2}}\right)\) then the value of \(n\) is.

Equation of a plane progressive wave is given by \(y=0.6 \sin 2 \pi\left(t-\frac{x}{2}\right)\). On reflection from a denser medium, its amplitude becomes \(\frac{2}{3}\) of the amplitude of the incident wave. The equation of the reflected wave is (A) \(y=0.6 \sin 2 \pi\left(t+\frac{x}{2}\right)\) (B) \(y=-0.4 \sin 2 \pi\left(t+\frac{x}{2}\right)\) (C) \(y=0.4 \sin 2 \pi\left(t+\frac{x}{2}\right)\) (D) \(y=-0.4 \sin 2 \pi\left(t-\frac{x}{2}\right)\)

A particle starts SHM at time \(t=0 .\) Its amplitude is \(A\) and angular frequency is \(\omega .\) At time \(t=0\), its kinetic energy is \(\frac{E}{4}\), where \(E\) is total energy. Assuming potential energy to be zero at mean position, the displacement-time equation of the particle can be written as (A) \(x=A \cos \left(\omega t+\frac{\pi}{6}\right)\) (B) \(x=A \sin \left(\omega t+\frac{\pi}{3}\right)\) (C) \(x=A \sin \left(\omega t-\frac{2 \pi}{3}\right)\) (D) \(x=A \cos \left(\omega t-\frac{\pi}{6}\right)\)
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