Question

A wire having length \(\mathrm{L}\) is kept under tension between \(\mathrm{x}=0\) and \(\mathrm{x}=\mathrm{L}\). In one experiment, the equation of the wave and energy is given by \(\mathrm{y}_{1}=\mathrm{A} \sin (\pi \mathrm{x} / \mathrm{L}) \sin \omega \mathrm{t}\) and \(\mathrm{E}_{1}\) respectively. In another experiment, it is \(\mathrm{y}_{2}=\mathrm{A} \sin\) \(\\{(2 \pi \mathrm{x}) / \mathrm{L}\\} \sin 2 \omega \mathrm{t}\) and \(\mathrm{E}_{2}\). Then........ (A) \(E_{2}=E_{1}\) (B) \(E_{2}=2 \mathrm{E}_{1}\) (C) \(\mathrm{E}_{2}=4 \mathrm{E}_{1}\) (D) \(E_{2}=16 \mathrm{E}_{1}\)

Short answer

The correct answer is (C) \(E_2 = 4E_1\).

Step-by-step solution

Calculate energy expression for the first wave

We are given the wave equation for the first experiment as: \(y_1 = A \sin(\frac{\pi x}{L}) \sin(\omega t)\) To find the energy expression for this wave, we know that the wave energy can be expressed as: \(E = \frac{1}{2} \mu \int_0^L (\frac{\partial y}{\partial t})^2 dx\) First, we need to find the time derivative of the wave equation: \(\frac{\partial y_1}{\partial t} = A \sin(\frac{\pi x}{L}) \cos(\omega t) \cdot \omega\) Now, we will square the time derivative: \((\frac{\partial y_1}{\partial t})^2 = \omega^2 A^2 \sin^2(\frac{\pi x}{L})\cos^2(\omega t)\) Now we can find the energy expression for the first experiment by plugging this value into the energy expression formula: \(E_1 = \frac{1}{2} \mu \int_0^L \omega^2 A^2 \sin^2(\frac{\pi x}{L})\cos^2(\omega t) dx\)

Calculate energy expression for the second wave

We are given the wave equation for the second experiment as: \(y_2 = A \sin(\frac{2 \pi x}{L}) \sin(2\omega t)\) Now, similar to the first wave, we will find the time derivative of the wave equation: \(\frac{\partial y_2}{\partial t} = A \sin(\frac{2 \pi x}{L}) \cos(2\omega t) \cdot 2\omega\) Then we square the time derivative: \((\frac{\partial y_2}{\partial t})^2 = 4\omega^2 A^2 \sin^2(\frac{2\pi x}{L})\cos^2(2\omega t)\) Now we can find the energy expression for the second experiment: \(E_2 = \frac{1}{2} \mu \int_0^L 4\omega^2 A^2 \sin^2(\frac{2\pi x}{L})\cos^2(2\omega t) dx\)

Compare energy expressions of both waves to find the relationship

Now we have the energy expressions for both the first and second experiments: \(E_1 = \frac{1}{2} \mu \int_0^L \omega^2 A^2 \sin^2(\frac{\pi x}{L})\cos^2(\omega t) dx\) \(E_2 = \frac{1}{2} \mu \int_0^L 4\omega^2 A^2 \sin^2(\frac{2\pi x}{L})\cos^2(2\omega t) dx\) We can see that the only difference between these two expressions are the coefficients of \(\omega^2 A^2\) in the integral: First integral has \(\omega^2 A^2\), while the second integral has \(4\omega^2 A^2\). Therefore, the energy expressions are related to each other as: \(E_2 = 4E_1\) Thus, the correct answer is: (C) \(E_2=4E_1\)

Key concepts

Harmonic Waves

Harmonic waves are a fundamental concept in wave mechanics. These waves are periodic, meaning they repeat their shape at regular intervals along their axis. In simple terms, think of a harmonic wave as a perfectly shaped sine or cosine wave that travels through a medium. Each point on the wave oscillates in a simple harmonic motion, with a fixed amplitude and frequency.

These waves are described mathematically by functions like sine or cosine, which capture their repetitive nature. This periodic behavior is common in many physical situations, such as sound waves in air or waves on a string.

In the context of the problem, two types of harmonic waves are presented, one with a frequency \( \omega \) and another with \( 2 \omega \). This doubling of frequency results in different wave behaviors and energy, as we'll explore further.

Wave Equation

The wave equation is a key mathematical tool in describing how waves propagate through different mediums. It typically has the form \( y(x, t) = A \sin(kx - \omega t)\), where \( y \) is the displacement, \( A \) is the amplitude, \( k \) is the wave number, \( x \) is the position, \( \omega \) is the angular frequency, and \( t \) is the time.

For the given exercises, the wave equations are given for waves on a wire. We see that they take the form \( y_1 = A \sin(\frac{\pi x}{L}) \sin(\omega t) \) for the first experiment and \( y_2 = A \sin(\frac{2 \pi x}{L}) \sin(2 \omega t) \) for the second.

• This reflects the wavelength and frequency differences between the two experiments.
• These changes alter not only the wave propagation but also how energy is distributed and calculated.

Understanding the wave equation allows us to predict wave movements and interactions, crucial for both theoretical and practical applications.

Wave Energy

Wave energy is the measure of the energy carried by a wave as it propagates through a medium. In general, a wave’s energy is related to the square of its amplitude and its frequency.

The problem involves comparing the energy of two different waves, where the second wave possesses a higher frequency. The energy expressions are derived from the wave equations by considering the time derivatives and squaring them.

Here's the process to find the wave energy:

• First, calculate the partial derivative of the wave with respect to time to get the velocity of the particles in the medium.
• Square this derivative to relate it to the energy density.
• Integrate over the period to find the total energy.

In this problem, since the frequency and wave number increase for the second experiment, its energy becomes \( E_2 = 4E_1 \), meaning the energy quadruples compared to the first.

Tension in Wire

The tension in a wire plays a crucial role in wave propagation. It affects both the speed and behavior of waves traveling through it. When a wire is tightened or the tension is increased, the wave speed increases, and the frequency of the wave can also be altered.

For harmonic waves along a wire, the wave speed \( v \) is related to the tension \( T \) and the mass per unit length \( \mu \) of the wire through the relationship \( v = \sqrt{\frac{T}{\mu}} \). This equation shows how an increase in tension results in higher wave speeds.

• High tension means waves travel faster, leading to higher frequencies or shorter wavelengths.
• This impacts how the wave energy is distributed and affects the overall wave dynamics on the wire.

By understanding the tension in the wire and its effects, we can control and predict wave behavior in many practical applications, from musical instruments to engineering projects.