The ends of a stretched wire of length L are fixed at x = 0 and x = L. In one experiment the displac

Q: Why can we use the maximum kinetic energy to find the total energy of the string?

By conservation of mechanical energy, the total energy of a conservative system undergoing oscillation is constant. When the string passes through its mean position ($y=0$), its elastic potential energy is zero, meaning all of its energy is in the form of kinetic energy. Thus, the maximum kinetic energy equals the total mechanical energy.

Q: Does the shape of the wave envelope affect the energy integral?

Yes, the spatial distribution of the amplitude affects the integral. However, for sinusoidal envelopes like $\sin(\pi x/L)$ and $\sin(2\pi x/L)$, the integral of $\sin^2$ over their respective integer half-cycles always yields $L/2$. Thus, the spatial part contributes equally in both cases.

Q: Why is the frequency of the second wave $2\omega$ instead of $\omega$?

The time-dependent term in the equation for $y_2$ is given as $\sin(2\omega t)$, which explicitly defines its angular frequency as $2\omega$. This is physically consistent with the second harmonic of a string fixed at both ends, where the frequency is twice the fundamental frequency.

Q: What if the amplitude of the second wave was also different, say $2A$?

Since energy scales quadratically with both amplitude and frequency ($E \propto A^2 \omega^2$), if the amplitude doubled ($2A$) and the frequency doubled ($2\omega$), the energy would increase by a factor of $2^2 \times 2^2 = 16$ times.

Q: Is this quadratic energy scaling valid for traveling waves as well?

Yes, for traveling waves, the energy density and intensity (power per unit area) also scale quadratically with both amplitude and frequency ($I \propto A^2 \omega^2$). This is a fundamental property of all wave motion.

Question

The ends of a stretched wire of length $L$ are fixed at $x = 0$ and $x = L$. In one experiment the displacement of the wire is $y_1 = A \sin\left(\frac{\pi x}{L}\right) \sin\omega t$ and energy is $E_1$ and in other experiment its displacement is $y_2 = A \sin\left(\frac{2\pi x}{L}\right) \sin 2\omega t$ and energy is $E_2$. Then

Options:

(A) $E_2 = E_1$

(B) $E_2 = 2E_1$

(C) $E_2 = 4E_1$

(D) $E_2 = 16E_1$

Sigma Percentile · Accepted Answer

### Introduction

Imagine a guitar string plucked and vibrating. The beautiful sound we hear is the result of **standing waves** trapped between the fixed ends of the string. But beyond the sound, these waves carry a physical quantity of immense importance: **energy**.

In this problem, we explore how the energy of a standing wave scales when we transition from the fundamental mode to a higher harmonic. By analyzing the mathematical descriptions of two different vibration experiments on the same stretched string, we will uncover a fundamental scaling law of wave mechanics.

---

### The Physics of Energy in Standing Waves

To understand how energy is distributed in a standing wave, we must look at a small element of the string of mass $dm$. As the string vibrates, each element undergoes simple harmonic motion (SHM) perpendicular to the length of the string.

The total mechanical energy $E$ of any system undergoing SHM is constant and is equal to its maximum kinetic energy. This maximum kinetic energy occurs when the string passes through its equilibrium (mean) position, where its potential energy is zero.

For an element of length $dx$ and mass $dm = \mu dx$ (where $\mu$ is the linear mass density), vibrating with a local amplitude $A(x)$ and angular frequency $\omega$, the maximum velocity is:

$$v_{	ext{max}}(x) = A(x) \omega$$

Thus, the maximum kinetic energy of this small element is:

$$dE = \frac{1}{2} dm (v_{	ext{max}})^2 = \frac{1}{2} (\mu dx) [A(x) \omega]^2$$

To find the total energy of the entire string, we integrate this expression over the length of the string from $x = 0$ to $x = L$:

$$E = \int_0^L \frac{1}{2} \mu \omega^2 [A(x)]^2 dx$$

This is our **master equation** for the energy of a standing wave.

---

### Analyzing the First Experiment

In the first experiment, the displacement of the wire is given by:

$$y_1 = A \sin\left(\frac{\pi x}{L}ight) \sin\omega t$$

From this equation, we can identify:
- The spatial amplitude function: $A_1(x) = A \sin\left(\frac{\pi x}{L}ight)$
- The angular frequency: $\omega_1 = \omega$

Substituting these into our master energy equation gives:

$$E_1 = \frac{1}{2} \mu \omega^2 \int_0^L A^2 \sin^2\left(\frac{\pi x}{L}ight) dx$$

Using the standard trigonometric integral identity $\int_0^L \sin^2\left(\frac{\pi x}{L}ight) dx = \frac{L}{2}$, we compute the energy $E_1$:

$$E_1 = \frac{1}{2} \mu \omega^2 A^2 \left(\frac{L}{2}ight) = \frac{1}{4} \mu A^2 L \omega^2$$

This represents the energy of the string vibrating in its **fundamental mode** (first harmonic).

---

### Analyzing the Second Experiment

In the second experiment, the displacement is given by:

$$y_2 = A \sin\left(\frac{2\pi x}{L}ight) \sin 2\omega t$$

Here, we identify:
- The spatial amplitude function: $A_2(x) = A \sin\left(\frac{2\pi x}{L}ight)$
- The angular frequency: $\omega_2 = 2\omega$

Substituting these values into our master energy equation yields:

$$E_2 = \frac{1}{2} \mu (2\omega)^2 \int_0^L A^2 \sin^2\left(\frac{2\pi x}{L}ight) dx$$

Simplifying the terms:

$$E_2 = \frac{1}{2} \mu (4\omega^2) A^2 \int_0^L \sin^2\left(\frac{2\pi x}{L}ight) dx$$

Since the integral of $\sin^2$ over two full cycles (from $0$ to $L$) is also $\frac{L}{2}$, we get:

$$E_2 = \frac{1}{2} \mu (4\omega^2) A^2 \left(\frac{L}{2}ight) = \mu A^2 L \omega^2$$

This is the energy of the string vibrating in its **second harmonic**.

---

### The Grand Comparison

Now, let us compare the two energies by taking their ratio:

$$\frac{E_2}{E_1} = \frac{\mu A^2 L \omega^2}{\frac{1}{4} \mu A^2 L \omega^2} = 4$$

Thus, we find:

$$E_2 = 4E_1$$

This elegant result shows that the energy of the second harmonic is exactly **four times** the energy of the fundamental mode when the maximum amplitude $A$ is kept constant. This perfectly matches **option (c)**.

---

### Deep Intuition & Scaling Laws

Why did the energy quadruple?

Notice that the spatial integral for both modes yielded the same value, $\frac{L}{2}$. This is because, although the second harmonic has twice as many loops, the total spatial "average" of the squared amplitude over the entire string remains identical.

Therefore, the only factor that caused the energy to change was the **frequency**. Since energy is proportional to the square of the frequency ($E \propto \omega^2$), doubling the frequency from $\omega$ to $2\omega$ increases the energy by a factor of $2^2 = 4$.

This quadratic scaling with frequency is a universal hallmark of wave energy, whether in mechanical strings, sound waves, or electromagnetic radiation!

Select Answer:

Visualized Solution

Visual Anchor (Orient)

Logic Bridge (Tool)

Raw Setup (Integration)

Analyze Experiment 1

Compute Energy $E_{1}$

Analyze Experiment 2

Compute Energy $E_{2}$

Compare Energies

Final Answer

The Way Forward

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction

The Physics of Energy in Standing Waves

Analyzing the First Experiment

Analyzing the Second Experiment

The Grand Comparison

Deep Intuition & Scaling Laws

Similar Questions

A wire of length $L$ and mass per unit length $6.0 \times 1 0^{- 3} kgm^{- 1}$ is put under tension of $540 N$ . Two consecutive frequencies that it resonates at are : $420 Hz$ and $490 Hz$ . Then, $L$ in metres is

A wave $y = a sin (ω t - k x)$ on a string meets with another wave producing a node at $x = 0$ . Then, the equation of the unknown wave is

A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is $Y = 0.3 sin (0.157 x) cos (200 π t)$ . The length of the string is (All quantities are in SI units)