A string of mass per unit length μ is clamped at both ends such that one end of the string is at x =

Q: Why is the total energy calculated using only the maximum kinetic energy?

In simple harmonic motion, energy is continuously transferred between kinetic and potential forms. At the instant the string passes through its equilibrium position, its potential energy is zero, and its kinetic energy is at a maximum. Therefore, the maximum kinetic energy of the system is exactly equal to its total mechanical energy.

Q: Why does the mass density $\mu$ cancel out in the final energy expression?

While a heavier string (higher $\mu$) has more mass to move (increasing kinetic energy), it also vibrates at a lower frequency for a given tension ($f \propto 1/\sqrt{\mu}$). Since energy depends on $f^2$, the frequency reduction perfectly offsets the mass increase, making the total energy independent of $\mu$.

Q: What would change if the string vibrated in the second harmonic instead of the fundamental mode?

In the second harmonic, the wavelength becomes $\lambda = l$, and the frequency doubles. If the maximum amplitude remains $a$, the total energy would increase by a factor of 4 because energy is proportional to the square of the frequency.

Q: How do we integrate $\sin^2(\pi x / l)$ over the interval $[0, l]$?

We use the identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$. Integrating this from $0$ to $l$ gives $\int_0^l \frac{1}{2} dx - \int_0^l \frac{\cos(2\pi x / l)}{2} dx$. The cosine term integrates to zero over a full cycle, leaving only $\frac{l}{2}$.

Q: Is the tension $T$ assumed to be constant throughout the string's vibration?

Yes, for small-amplitude oscillations, the change in the length of the string is negligible, so the tension $T$ remains constant and uniform along the string.

Question

A string of mass per unit length $\mu$ is clamped at both ends such that one end of the string is at $x = 0$ and the other is at $x = l$. When string vibrates in fundamental mode, amplitude of the mid-point $O$ of the string is $a$, and tension in the string is $T$. Find the total oscillation energy stored in the string.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a uniform string of length $l$ tightly clamped at both ends, $x = 0$ and $x = l$.

When we pluck this string, waves travel back and forth, reflecting at the boundaries.

Because the ends are clamped, they must remain completely stationary, forming **nodes** at $x = 0$ and $x = l$.

When the string vibrates in its **fundamental mode** (also known as the first harmonic), it forms a single, beautiful vibrating loop.

This physical constraint dictates that the length of the string $l$ is exactly equal to half of the wavelength $\lambda$:

$$l = \frac{\lambda}{2} \implies \lambda = 2l$$

From this, we can immediately determine the wave number $k$, which represents the spatial frequency of the wave:

$$k = \frac{2\pi}{\lambda} = \frac{2\pi}{2l} = \frac{\pi}{l}$$

### The Amplitude Equation

For any standing wave, the amplitude of oscillation is not uniform; it varies sinusoidally along the length of the string.

We can write the amplitude at any position $x$ as:

$$A(x) = A_0 \sin(kx)$$

We are given that the amplitude at the midpoint $O$ (where $x = l/2$) is $a$.

Let's substitute this boundary condition to find the maximum amplitude $A_0$:

$$A\left(\frac{l}{2}ight) = A_0 \sin\left(\frac{\pi}{l} \cdot \frac{l}{2}ight) = A_0 \sin\left(\frac{\pi}{2}ight) = A_0 = a$$

This tells us that the peak amplitude of the standing wave is indeed $a$.

Thus, the amplitude at any point $x$ along the string is:

$$A(x) = a \sin\left(\frac{\pi x}{l}ight)$$

### Energy of an Infinitesimal Element

To find the total energy stored in the vibrating string, we must look at a tiny, infinitesimal segment of length $dx$ located at position $x$.

The mass of this tiny element is given by:

$$dm = \mu dx$$

where $\mu$ is the mass per unit length of the string.

As this element oscillates up and down in simple harmonic motion, its energy continuously converts between kinetic and potential energy.

The total mechanical energy $dE$ of this element is equal to its maximum kinetic energy, which occurs when it passes through the equilibrium position with maximum velocity:

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

where $\omega = 2\pi f$ is the angular frequency of vibration.

### Integrating for the Total Energy

Now, let's substitute the amplitude function $A(x)$ into our energy equation:

$$dE = \frac{1}{2} \mu \omega^2 \left[a \sin\left(\frac{\pi x}{l}ight)ight]^2 dx = 2\pi^2 \mu f^2 a^2 \sin^2\left(\frac{\pi x}{l}ight) dx$$

To find the total energy $E$ stored in the entire string, we must sum (integrate) the energy of all such infinitesimal elements from $x = 0$ to $x = l$:

$$E = \int_0^l 2\pi^2 \mu f^2 a^2 \sin^2\left(\frac{\pi x}{l}ight) dx$$

Before integrating, let's express the frequency $f$ in terms of the physical properties of the string: tension $T$ and mass density $\mu$.

The velocity of a transverse wave in a stretched string is $v = \sqrt{T/\mu}$.

For the fundamental mode, the frequency is:

$$f = \frac{v}{\lambda} = \frac{1}{2l}\sqrt{\frac{T}{\mu}} \implies f^2 = \frac{T}{4\mu l^2}$$

Substituting this back into our integral:

$$E = \int_0^l 2\pi^2 \mu \left(\frac{T}{4\mu l^2}ight) a^2 \sin^2\left(\frac{\pi x}{l}ight) dx$$

Notice how the mass density $\mu$ cancels out beautifully! This is a wonderful physical simplification.

Pulling the constants out of the integral, we get:

$$E = \frac{\pi^2 a^2 T}{2l^2} \int_0^l \sin^2\left(\frac{\pi x}{l}ight) dx$$

### Final Calculation

To evaluate the integral, we use the standard trigonometric identity:

$$\int_0^l \sin^2\left(\frac{\pi x}{l}ight) dx = \frac{l}{2}$$

Substituting this result back into our energy equation:

$$E = \frac{\pi^2 a^2 T}{2l^2} \cdot \left(\frac{l}{2}ight) = \frac{\pi^2 a^2 T}{4l}$$

This is our final, elegant result!

The total oscillation energy stored in the string is **$\frac{\pi^2 a^2 T}{4l}$**.

Visualized Solution

Visualizing the Fundamental Mode

Determining the Wave Number $k$

Formulating the Amplitude Equation

Energy of an Infinitesimal Element

Expressing Frequency in terms of Tension

Substituting into the Energy Element Equation

Integrating to Find Total Energy

Evaluating the Definite Integral

Final Calculation of Total Energy

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Amplitude Equation

Energy of an Infinitesimal Element

Integrating for the Total Energy

Final Calculation

Similar Questions

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A string of length $1 m$ and mass $2 \times 1 0^{- 5} kg$ is under tension $T$ . when the string vibrates, two successive harmonics are found to occur at frequencies $750 Hz$ and $1000 Hz$ . The value of tension $T$ is _______ Newton.

Length of a string tied to two rigid supports is $40 cm$ . Maximum length (wavelength in cm) of a stationary wave produced on it, is

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Visualized Solution

Visualizing the Fundamental Mode

Determining the Wave Number .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Formulating the Amplitude Equation

Energy of an Infinitesimal Element

Expressing Frequency in terms of Tension

Substituting into the Energy Element Equation

Integrating to Find Total Energy

Evaluating the Definite Integral

Final Calculation of Total Energy

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Amplitude Equation

Energy of an Infinitesimal Element

Integrating for the Total Energy

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Length of a string tied to two rigid supports is 40 cm. Maximum length (wavelength in cm) of a stationary wave produced on it, is

List-I

List-II

Determining the Wave Number $k$

Length of a string tied to two rigid supports is $40 cm$ . Maximum length (wavelength in cm) of a stationary wave produced on it, is