Question

A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, \(y(x, t)=(0.01 \mathrm{~m}) \sin \left[\left(62.8 \mathrm{~m}^{-1}\right) x\right] \cos \left[\left(628 s^{-1}\right) t\right]\). Assuming \(\pi=3.14\), the correct statement(s) is (are) (A) The number of nodes is 5 . (B) The length of the string is \(0.25 \mathrm{~m}\). (C) The maximum displacement of the midpoint of the string, from its equilibrium position is \(0.01 \mathrm{~m}\). (D) The fundamental frequency is \(100 \mathrm{~Hz}\).

Short answer

The number of nodes is 6 (not 5), the length of the string is 0.5 m, the maximum displacement of the midpoint of the string is 0.01 m, and the fundamental frequency is 20 Hz.

Step-by-step solution

Analyze the Given Wave Equation

Consider the given equation of the wave on the string: \(y(x, t)=(0.01 \text{ m}) \sin\left[\left(62.8 \text{ m}^{-1}\right)x\right]\cos\left[\left(628 \text{ s}^{-1}\right)t\right]\). The term within the sine function represents the spatial part, while the term within the cosine function represents the temporal part of the wave.

Determine the Number of Nodes

The number of nodes in a standing wave is determined by the number of points along the length of the string that remain at rest (zero displacement). For the nth harmonic, there will be \(n+1\) nodes. Since the wave is vibrating in its fifth harmonic, there will be \(5+1=6\) nodes, not 5, making statement (A) incorrect.

Calculate the Length of the String

The wavenumber \(k\) in the spatial part of the wave is related to the number of half wavelengths \(\lambda/2\) that fit into the length of the string \(L\) by: \(k = \frac{2\pi}{\lambda} = \frac{(n)(2\pi)}{L}\). Given \(k = 62.8 \text{ m}^{-1}\), we can find \(L\) for the fifth harmonic, where \(n=5\). \(L = \frac{(n)(2\pi)}{k} = \frac{5(2)(3.14)}{62.8} = 0.5 \text{ m}\), so statement (B) is incorrect.

Find the Maximum Displacement of the Midpoint

The maximum displacement of a point on the string occurs at the antinodes, where the sine part of the wave function is \(\pm1\). The midpoint of the string in the fifth harmonic will be at an antinode. Plug in \(x = L/2\) into the spatial part to find the maximum displacement: \(\sin\left[\frac{5\pi}{L} \times \frac{L}{2}\right] = \sin(\frac{5\pi}{2}) = \pm1\). The maximum amplitude is the coefficient of the sine function, \(0.01 \text{ m}\), making statement (C) correct.

Determine the Fundamental Frequency

The frequency of the nth harmonic is related to the fundamental frequency \(f\) by: \(f_n = n\cdot f\). The given angular frequency \(\omega = 628 \text{ s}^{-1}\) is related to the frequency of the fifth harmonic by \(\omega = 2\pi f_5\). The fundamental frequency is then \(f = \frac{f_5}{n} = \frac{\omega}{2\pi n} = \frac{628}{2(3.14)5} = 20 \text{ Hz}\), making statement (D) incorrect.

Key concepts

Wave Equation

The wave equation is a mathematical model that describes how waves propagate through different media. It is a second-order partial differential equation that brings together the spatial and temporal components of a wave. In the case of a horizontal stretched string vibrating in its fifth harmonic, the wave equation given by the sinusoidal functions encapsulates these components.

Consider the given wave equation:
\[ y(x, t) = (0.01 \, \text{m}) \sin\left[\left(62.8 \, \text{m}^{-1}\right) x\right] \cos\left[\left(628 \, \text{s}^{-1}\right) t\right] \].

The wave equation represents the displacement (\(y\)) of a point on the string at a position (\(x\)) and at a time (\(t\)). The sine function (\(\sin\)) relates to the spatial variation along the string, while the cosine function (\(\cos\)) describes the temporal variation over time. The product of these sinusoidal functions models the standing wave pattern we observe on the string.

Nodes and Antinodes in Standing Waves

In a standing wave, nodes and antinodes are points of zero and maximum displacement, respectively. Nodes are the locations where the string does not move at all, remaining at rest as the wave oscillates, while antinodes are where the displacement reaches a peak - the points of greatest amplitude.

For a string vibrating in its nth harmonic, the pattern consists of \(n+1\) nodes, including the fixed ends. This is because each harmonic introduces one additional wavelength to the standing wave pattern, which corresponds to an extra node in between the ends.

For the exercise provided, the equation hints at the string vibrating in its fifth harmonic, implying there would be
\[6 \text{ nodes} (5 + 1)\].

The midpoint of the string will always be an antinode in odd harmonics since it corresponds to the point of maximum fluctuation. Therefore, the midpoint in a fifth harmonic standing wave is indeed at an antinode, validating the presence of maximum displacement there.

Wavelength and Frequency Relationship

The relationship between wavelength and frequency is fundamental to understanding wave phenomena. As the speed of a wave is the product of its wavelength (\(\lambda\)) and frequency (\(f\)), this relationship is especially important in a fixed medium where the speed of the wave is constant.

\[ v = \lambda \times f \]

In the context of a string fixed at both ends, like in the exercise, the standing wave can only support wavelengths that fit precisely into the length of the string, leading to the creation of distinct patterns or harmonics.

The fundamental frequency is the lowest frequency at which a string can vibrate, associated with the first harmonic (\(n=1\)). In the solution provided, the fundamental frequency was found to be inversely proportional to the harmonic number, given by
\[ f = \frac{f_n}{n} \].

Where \(f_n\) is the frequency at the nth harmonic. Thus, understanding how the length of the string imposes conditions on wavelength, and how this affects the frequency, is crucial to grasping the spectrum of harmonics that a string can support.