Question

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation \(y(x, t) = (5.60 \, \mathrm{cm}) \mathrm{sin} [(0.0340 \mathrm{rad/cm})x] \mathrm{sin} [(150.0 \, \mathrm{rad/s})t]\), where the origin is at the left end of the string, the x-axis is along the string, and the \(y\)-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation \(y(x,t)\) for this string if it were vibrating in its eighth harmonic?

Short answer

(a) Sketch nodes at x = 0, L/3, 2L/3, L. (b) 2.80 cm. (c) 276.09 cm. (d) \( \lambda = 184.06 \, \mathrm{cm}, f = 23.87 \, \mathrm{Hz}, v = 439.40 \, \mathrm{cm/s} \). (e) 840 cm/s. (f) \( y(x,t) = 5.60 \sin[(0.091)x]\sin[150.0t] \).

Step-by-step solution

Understanding the Standing Wave

A standing wave is typically formed by the superposition of two traveling waves moving in opposite directions. The given equation \( y(x, t) = (5.60 \, \mathrm{cm}) \sin[(0.0340 \, \mathrm{rad/cm})x] \sin[(150.0 \, \mathrm{rad/s})t] \) represents a standing wave.

Sketching the Standing Wave Pattern

The given equation describes the third harmonic, which has nodes at both ends and two additional nodes in between. Sketch the wave by marking nodes at points \( x = 0, L/3, 2L/3, L \) and antinodes at points \( x = L/6, L/2, 5L/6 \).

Amplitude of Traveling Waves

The amplitude of the standing wave is 5.60 cm. For a standing wave formed by two identical traveling waves moving in opposite directions, the amplitude of each traveling wave is half of the standing wave's amplitude. Thus, each traveling wave has an amplitude of \( 5.60 \, \mathrm{cm} / 2 = 2.80 \, \mathrm{cm} \).

Determining the Length of the String

In the third harmonic, the length of the string is \( L = \frac{3}{2} \lambda \). From the wave equation, the wave number \( k = 0.0340 \, \mathrm{rad/cm} \), so the wavelength \( \lambda = \frac{2\pi}{k} = \frac{2\pi}{0.0340} \approx 184.06 \, \mathrm{cm} \). The length \( L = \frac{3}{2} \times 184.06 \approx 276.09 \, \mathrm{cm} \).

Calculating Wavelength, Frequency, and Speed

Wavelength \( \lambda = \frac{2\pi}{0.0340} = 184.06 \, \mathrm{cm} \). The angular frequency \( \omega = 150.0 \, \mathrm{rad/s} \) gives the frequency \( f = \frac{\omega}{2\pi} = \frac{150.0}{2\pi} \approx 23.87 \, \mathrm{Hz} \). The speed \( v = f \lambda = 23.87 \times 1.8406 \approx 439.40 \, \mathrm{cm/s} \).

Maximum Transverse Speed

The maximum transverse speed of a point on the string is given by the product of the amplitude and the angular frequency: \( v_{max} = A \omega = 5.60 \, \mathrm{cm} \times 150.0 \, \mathrm{rad/s} = 840 \, \mathrm{cm/s} \).

Equation for the Eighth Harmonic

For the eighth harmonic, the relationship between length and wavelength is \( L = 4\lambda \). So, \( \lambda = \frac{L}{4} = \frac{276.09}{4} = 69.0225 \, \mathrm{cm} \). The new wave number \( k = \frac{2\pi}{\lambda} \approx 0.091 \, \mathrm{rad/cm} \). The wave equation becomes \( y(x, t) = (5.60 \, \mathrm{cm}) \sin[(0.091 \, \mathrm{rad/cm})x] \sin[(150.0 \, \mathrm{rad/s})t] \).

Key concepts

Harmonics

In the world of waves, harmonics are special natural patterns that occur on a string fixed at both ends. When a string is vibrated, it doesn't just generate random waves; instead, specific patterns called harmonics are formed. The harmonics indicate the different ways the string can vibrate. Each harmonic corresponds to a standing wave, with the first harmonic being the fundamental frequency. The third harmonic, like in our exercise, means that the string oscillates with three segments. There are nodes at each end and additional nodes between these segments, where the string doesn't move, while the sections in between these nodes, called antinodes, experience the maximum displacement. Understanding harmonics helps us determine the characteristic frequencies at which a system can naturally oscillate.

Wave Equation

The wave equation is a mathematical way to describe how the shape of a wave changes over time and space. It's an essential part of understanding wave behavior. In general, a wave equation can be expressed in terms of a sine function due to its repetitive nature, showing how the wave repeats in cycles.
The equation given in the exercise, \(y(x, t) = (5.60 \, \mathrm{cm}) \sin[(0.0340 \, \mathrm{rad/cm})x] \sin[(150.0 \, \mathrm{rad/s})t]\) is specific to the standing wave on the string in its third harmonic. Here, the two sine functions represent the spatial and temporal components that describe the wave's shape as it oscillates over time. The wave number \(k = 0.0340 \, \mathrm{rad/cm}\) provides information about the wavelength, and the angular frequency \(\omega = 150.0 \, \mathrm{rad/s}\) relates to how fast the wave oscillates. Together, these parameters give a complete picture of the wave's behavior.

Transverse Speed

In waves on a string, the transverse speed is about how fast a point on the string moves up and down (perpendicular to the direction of the wave). Think of a surfer riding waves; the up and down motion is similar to the transverse speed. For any point on the string, the transverse speed can change as the wave passes through it. The maximum transverse speed is the highest speed at which any point moves vertically.
To calculate the maximum transverse speed, we use the formula \(v_{\text{max}} = A \omega\), where \(A\) is the amplitude of the wave and \(\omega\) is the angular frequency. From the exercise's wave equation, the amplitude \(A\) is \(5.60 \, \mathrm{cm}\), and \(\omega\) is \(150.0 \, \mathrm{rad/s}\). Hence, the maximum transverse speed \(v_{\text{max}}\) comes out to be \(840 \, \mathrm{cm/s}\). This gives us an understanding of the dynamic movement of the string as the waves travel along it.