Problem 49 - University Physics with Modern Physics

Question

A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels from left to right along a long, horizontal, stretched string with a speed of 36.0 m/s. Take the origin at the left end of the undisturbed string. At time \(t = 0\) the left end of the string has its maximum upward displacement. (a) What are the frequency, angular frequency, and wave number of the wave? (b) What is the function \(y(x, t)\) that describes the wave? (c) What is \(y(t)\) for a particle at the left end of the string? (d) What is \(y(t)\) for a particle 1.35 m to the right of the origin? (e) What is the maximum magnitude of transverse velocity of any particle of the string? (f) Find the transverse displacement and the transverse velocity of a particle 1.35 m to the right of the origin at time \(t = 0.0625\) s.

Step-by-Step Solution

Answer

Frequency: 20 Hz; Angular Frequency: 40\(\pi\) rad/s; Wave Number: \(\frac{10\pi}{9}\) m\(^{-1}\).

Step 1: Calculate the Frequency

To find the frequency, we use the wave speed formula: \( v = f \lambda \). Here, \( v = 36.0 \) m/s and \( \lambda = 1.80 \) m. Solving for \( f \), we get \( f = \frac{v}{\lambda} = \frac{36.0}{1.80} = 20.0 \) Hz.

Step 2: Determine the Angular Frequency

The angular frequency, \( \omega \), is related to the frequency by \( \omega = 2\pi f \). Substituting \( f = 20.0 \) Hz, we get \( \omega = 2\pi \times 20.0 = 40\pi \) rad/s.

Step 3: Find the Wave Number

The wave number, \( k \), is given by \( k = \frac{2\pi}{\lambda} \). Substituting \( \lambda = 1.80 \) m, we calculate \( k = \frac{2\pi}{1.80} = \frac{10\pi}{9} \) m\(^{-1}\).

Step 4: Write the Wave Function

The wave function is given by \( y(x, t) = A \sin(kx - \omega t + \phi) \). Since at \( t = 0 \), \( x = 0 \), and the displacement is maximum, \( \phi = \frac{\pi}{2} \). Thus, \( y(x, t) = 2.5 \sin(\frac{10\pi}{9}x - 40\pi t + \frac{\pi}{2}) \).

Step 5: Find the Function at the Left End

For a particle at the left end, \( x = 0 \). Thus, the function simplifies to \( y(0, t) = 2.5 \sin(-40\pi t + \frac{\pi}{2}) \).

Step 6: Function for Particle 1.35 m to the Right

For a particle 1.35 m to the right, \( x = 1.35 \) m. Substitute into the wave function: \( y(1.35, t) = 2.5 \sin(\frac{10\pi}{9} \times 1.35 - 40\pi t + \frac{\pi}{2}) \).

Step 7: Maximum Transverse Velocity

The maximum transverse velocity, \( v_{max} \), is given by \( A \omega \). Using \( A = 2.5 \) mm = 0.0025 m and \( \omega = 40\pi \) rad/s, we get \( v_{max} = 0.0025 \times 40\pi = 0.1\pi \) m/s.

Step 8: Displacement and Velocity at Specific Time

At \( t = 0.0625 \) s and \( x = 1.35 \) m, substitute into the wave function to find displacement: \( y(1.35, 0.0625) = 2.5 \sin(\frac{10\pi}{9} \times 1.35 - 40\pi \times 0.0625 + \frac{\pi}{2}) \). To find the velocity, differentiate \( y(x, t) \) with respect to \( t \) and substitute \( t = 0.0625 \), \( x = 1.35 \).

Key Concepts

Wave Frequency

Wave frequency is a fundamental aspect of wave physics, representing how many waves pass a point per second. In our problem, we calculated the frequency using the formula:

• Formula: \( v = f \lambda \)
• Where \( v \) is wave speed, \( f \) is frequency, and \( \lambda \) is wavelength.

Inserting the given values, \( v = 36.0 \text{ m/s} \) and \( \lambda = 1.80 \text{ m} \), we solve for frequency:\[ f = \frac{36.0}{1.80} = 20.0 \text{ Hz} \]This outcome tells us that 20 waves pass by any point on the string every second. Understanding wave frequency helps in deciphering how fast a wave is oscillating, which is crucial in predicting wave behaviors in different environments.

Angular Frequency

Angular frequency, denoted as \( \omega \), relates to how quickly the wave oscillates in terms of radians per second. Unlike the normal frequency, it uses radians to describe oscillations rather than cycles.

• Formula: \( \omega = 2\pi f \)
• \( f \) is the wave frequency.

By substituting the frequency \( f = 20.0 \text{ Hz} \), the angular frequency \( \omega \) is calculated as:\[ \omega = 2\pi \times 20.0 = 40\pi \text{ rad/s} \]This means the wave completes \( 40\pi \) radians every second. Angular frequency is essential in wave studies as it provides a more precise measure of wave oscillation for calculations involving complex wave interactions.

Wave Number

The wave number \( k \) defines the number of wave cycles in a unit distance and is essential for understanding wave propagation. The formula for determining the wave number is:

• Formula: \( k = \frac{2\pi}{\lambda} \)
• \( \lambda \) is the wavelength.

Given \( \lambda = 1.80 \text{ m} \), the wave number is:\[ k = \frac{2\pi}{1.80} = \frac{10\pi}{9} \text{ m}^{-1} \]This result indicates there are \( \frac{10\pi}{9} \) cycles per meter along the string. Understanding the wave number is important for analyzing spatial properties of waves, such as wave fronts and interference patterns.

Wave Function

The wave function describes how each point on a medium vibrates over time in a wave. For transverse waves like this one, the wave function is given by:

• Formula: \( y(x, t) = A \sin(kx - \omega t + \phi) \)
• \( A \) is the amplitude, \( k \) is the wave number, \( \omega \) is angular frequency, and \( \phi \) is phase constant.

In this context, the wave function becomes:\[ y(x, t) = 2.5 \sin\left(\frac{10\pi}{9}x - 40\pi t + \frac{\pi}{2}\right) \]This expression shows the displacement of any point at a location \( x \) at a time \( t \). It's a powerful tool to predict wave behavior across different regions of the string, crucial for various applications like generating and interpreting wave patterns.

Transverse Velocity

Transverse velocity describes how fast a single point on the string moves up and down as the wave travels along it. The maximum transverse velocity, \( v_{max} \), happens when the rate of change of wave function is greatest.

• Formula: \( v_{max} = A \omega \)
• \( A \) is amplitude and \( \omega \) is angular frequency.

Substituting the parameters \( A = 2.5 \text{ mm} = 0.0025 \text{ m} \) and \( \omega = 40\pi \text{ rad/s} \), we find:\[ v_{max} = 0.0025 \times 40\pi = 0.1\pi \text{ m/s} \]The transverse velocity is important in determining how kinetic energy is transferred through the medium. Understanding this helps in applications like acoustics and predicting how materials will respond to wave-induced vibrations.