Problem 36 - University Physics with Modern Physics

Question

15.36 .. CALC Adjacent antinodes of a standing wave on a string are 15.0 \(\mathrm{cm}\) apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 \(\mathrm{cm}\) and period 0.0750 s. The string lies along the \(+x\) -axis and is fixed at \(x=0 .\) (a) How far apart are the adjacent nodes? (b) What are the wavelength, amplitude, and speed of the two traveling waves that form this pattern?(c) Find the maximum and minimum transverse speeds of a point at an antinode. (d) What is the shortest distance along the string between a node and an antinode?

Step-by-Step Solution

Answer

(a) 15.0 cm, (b) 30.0 cm, 0.850 cm, 400 cm/s, (c) 71.21 cm/s and 0 cm/s, (d) 7.5 cm.

Step 1: Understanding the Problem

We need to analyze a standing wave on a string to solve for various parameters, including the distance between nodes, wavelength, amplitude, speed of traveling waves, and various speeds of a point at an antinode.

Step 2: Relationship Between Antinodes and Nodes

Antinodes in a standing wave are located at half the wavelength (\( \frac{\lambda}{2} \)) from each other. Thus, the wavelength \( \lambda \) can be calculated using the given distance between antinodes, 15.0 cm: \[ \lambda = 2 \times 15.0 \text{ cm} = 30.0 \text{ cm} \] The distance between adjacent nodes is half the wavelength: \[ \text{Distance between nodes} = \frac{\lambda}{2} = 15.0 \text{ cm} \]

Step 3: Calculate Wavelength, Amplitude, and Speed of the Waves

Reaffirming from Step 2, the wavelength is 30.0 cm. The amplitude of each traveling wave creating the standing wave is the same as the amplitude of the oscillation of a particle at an antinode, 0.850 cm. The speed \( v \) of the waves can be calculated from the wavelength and period \( T \): \[ v = \frac{\lambda}{T} = \frac{30.0 \text{ cm}}{0.0750 \text{ s}} = 400 \text{ cm/s} \]

Step 4: Calculate Maximum and Minimum Transverse Speeds at an Antinode

The maximum transverse speed \( v_{max} \) of a particle at an antinode is given by: \[ v_{max} = \omega A \] where \( \omega = \frac{2\pi}{T} \) and \( A \) is the amplitude. Using \( \omega \): \[ \omega = \frac{2\pi}{0.0750} \approx 83.78 \, \text{rad/s} \]\[ v_{max} = 83.78 \times 0.850 = 71.21 \text{ cm/s} \]Since the minimum speed of a particle at an antinode is simply when it's at its turning point, it momentarily stops, so \( v_{min} = 0 \text{ cm/s} \).

Step 5: Shortest Distance Between Node and Antinode

The shortest distance along the string between a node and an adjacent antinode is one quarter of the wavelength, since \( \frac{\lambda}{4} \) of a complete wave cycle is from a node to an antinode:\[ \frac{\lambda}{4} = \frac{30.0 \text{ cm}}{4} = 7.5 \text{ cm} \]

Key Concepts

Wavelength Calculation

When dealing with standing waves on a string, understanding how to calculate the wavelength is pivotal. A standing wave is formed by two traveling waves moving in opposite directions at the same speed, and the location where the amplitude is maximized is known as an antinode. In the problem given, the antinodes are 15.0 cm apart. Since antinodes are positioned at half the wavelength (\( \frac{\lambda}{2} \)) apart, you can easily find the wavelength (\( \lambda \)) by doubling the distance between antinodes:

• Given that the distance between antinodes is 15.0 cm, the wavelength is calculated as:\[ \lambda = 2 \times 15.0 \text{ cm} = 30.0 \text{ cm} \]

The wavelength is an essential measurement, as it describes the length of one complete wave cycle between repeating points, such as from crest to crest.

Simple Harmonic Motion

Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement. This motion is characteristic of the oscillation seen at the antinode in a standing wave. In this exercise, a particle located at an antinode oscillates with an amplitude of 0.850 cm and a period of 0.0750 seconds. These two values are crucial aspects:

• Amplitude: This is the maximum displacement of the particle from its resting position—here, it's 0.850 cm.
• Period: This is the time taken for one complete cycle of motion—in this problem, 0.0750 seconds define this cycle.

During this type of motion, particles at the antinode move back and forth with maximum energy. The formula for angular frequency (\( \omega \)) is \( \omega = \frac{2\pi}{T} \) where \( T \) is the period, allowing us to calculate how fast the particles are oscillating in radians per second.

Wave Speed Calculation

Wave speed (\( v \)) indicates how fast a wave travels along a string. For standing waves, the wave speed can be found using the relationship between wavelength and period:

• The formula used is \[ v = \frac{\lambda}{T} \] where \( \lambda \) is the wavelength and \( T \) is the period.

In this case, the calculated values were:

• Wavelength: 30.0 cm
• Period: 0.0750 s

By substituting these into the formula, the wave speed is:

• \[ v = \frac{30.0 \text{ cm}}{0.0750 \text{ s}} = 400 \text{ cm/s} \]

This speed reflects how quickly each point of the wave (like the crest or trough) moves across the string.

Node and Antinode Distance

Nodes and antinodes are key features of standing waves. A node is a point on the string where there is no movement, while an antinode is where the maximum movement occurs. Knowing the distance between these points helps to understand the wave's behavior. In the given exercise, the shortest distance from a node to an antinode is one quarter of the wavelength:

• The formula is: \[ \text{Distance} = \frac{\lambda}{4} \]

Thus, for a wavelength of 30.0 cm, the distance is:

• \[ \frac{30.0 \text{ cm}}{4} = 7.5 \text{ cm} \]

This distance tells us how quickly maximum displacement transitions into no displacement along the string. Understanding this concept provides insight into the spatial structure of standing waves.