Problem 60
Question
A continuous succession of sinusoidal wave pulses are produced at one end of a very long string and travel along the length of the string. The wave has frequency 70.0 Hz, amplitude 5.00 mm, and wavelength 0.600 m. (a) How long does it take the wave to travel a distance of 8.00 m along the length of the string? (b) How long does it take a point on the string to travel a distance of 8.00 m, once the wave train has reached the point and set it into motion? (c) In parts (a) and (b), how does the time change if the amplitude is doubled?
Step-by-Step Solution
Verified Answer
(a) 0.1905 s, (b) 5.71 s, and doubling amplitude doesn't affect (a) but reduces (b) to 2.86 s.
1Step 1: Understanding wave speed
To find the speed of a wave traveling on a string, use the formula \( v = f \lambda \), where \( v \) is the wave speed, \( f = 70.0 \text{ Hz} \) is the frequency, and \( \lambda = 0.600 \text{ m} \) is the wavelength.
2Step 2: Calculate wave speed
Substitute the given values into the formula to find \( v \):\[ v = 70.0 \, \text{Hz} \times 0.600 \, \text{m} = 42.0 \, \text{m/s} \].
3Step 3: Time for the wave to travel 8.00 m
Use the formula \( t = \frac{d}{v} \), where \( d = 8.00 \text{ m} \) is the distance the wave travels. \[ t = \frac{8.00 \, \text{m}}{42.0 \, \text{m/s}} \approx 0.1905 \, \text{s} \].
4Step 4: Time for a point on the string
A point on the string oscillates with the frequency of the wave. To determine the time for a point to travel 8.00 m, we calculate the time based on the oscillations:Since the path traveled by the point is along the wave, use the total distance over the circumference of the wave segment: The time taken is \( t = \frac{d}{v_{point}} \), where each oscillation completes a cycle of length 4 times the amplitude,\( d = 8.00 \text{ m} \), \, \( v_{point} = 4 \cdot A \cdot f = 4 \times 5.00 \times 10^{-3} \times 70.0 \).
5Step 5: Calculate point travel time
First, find the speed of a point on the string:\[ v_{point} = 4 \times (5.00 \, \text{mm}) \times 70.0 \, \text{Hz} = 1.4 \, \text{m/s} \].Then, calculate the time:\[ t = \frac{8.00 \, \text{m}}{1.4 \, \text{m/s}} = 5.7143 \, \text{s} \].
6Step 6: Effect of doubling the amplitude on wave speed (part (a))
The wave speed \( v = f \lambda \) is independent of the amplitude, so doubling the amplitude has no effect on the time calculated in part (a).
7Step 7: Effect of doubling the amplitude on point speed (part (b))
If the amplitude is doubled, the speed of a point on the string \( v_{point} \) would become:\[ v_{point} = 4 \times (10.00 \, \text{mm}) \times 70.0 \, \text{Hz} = 2.8 \, \text{m/s} \],which leads to a shorter time:\[ t = \frac{8.00 \, \text{m}}{2.8 \, \text{m/s}} = 2.8571 \, \text{s} \].
Key Concepts
wave speedsinusoidal wavesoscillation frequencyamplitude effect
wave speed
Wave speed is a crucial concept in understanding wave motion. It is defined as the rate at which a wave travels through a medium, which in this case is the string. The formula to determine wave speed \( v \) is given by \( v = f \lambda \), where \( f \) represents the frequency of the wave and \( \lambda \) is the wavelength.
In the given exercise, the frequency \( (f) \) is 70.0 Hz, and the wavelength \( (\lambda) \) is 0.600 m. By substituting these values into the formula, we calculate the wave speed \( v = 70.0 \text{ Hz} \times 0.600 \text{ m} = 42.0 \text{ m/s} \). This speed tells us how fast the wavefront moves along the string.
The concept of wave speed is independent of amplitude, meaning changes in the wave's height do not affect its speed along the string.
In the given exercise, the frequency \( (f) \) is 70.0 Hz, and the wavelength \( (\lambda) \) is 0.600 m. By substituting these values into the formula, we calculate the wave speed \( v = 70.0 \text{ Hz} \times 0.600 \text{ m} = 42.0 \text{ m/s} \). This speed tells us how fast the wavefront moves along the string.
The concept of wave speed is independent of amplitude, meaning changes in the wave's height do not affect its speed along the string.
sinusoidal waves
Sinusoidal waves are a type of periodic wave, characterized by their smooth and consistent oscillation pattern. These waves can be modeled mathematically using the sine function, which provides a precise description of the wave's oscillation over time.
Because they have a regular, repeating form, sinusoidal waves are particularly useful in various applications, including sound waves, radio waves, and even in analyzing wave patterns on strings, as seen in the exercise.
Sinusoidal waves are often described by several key attributes:
Because they have a regular, repeating form, sinusoidal waves are particularly useful in various applications, including sound waves, radio waves, and even in analyzing wave patterns on strings, as seen in the exercise.
Sinusoidal waves are often described by several key attributes:
- Amplitude: This describes the height of the wave, which impacts its energy but not necessarily its speed.
- Wavelength: The distance between successive points that are equivalent on the wave, like two peaks.
- Frequency: The number of oscillations that occur in a second, typically measured in Hertz (Hz).
oscillation frequency
Oscillation frequency is a fundamental aspect of wave motion, describing how frequently the wave oscillates over a given period. In simpler terms, it is the number of complete wave cycles that pass a point per second, measured in Hertz (Hz).
For this exercise, the given frequency of 70.0 Hz indicates that each second, 70 complete cycles of the wave occur at any stationary point along the string. This rapid oscillation is crucial for energy transfer and affects how quickly a point on the string moves with the wave.
The oscillation frequency is directly involved in determining wave speed, through the relation \( v = f \lambda \), where \( v \) is the wave's speed, \( f \) is the frequency, and \( \lambda \) is the wavelength. Thus, changes in frequency will linearly affect the wave's speed across the medium.
For this exercise, the given frequency of 70.0 Hz indicates that each second, 70 complete cycles of the wave occur at any stationary point along the string. This rapid oscillation is crucial for energy transfer and affects how quickly a point on the string moves with the wave.
The oscillation frequency is directly involved in determining wave speed, through the relation \( v = f \lambda \), where \( v \) is the wave's speed, \( f \) is the frequency, and \( \lambda \) is the wavelength. Thus, changes in frequency will linearly affect the wave's speed across the medium.
amplitude effect
Amplitude relates to the displacement of the wave from its rest position. In terms of wave motion, it signifies the wave's height, or how 'intense' it appears. The amplitude also indicates the energy transported by the wave, where larger amplitudes correspond to greater energy.
In the current exercise, the amplitude is initially 5.00 mm. Doubling the amplitude, as explored in the solution, affects the motion of individual points on the string but does not change the wave speed itself. Wave speed, calculated as \( v = f \lambda \), remains unaffected by amplitude changes.
However, the speed of points on the string, which is calculated by \( v_{point} = 4 \cdot A \cdot f \), does change with amplitude. Doubling the amplitude leads to an increase in \( v_{point} \), causing the points to move more swiftly and completing their oscillatory path in less time. This illustrates that while wave propagation speed remains constant, the local effects of amplitude changes can influence the movement efficiency of points on the string.
In the current exercise, the amplitude is initially 5.00 mm. Doubling the amplitude, as explored in the solution, affects the motion of individual points on the string but does not change the wave speed itself. Wave speed, calculated as \( v = f \lambda \), remains unaffected by amplitude changes.
However, the speed of points on the string, which is calculated by \( v_{point} = 4 \cdot A \cdot f \), does change with amplitude. Doubling the amplitude leads to an increase in \( v_{point} \), causing the points to move more swiftly and completing their oscillatory path in less time. This illustrates that while wave propagation speed remains constant, the local effects of amplitude changes can influence the movement efficiency of points on the string.
Other exercises in this chapter
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