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Chapter 16: Problem 59

The speed of a transverse wave on a string is 300.00 \(\mathrm{m} / \mathrm{s},\) its wavelength is \(0.50 \mathrm{m},\) and the amplitude is 20.00 cm. How much time is required for a particle on the string to move through a distance of \(5.00 \mathrm{km}\) ?

Short Answer

Expert verified
The time required for a particle on the string to move through a distance of 5.00 km is \( \frac{2500}{24\pi} \) seconds.

Step by step solution

01

Convert units

First, we need to make sure all the quantities are in the same unit system. Convert the given distance and amplitude to meters: - Distance = 5.00 km = 5,000 m - Amplitude = 20.00 cm = 0.20 m
02

Calculate frequency

Using the wave speed equation, we can find the frequency (f) of the wave: Wave speed = v = 位 脳 f We want to solve for f: f = v / 位 Given v = 300.00 m/s and 位 = 0.50 m: f = 300.00 / 0.50 = 600 Hz
03

Calculate the period

Next, find the period (T) of the wave using the relationship between frequency and period: Period = T = 1 / f Given f = 600 Hz: T = 1 / 600 = (1/600) seconds
04

Calculate wave particle speed

Now, we calculate the speed of the particle (v_p) on the string using the formula: Particle speed (v_p) = A 脳 2蟺 脳 f Given A = 0.20 m and f = 600 Hz: v_p = 0.20 脳 2蟺 脳 600 = 240蟺 meters per second
05

Calculate time required for the particle to move

Finally, we will use the formula speed equals distance over time (v = d/t) to find the time required for the particle to move through the 5,000 meters distance: Particle speed = v_p = Distance / Time Solving for Time: Time = Distance / Particle speed Given Distance = 5,000 m and Particle speed = 240蟺 m/s: Time = 5000 / (240蟺) = \( \frac{2500}{24\pi} \) seconds This is the time required for a particle on the string to move through a distance of 5.00 km.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transverse Waves
Transverse waves are a type of wave where the oscillation or movement of the medium is perpendicular to the direction of the wave. Imagine a wave moving along a string 鈥� the string itself moves up and down while the wave travels horizontally.
This kind of wave is common in scenarios like waves on a water surface or light waves. They contrast with longitudinal waves, where the movement is parallel to the wave direction, like sound waves in air.
  • The peaks and troughs are the highest and lowest points in a transverse wave, respectively.
  • Transverse waves can be visualized as a series of crests and troughs.
Understanding transverse waves helps in visualizing and analyzing the properties of waves, such as amplitude and frequency.
Wave Speed
Wave speed is the rate at which the wave propagates through the medium. It is calculated using the formula: \[\text{Wave speed} (v) = \text{wavelength} (\lambda) \times \text{frequency} (f)\] This tells us how quickly a wave travels from one point to another.
In the provided exercise, the wave speed is given as 300.00 meters per second, showing the rapid movement of energy along the string.
  • Wave speed depends on the medium's properties, like tension in a string or air pressure for sound waves.
  • It is usually constant for a given medium, as long as conditions don't change.
Wave speed is crucial in determining how quickly signals or energy can be transmitted.
Frequency
Frequency represents the number of wave cycles passing a point per second, measured in Hertz (Hz). In the context of the exercise, frequency is derived using the wave speed and wavelength: \[f = \frac{v}{\lambda}\] Given the wave speed as 300.00 m/s and the wavelength as 0.50 m, the frequency is 600 Hz.
A higher frequency means more cycles per second, often resulting in a higher pitch sound or finer resolution in imaging.
  • Frequency is the reciprocal of the wave period.
  • It determines the color or note of the wave, like the pitch in sound waves.
Mastering frequency concepts aids in understanding how waves behave in different scenarios.
Wave Period
The wave period is the time it takes for one complete wave cycle to pass a given point. It relates closely to frequency through the equation: \[\text{Period} (T) = \frac{1}{f}\] In our example, with a frequency of 600 Hz, the period of the wave is approximately \(\frac{1}{600}\) seconds.
The period tells us about the timing of the wave motion, whether a sound wave or an oscillating string.
  • A shorter period means a higher frequency.
  • It is an important aspect of understanding wave timing and synchronization.
Period analysis is valuable for coordinating events in wave dynamics.
Amplitude
Amplitude is the maximum displacement of a point on the wave from its rest position, which determines the wave's energy and intensity. The given amplitude in the exercise is 0.20 meters (after converting from centimeters).
Amplitude affects the wave's visual height and perceived loudness in sound waves. A larger amplitude usually means more energy in the wave.
  • Amplitude is crucial for understanding energy transfer in waves.
  • Higher amplitudes mean stronger waves, such as louder sounds or brighter lights.
By understanding amplitude, we can gauge the power and impact of a wave in various physical contexts.

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Most popular questions from this chapter

Two sinusoidal waves are moving through a medium in the positive \(x\) -direction, both having amplitudes of 6.00 \(\mathrm{cm},\) a wavelength of \(4.3 \mathrm{m},\) and a period of \(6.00 \mathrm{s},\) but one has a phase shift of an angle \(\phi=0.50\) rad. What is the height of the resultant wave at a time \(t=3.15 \mathrm{s}\) and a position \(x=0.45 \mathrm{m} ?\).

Two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the resultant wave is $$y_{R}(x, t)=0.70 \mathrm{m} \sin \left(3.00 \mathrm{m}^{-1} x-6.28 \mathrm{s}^{-1} t+\pi / 16 \mathrm{rad}\right)$$,What are the angular frequency, wave number, amplitude, and phase shift of the individual waves?

Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where low frequency sound is originating from. (a) Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 18 \(\mathrm{cm}\) apart, and the speed of sound generated is \(340 \mathrm{m} / \mathrm{s}\). How long is the interval between when the sound arrives at the right ear and the sound arrives at the left ear? (b) Assume the same person was scuba diving and a low- frequency sound source was to the right of the scuba diver. How long is the interval between when the sound arrives at the right ear and the sound arrives at the left ear, if the speed of sound in water is \(1500 \mathrm{m} / \mathrm{s}\) ? (c) What is significant about the time interval of the two situations?

The energy of a ripple on a pond is proportional to the amplitude squared. If the amplitude of the ripple is \(0.1 \mathrm{cm}\) at a distance from the source of 6.00 meters, what was the amplitude at a distance of 2.00 meters from the source?

A sinusoidal wave travels down a taut, horizontal string with a linear mass density of \(\mu=0.060 \mathrm{kg} / \mathrm{m} .\) The \(\begin{array}{lllllll}\text { maximum } & \text { vertical } & \text { speed } & \text { of } & \text { the } & \text { wave } & \text { is }\end{array}\),\(v_{y \max }=0.30 \mathrm{cm} / \mathrm{s} .\) The wave is modeled with the wave equation \(\quad y(x, t)=A \sin \left(6.00 \mathrm{m}^{-1} x-24.00 \mathrm{s}^{-1} t\right)\).What is the amplitude of the wave? (b) What is the tension in the string?
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