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A simple harmonic oscillator at the point \(x = 0\) generates a wave on a rope. The oscillator operates at a frequency of 40.0 Hz and with an amplitude of 3.00 cm. The rope has a linear mass density of 50.0 g/m and is stretched with a tension of 5.00 N. (a) Determine the speed of the wave. (b) Find the wavelength. (c) Write the wave function \(y(x, t)\) for the wave. Assume that the oscillator has its maximum upward displacement at time \(t = 0\). (d) Find the maximum transverse acceleration of points on the rope. (e) In the discussion of transverse waves in this chapter, the force of gravity was ignored. Is that a reasonable approximation for this wave? Explain.

Short Answer

Expert verified
(a) 10.0 m/s, (b) 0.25 m, (c) \(y(x, t) = 0.030 \sin(8\pi x - 80\pi t + \frac{\pi}{2})\), (d) 1894 m/s², (e) Yes, gravity's effect is negligible due to high tension.

Step by step solution

01

Convert Units

First, convert the given units to standard SI units. The amplitude is given as 3.00 cm, which is 0.030 m. The linear mass density is given as 50.0 g/m, which is 0.050 kg/m.
02

Calculate Wave Speed

The wave speed can be calculated using the formula for wave speed in a string, given by:\[ v = \sqrt{\frac{T}{\mu}} \]where \( T = 5.00 \) N is the tension and \( \mu = 0.050 \) kg/m is the linear mass density. Substituting these values, we get:\[ v = \sqrt{\frac{5.00}{0.050}} = \sqrt{100} = 10.0 \text{ m/s} \]
03

Find Wavelength

The wavelength \( \lambda \) can be found using the relationship between wave speed, frequency, and wavelength, given by:\[ v = f \lambda \]Rearranging for wavelength \( \lambda \), we have:\[ \lambda = \frac{v}{f} \]Given the frequency \( f = 40.0 \) Hz and the wave speed \( v = 10.0 \) m/s, substituting these values gives:\[ \lambda = \frac{10.0}{40.0} = 0.25 \text{ m} \]
04

Write the Wave Function

The wave function is of the form:\[ y(x, t) = A \sin(kx - \omega t + \phi) \]where \( A = 0.030 \) m is the amplitude, \( k = \frac{2\pi}{\lambda} \) is the wave number, and \( \omega = 2\pi f \) is the angular frequency. Phi \( \phi \) is the phase constant.- From Step 3, \( \lambda = 0.25 \) m, so \(k = \frac{2\pi}{0.25} = 8\pi \) rad/m.- The angular frequency \( \omega = 2\pi \times 40.0 = 80\pi \) rad/s.- Since the maximum displacement is upwards at \( t = 0 \), \( \phi = \frac{\pi}{2} \).The final wave function is:\[ y(x, t) = 0.030 \sin(8\pi x - 80\pi t + \frac{\pi}{2}) \]
05

Find Maximum Transverse Acceleration

The maximum transverse acceleration \( a_{max} \) of the points on the rope is given by the magnitude of:\[ a_{max} = A\omega^2 \]Substitute \( A = 0.030 \) m and \( \omega = 80\pi \) rad/s:\[ a_{max} = 0.030 \times (80\pi)^2 = 0.030 \times 6400\pi^2 \approx 1894 \text{ m/s}^2 \]
06

Evaluate Influence of Gravity

Considering the effect of gravity, the force due to tension (5.00 N) is significantly larger than the gravitational force on 1 m of the rope (0.50 N), given by \( F_g = mg = 0.050 \times 9.81 = 0.491 \) N. Thus, ignoring gravity is reasonable as its influence is minor compared to the tension in the rope.

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Key Concepts

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

Wave Speed Calculation
The speed of a wave on a rope is a crucial aspect when studying waves. To calculate the wave speed, we use the formula for wave speed on a string, which is given by:
  • \( v = \sqrt{\frac{T}{\mu}} \)
Here, \( T \) represents the tension in the rope, and \( \mu \) is the linear mass density of the rope. In this exercise, the tension \( T \) is 5.00 N, and \( \mu \) is 0.050 kg/m.The unit 'N' stands for Newtons and represents the force applied to the rope, while 'kg/m' is the mass per meter.Substituting the given values, we find the wave speed to be:
  • \( v = \sqrt{\frac{5.00}{0.050}} = \sqrt{100} = 10.0 \text{ m/s} \)
This formula shows how the tension and the linear mass density directly influence the speed of the wave moving along the rope. A higher tension results in a faster wave speed, while a higher mass density slows it down.
Wave Function in Physics
In physics, wave functions describe the displacement of points on a wave at any given time and position. For waves created by simple harmonic oscillators, we write the wave function as:
  • \( y(x, t) = A \sin(kx - \omega t + \phi) \)
In this function:- \( A \) is the amplitude, which indicates the maximum displacement of the wave. It is the height from the equilibrium position, given as 0.030 m in this problem.- \( k \) is the wave number, calculated as \( k = \frac{2\pi}{\lambda} \). For our exercise, \( \lambda = 0.25 \) m, resulting in \( k = 8\pi \) rad/m.- \( \omega \) is the angular frequency, \( \omega = 2\pi f \), where \( f \) is the frequency. Here, \( f = 40.0 \) Hz, thus \( \omega = 80\pi \) rad/s.- \( \phi \) is the phase constant, which adjusts the wave function based on initial conditions. Since the displacement is maximum at \( t = 0 \), \( \phi = \frac{\pi}{2} \).Putting it together, the wave function becomes:
  • \( y(x, t) = 0.030 \sin(8\pi x - 80\pi t + \frac{\pi}{2}) \)
This equation allows us to calculate the displacement of any point on the wave over time.
Transverse Waves
Transverse waves are waves where the motion of the medium is perpendicular to the direction of the wave. Imagine a wave traveling along a rope; while the wave moves horizontally, the points on the rope move up and down vertically.
This up-and-down motion is what characterizes a transverse wave. In our scenario, a simple harmonic oscillator produces transverse waves on the rope, with particles on the rope moving perpendicular to wave propagation.
Key characteristics include:
  • Amplitude: The maximum height of the wave crest or depth of the trough.
  • Wavelength: The distance over which the wave's shape repeats, found here as 0.25 m.
  • Frequency: The number of cycles per second, which affects how fast the waves appear to move past a point.
  • Speed: Determines how quickly the wave travels through the medium; here calculated at 10.0 m/s.
Each of these factors helps in fully describing the behavior of transverse waves on ropes or strings.
Tension in a Rope
Tension plays a critical role when considering waves moving along ropes or strings. Tension is essentially the force applied to stretch the rope, directly affecting the wave speed.
In our problem, the rope's tension is 5.00 N. It's important because:
  • Higher tension increases the wave speed. This happens because a tighter rope transmits forces faster.
  • Less tension would mean a slower wave speed, as the wave would encounter more resistance from the rope's mass.
When calculating wave speed using \( v = \sqrt{\frac{T}{\mu}} \), a higher \( T \) results in a larger speed \( v \). Thus, controlling the tension allows us to manage how fast waves travel through a rope.
Effects of Gravity on Waves
Gravity influences how waves on a rope behave, but in many situations, its effect can be minor compared to other forces like tension.
Gravity pulls each segment of the rope downwards with a force given by \( F_g = mg \), where \( m = 0.050 \) kg/m is the mass per meter and \( g = 9.81 \) m/s² is the gravitational acceleration. Thus, \( F_g = 0.50 \) N per meter of the rope.
In comparison, the given tension force is 5.00 N, which is much larger than the gravitational force on the rope's mass. This means:
  • Gravity does not significantly affect wave speed, as tension is the dominant force.
  • For practical wave calculations, especially where tension is high, gravity can often be neglected.
It's a reasonable simplification for this exercise to ignore gravity, focusing more on how tension and the wave's intrinsic properties shape its motion and behavior.

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Most popular questions from this chapter

A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m long and has a mass of 3.00 g. (a) What is the frequency of its fundamental mode of vibration? (b) What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to 10,000 Hz?

For a string stretched between two supports, two successive standing-wave frequencies are 525 Hz and 630 Hz. There are other standing-wave frequencies lower than 525 Hz and higher than 630 Hz. If the speed of transverse waves on the string is 384 m/s, what is the length of the string? Assume that the mass of the wire is small enough for its effect on the tension in the wire to be ignored.

A strong string of mass 3.00 g and length 2.20 m is tied to supports at each end and is vibrating in its fundamental mode. The maximum transverse speed of a point at the middle of the string is 9.00 m/s. The tension in the string is 330 N. (a) What is the amplitude of the standing wave at its antinode? (b) What is the magnitude of the maximum transverse acceleration of a point at the antinode?

A wave on a string is described by \(y(x, t) = A \mathrm{cos}(kx - \omega t)\). (a) Graph \(y, v_y\), and \(a_y\) as functions of \(x\) for time \(t = 0\). (b) Consider the following points on the string: (i) \(x =\) 0; (ii) \(x = \pi/4k\); (iii) \(x = \pi/2k\); (iv) \(x = 3\pi/4k\); (v) \(x = \pi k\); (vi) \(x = 5\pi/4k\); (vii) \(x = 3\pi/2k\); (viii) \(x = 7\pi/4k\). For a particle at each of these points at \(t = 0\), describe in words whether the particle is moving and in what direction, and whether the particle is speeding up, slowing down, or instantaneously not accelerating.

A 1750-N irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires (\(A\) and \(B\)), each 1.25 m long and weighing 0.290 N. The center of gravity of this beam is one-third of the way along the beam from the end where wire A is attached. If you pluck both strings at the same time at the beam, what is the time delay between the arrival of the two pulses at the ceiling? Which pulse arrives first? (Ignore the effect of the weight of the wires on the tension in the wires.)

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