A string oscillates according to the equation y^'=(0.50 cm) sin[((π)/(3) cm^-1) x] cos[(40 πs^-1

A string oscillates according to the equation y^'=(0.50 cm)...

A string oscillates according to the equation $$y^{\prime}=(0.50 \mathrm{cm}) \sin \left[\left(\frac{\pi}{3} \mathrm{cm}^{-1}\right) x\right] \cos \left[\left(40 \pi \mathrm{s}^{-1}\right) t\right]$$ What are the (a) amplitude and (b) speed of the two waves (identical except for direction of travel) whose superposition gives this oscillation? (c) What is the distance between nodes? (d) What is the transverse speed of a particle of the string at the position $x=1.5 \mathrm{cm}$ when $t=\frac{2}{8} s ?$

A string oscillates according to the equation
$$y^{\prime}=(0.50 \mathrm{cm}) \sin \left[\left(\frac{\pi}{3} \mathrm{cm}^{-1}\right) x\right] \cos \left[\left(40 \pi \mathrm{s}^{-1}\right) t\right]$$ What are the (a) amplitude and (b) speed of the two waves
(identical except for direction of travel) whose superposition
gives this oscillation? (c) What is the distance between nodes?
(d) What is the transverse speed of a particle of the string at the
position $x=1.5 \mathrm{cm}$ when $t=\frac{2}{8} s ?$

Key Concepts

Amplitude of a Wave

Amplitude refers to the maximum displacement of the wave from its equilibrium position and is a measure of the energy carried by the wave. In any oscillation, the amplitude tells you how far the medium is displaced at the peak of the wave, and it is an intrinsic property of the wave itself.

Superposition Principle and Standing Waves

The superposition principle states that when two or more waves traverse the same space, the net displacement is the sum of the individual displacements. When two identical waves traveling in opposite directions are superimposed, they form a standing wave, characterized by fixed nodes (points of zero amplitude) and antinodes (points of maximum amplitude). This concept is crucial for understanding patterns that emerge in string vibrations and other oscillatory systems.

Wave Speed

Wave speed is the rate at which a wave travels through a medium, and it is related to the wave's angular frequency and wave number. Mathematically, this relationship is given by the ratio of the angular frequency to the wave number, providing a fundamental link between the temporal and spatial characteristics of waves.

Nodes in Standing Waves

Nodes are points along a standing wave where the medium remains at rest regardless of time. The distance between two consecutive nodes is directly related to the wavelength of the component traveling waves. Understanding the formation and spacing of nodes is essential in analyzing the stationary patterns that result from the superposition of waves.

Transverse Velocity of a Particle

The transverse velocity of a particle on a string refers to the rate of change of its displacement in the direction perpendicular to the wave propagation. It is obtained by differentiating the displacement function with respect to time. This concept is key for understanding the dynamic behavior of the medium at a given point and instant, particularly in vibrating systems.

Transcript

00:01 So in this question, we have this standing wave, and we want to figure out what is the amplitude and speed of the two waves whose position gives the standing wave.

00:10 Then there are a couple other things we want to figure out, like, what is distance between knots? distance between the nodes is like just what is the wavelength is what is asking.

00:19 And then finally, what is the transverse speed of a particle at a certain position of the wave? let's first look at part a and b.

00:29 What is the amplitude? and what is speed.

00:34 So when we have two waves together, if we have y, sine, kx plus omega -t, and y -m -y -m -sign k -x minus omega -t, their superposition will become two times y -m, cosine, sine kx, cosine omega -t.

01:06 So that is when we have a standing wave.

01:09 So in this case, when we say what is the amplitude of individual wave, the individual wave will be half.

01:17 The individual waves is ym, and right now we have 0 .5 cm is this, is 2ym.

01:26 So the ym is half of 0 .5 centimeter, which is 0 .25 centimeter.

01:34 Part b asks us what is the speed? well, again, let's look at this form and this y prime that we're given.

01:42 So this is 2ym, sine kx, so this is k, cosine omega -t, so this is omega.

01:52 So we have both k and omega.

01:54 And if we have both k and omega, we can find what the speed is, because speed v equals omega over t.

02:01 And omega is 40 pi per second and omega, sorry, omega over k.

02:07 And k is pi over 3 per centimeter.

02:12 So the velocity is 1 .2 times 10 to the second centimeter per second.

02:23 Part c asks us what is the distance between each knot, and if we look at the distance between each node is half of the wayplanes, so it's asking what is half of the wavelengths.

02:35 So x equals wayplanes over 2, and how do you find the wavelength? we know that k equals 2 pi over lambda...

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