A sinusoidal wave moving along a string is shown twice in...

A sinusoidal wave moving along a string is shown twice in Fig. $16-33,$ as crest $A$ travels in the positive direction of an $x$ axis by distance $d=6.0 \mathrm{cm}$ in 4.0 $\mathrm{ms} .$ The tick marks along the axis are separated by $10 \mathrm{cm} ;$ height $H=6.00 \mathrm{mm} .$ The equation for the wave is in the form $y(x, t)=y_{m} \sin (k x \pm \omega t), \mathrm{so}$ what are (a) $y_{m}$ , (b) $k,$ (c) $\omega,$ and (d) the correct choice of sign in front of $\omega ?$

A sinusoidal wave moving along a string is shown
twice in Fig. $16-33,$ as crest $A$
travels in the positive direction of an $x$ axis by distance
$d=6.0 \mathrm{cm}$ in 4.0 $\mathrm{ms} .$ The
tick marks along the axis are
separated by $10 \mathrm{cm} ;$ height $H=6.00 \mathrm{mm} .$ The equation
for the wave is in the form
$y(x, t)=y_{m} \sin (k x \pm \omega t), \mathrm{so}$ what are (a) $y_{m}$ , (b) $k,$ (c) $\omega,$ and (d) the correct choice of sign in
front of $\omega ?$

Key Concepts

Sinusoidal Wave Equation

A sinusoidal wave is a type of periodic wave that can be mathematically described using a sine (or cosine) function. The general form y(x, t) = ym sin(kx ± ?t) captures both the spatial and temporal variations of the wave, where ym is the amplitude, k is the wave number, and ? is the angular frequency. This equation provides a complete description of how the wave oscillates in space and time.

Amplitude

Amplitude, represented as ym in the wave equation, is the maximum displacement of the wave from its equilibrium position. It indicates the strength or intensity of the wave and is a crucial parameter that directly relates to the energy carried by the wave.

Wave Number

The wave number, k, is defined as 2? divided by the wavelength (?). It expresses the spatial frequency of the wave, showing how quickly the wave oscillates as a function of distance. A higher k value indicates a shorter wavelength, meaning that the wave undergoes rapid spatial oscillations.

Angular Frequency

Angular frequency, ?, represents how fast the wave oscillates in time. It is related to the ordinary frequency (f) of the oscillation by the relation ? = 2?f. Angular frequency is essential for determining the temporal behavior of the wave and the rate at which the wave's phase evolves over time.

Wave Propagation Direction

The sign in front of the ?t term in the wave equation (kx ± ?t) determines the wave's direction of travel. A negative sign typically means the wave travels in the positive x-direction, so the phase of the wave moves to the right as time increases. This choice is made based on the observed motion of the wave's features, such as the crest in the given scenario.

Transcript

00:01 So in this question, we have this graph of a standing wave traveled into the right, and we know the peak -to -peak length is 6mm, and it travels by distance d.

00:14 And then, yeah, and then we are given a couple numbers.

00:20 We say, like, this peak a, you travel 6 centimeter in 4 .0 milliseconds.

00:27 So the first thing is that i have to tell you this question, has a tiny mistake or it's not necessarily a mistake but it's the graph is not accurate because from the graph in the book the amount of the tick mark actually is around 5 centimeter so like when a moves from here to here it travels a little bit more than one tick mark but then the question itself says that's the tick mark is 10 centimeter which doesn't correspond to the graph at all.

01:06 So this is a mistake that happens for this new edition.

01:10 So just be aware of that if you are confused about the graph.

01:15 But here i'll be traveling, i will be working on this question by sort of ignoring this graph and only use, i'll use this graph but like i won't look at how it's drawn.

01:29 I will only look at the tick marks because the only useful information we're supposed to get from the tick mark is that if each tick mark is 10 cm, so wavelength of this wave is 40 centimeter.

01:41 That is one important use of information we're supposed to get.

01:45 So i will just use that.

01:49 Let's first look at this wave.

01:52 It has a peak -to -peak height of 6mm and we're supposed to find when it's put in this form, plus or minus, what is, what are the parameters in this equation of the wave.

02:07 So the peak to peak lens is, because peak to peak is 6mm, the amplitude is half of that, it's 3mm.

02:16 Then in order to find k, we need to find k, the wave number is equal to 2 pi over lambda.

02:24 And we find that lambda is 40 centimeter.

02:29 So we can say this...

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