For a particular transverse standing wave on a long string,...

For a particular transverse standing wave on a long string, one of the antinodes is at $x=0$ and an adjacent node is at $x=0.10 \mathrm{m}$ . The displacement $y(t)$ of the string particle at $x=0$ is shown in Fig. $16-40$ where the scale of the $y$ axis is set by $y_{s}=4.0 \mathrm{cm} .$ When $t=0.50 \mathrm{s},$ what is the displacement of the string particle at $(\mathrm{a}) x=0.20 \mathrm{m}$ and ( b ) $x=0.30 \mathrm{m} ?$ What is the transverse velocity of the string particle at $x=0.20 \mathrm{m}$ at (c) $t=0.50$ s and $(\mathrm{d}) t=1.0 \mathrm{s} ?(\mathrm{e})$ Sketch the standing wave at $t=$ 0.50 s for the range $x=0$ to $x=0.40 \mathrm{m} .$

For a particular transverse standing wave on a long string, one
of the antinodes is at $x=0$ and an
adjacent node is at $x=0.10 \mathrm{m}$ . The
displacement $y(t)$ of the string particle at $x=0$ is shown in Fig. $16-40$ where the scale of the $y$ axis is set by
$y_{s}=4.0 \mathrm{cm} .$ When $t=0.50 \mathrm{s},$ what is
the displacement of the string particle
at $(\mathrm{a}) x=0.20 \mathrm{m}$ and ( b ) $x=0.30 \mathrm{m} ?$ What is the transverse velocity of the string particle at $x=0.20 \mathrm{m}$ at
(c) $t=0.50$ s and $(\mathrm{d}) t=1.0 \mathrm{s} ?(\mathrm{e})$ Sketch the standing wave at $t=$
0.50 s for the range $x=0$ to $x=0.40 \mathrm{m} .$

Key Concepts

Standing Waves

Standing waves are patterns that remain fixed in space resulting from the superposition of two waves traveling in opposite directions with the same frequency and amplitude. They are characterized by positions that always remain at rest (nodes) and positions that oscillate with maximum amplitude (antinodes), illustrating how interference can give rise to stationary wave patterns.

Nodes and Antinodes

Nodes are fixed points along a medium where the displacement is always zero due to destructive interference, while antinodes are points where the displacement reaches its maximum due to constructive interference. Understanding the location of nodes and antinodes is fundamental to analyzing standing waves, as these features determine the spatial structure of the oscillation.

Wavelength Determination

The spatial separation between nodes and antinodes is directly related to the wavelength of the wave. Specifically, the distance between a node and its adjacent antinode is one quarter of the wavelength, and consecutive nodes are separated by half a wavelength. This relationship is crucial for understanding and calculating the spatial characteristics of standing waves on a string.

Transverse Velocity

The transverse velocity of a point on a wave is defined as the time derivative of its displacement. In the context of a standing wave, this derivative provides insight into how quickly a particular point on the string moves up and down at a given time, which is essential for understanding the dynamic behavior of the wave and for solving problems related to the instantaneous motion of different points along the string.

Phase Relationships in Standing Waves

Standing waves exhibit both time-dependent and spatial phase variations. The phase difference between points on the wave explains why different positions (such as those corresponding to nodes, antinodes, or intermediate points) reach their maximum and minimum displacements at different times. A proper grasp of these phase relationships is crucial when predicting displacements and velocities at various points along the string for given times.

Transcript

00:01 16 .50.

00:03 So we have a particular wave on a long string.

00:08 We have an anti -node at x equals 0, and an adjacent node is at 0 .1 meters, presumably the next, the other adjacent node would be at negative 0 .1 meters.

00:22 Now at a time of 0 .5 seconds, we want to know what the displacement is at 0 .2 meters and at 0 .3 meters.

00:33 And then at 0 .2 meters we want to know the speed, the transverse velocity, at 0 .5 seconds and at 1 second.

00:45 And then we'd like to sketch everything at 0 .5 seconds between 0 and 0 .4 meters.

00:54 So there are two ways you can go about doing this.

01:01 One of them is to write down what we're about to write down here, which is we can we can work out what the actual form of the standing wave is and then just directly compute things from that.

01:21 And the other one is that a lot of this we can do just from using what we know about, you know, nodes and antinodes in a standing wave on a string.

01:49 So our amplitude is four centimeters.

02:02 This is sort of encapsulating what we know that there has to be an anti -node at x equals the negative sign is because at time greater than zero, we see it goes down initially.

02:22 And then we have a sign of omega -t, or omega, so of course, 2 pi divided by the period.

02:36 We can see here that the period is two seconds.

02:42 So two seconds here, we just get pi radians per second is our angular frequency.

02:49 And now we know that for the wave number, we know that there is a node at 0 .1 meters, from the anti -note at zero.

03:05 So k times 0 .1 meters has to be pi over 2 because that's the phase difference to go half a cycle.

03:22 I mean, this is not the plot of the relevant thing, but it's all sinus sinusoidal functions.

03:29 And so then this implies that the wave number is five radians per meter.

03:37 So we can use this form in these parameters to then compute all of these things, but we don't actually need to for the most part because of the way that this is set up.

03:57 And in fact, let's work slightly backwards.

04:06 Let's draw a sketch of this down here.

04:21 Here's our displacement.

04:24 Here's x...

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