Video: Calculating the Motion of a Point on a String’s Wave Perpendicular to the Wave Propagation Direction

A transverse wave moves along a string at 3.00 × 10² m/s. The wave’s amplitude is 20.0 cm and its wavelength is 0.500 m. How much time is required for a point on the string to move through a vertical distance of 5.00 km?

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Video Transcript

A transverse wave moves along a string at 3.00 times 10 to the two metres per second. The wave’s amplitude is 20.0 centimetres and its wavelength is 0.500 metres. How much time is required for a point on the string to move through a vertical distance of 5.00 kilometres?

In this statement, we’re told that the speed of the wave along the string is 3.00 times 10 to the two metres per second. We’ll call that 𝑣. We’re then told that the wave’s amplitude is 20.0 centimetres, which we’ll call capital 𝐴. And the wavelength of the wave is 0.500 metres; we’ll represent that using the Greek letter 𝜆. We’ll then asked to solve for the time needed for a point on the string to move through a vertical distance of 5.00 kilometres. We’ll call that time that we’re looking to solve for 𝑡. And we’ll refer to the vertical distance as 𝑑 sub 𝑣.

Let’s start by drawing a picture of what the wave looks like from the side. In this diagram of our wave, we’ve modelled the wave moving left to right with the speed 𝑣 given in the problem statement. The amplitude 𝐴 of the wave is the distance from the horizontal midpoint, shown here with a dotted blue line, to the peak or crest of the wave. And the wavelength of the wave is the distance from one trough to an adjacent trough or equivalently one crest to an adjacent crest.

If we picked a certain point along the wave, say at the first crest of the wave that we see, and measured the vertical motion of that point as the wave passes through the point from left to right, we would see that point move up and down over and over as the wave passes through that point.

From the diagram, we can see that such a point would move a distance of two times the wave amplitude and going from a peak to a trough. And then on the return journey, the point would again go through a vertical distance of two times the wave amplitude. So that altogether when the point moves through one period of the wave, it would move through a vertical distance of four times the wave amplitude 𝐴.

To solve for the time 𝑡, it would take this point to move through a vertical distance of 5.00 kilometres. Let’s figure out how long one wave is and how much distance four times the amplitude is. First, the wave period. Let’s recall two wave relationship equations. The first is for wave speed: the speed of a wave 𝑣 is equal to the frequency of the wave 𝑓 multiplied by its wavelength 𝜆.

We also know that the frequency of a wave relates to the period of the wave inversely; that is, frequency 𝑓 is equal to one divided by period 𝑇. If we apply the wave speed equation to our scenario and substitute in one over 𝑇 for the frequency 𝑓, then we can rearrange this equation to solve for 𝑇, the period of the wave.

Let’s multiply both sides of the equation by 𝑇 divided by 𝑣. When we do this, the period 𝑇 on the right side of the equation cancels out. And the wave speed 𝑣 cancels out on the left side of our equation, leaving us with a relationship for 𝑇, which says that the wave period 𝑇 is equal to the wavelength 𝜆 divided by the wave speed 𝑣.

We’re given 𝜆 and 𝑣 and can insert those values into this equation. The period of this wave equals 0.500 metres divided by 3.00 times 10 to the two metres per second. This is equal to 1.667 times 10 to the negative third seconds.

Now that we know the wave period 𝑇, we can solve for the value of four times the wave amplitude, which is the vertical distance that a point on the string moves over one wave period of time. Four 𝐴 is four times 20.0 centimetres or four times 0.200 metres. Multiplying these two numbers together, we find that the vertical distance that a point on the string moves over one wave period is 0.800 metres. We now know the period and the vertical distance moved by a point on the string over one period.

Now, let’s figure out how many periods must pass by in order for the point on our string to move up and down at distance of 𝑑 sub 𝑣 5.00 kilometres. This will be equal to 𝑑 sub 𝑣 divided by the vertical distance the point on the string moves over one period, which we just found is 0.800 metres. That number of periods is equal to 5.00 kilometres or 5.00 times 10 to the third metres divided by four times the wave amplitude or 0.800 metres. When we calculate this fraction, we find a value of 6250 periods.

In order to solve for 𝘵, the time value for a point to move up and down this distance, now we need to multiply this number of periods by the total time that a period takes up, which we solved for as 1.667 times 10 to the negative third seconds. Therefore, 𝘵 is equal to 6250 times 1.667 times 10 to the negative third seconds.

Multiplying these two numbers together gives us a value of 10.4 seconds. This is how long to three significant figures it would take for a point on the string experiencing this wave to move a vertical distance of 5.00 kilometres.

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