### Video Transcript

A uniform oscillating wave is
traveling in a stretched rope of length 20 meters at a speed of 0.8 meters per
second. If 25 complete waves pass a certain
point in five seconds, what is the number of complete waves across the whole length
of the rope?

To begin, we should note that what
we want to find here, the number of complete waves across the rope, refers to the
number of total wave cycles that exist in the rope simultaneously. We know that the rope is 20 meters
long. So, if we can calculate the
wavelength of this wave — that is, the distance between each successive wave cycle —
we can easily tell how many complete wave cycles will fit along the rope’s
length. We’ll simply divide the total
length of the rope by the wave’s wavelength.

So, we first need to go about
finding the wavelength. To do this, we should recall the
wave speed formula, which states that the speed of a wave, 𝑣, equals 𝑓 times 𝜆,
where 𝑓 is the frequency and 𝜆 represents wavelength. Since we want to solve for the
wavelength, let’s rearrange this formula to make 𝜆 the subject by dividing both
sides by 𝑓. The frequency will cancel out of
the numerator and denominator, leaving 𝜆 by itself. Thus, the expression can be written
as 𝜆 equals 𝑣 divided by 𝑓.

We can go ahead and substitute this
into the denominator of our formula for the total number of cycles. Now, we already know that the wave
has a speed of 0.8 meters per second, so this is our value for 𝑣. But we don’t yet know the
frequency, so we’ll have to calculate it. Recall that frequency is the number
of complete wave cycles passing a point in one second. So, since we were told that 25
cycles pass a point in five seconds, we can calculate 𝑓 by simply dividing 25
cycles by five seconds, which equals a frequency of five cycles per second.

We’re now ready to substitute the
frequency into our main formula. We also know the values of 𝑣 and
the total length of the rope, so we’ll substitute those numeric values in too. Before we calculate, though, let’s
take a moment to check out the units here. We have meters divided by meters
per second over cycles per second. In the denominator, units of per
second will cancel each other out, as well as meters from the numerator and
denominator. That leaves only units of cycles in
a denominator within a denominator, which is equivalent to cycles in the overall
numerator.

Also, we should note that cycles
isn’t a physical dimension. It just helps us remember that this
quantity we’re solving for counts the number of wave cycles. We could have chosen to label the
units as “waves” instead.

Now, finally, evaluating this
expression, we find that the number of cycles, or waves, across the entire rope is
125. Thus, when asked for the number of
complete waves across the whole length of the rope, we know that the answer is 125
waves.