Question Video: Determining the Number of Complete Waves in a Rope | Nagwa Question Video: Determining the Number of Complete Waves in a Rope | Nagwa

# Question Video: Determining the Number of Complete Waves in a Rope Physics • Second Year of Secondary School

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A uniform oscillating wave is traveling in a stretched rope of length 20 m at a speed of 0.8 m/s. If 25 complete waves pass a certain point in 5 s, what is the number of complete waves across the whole length of the rope?

02:52

### 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.

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