The top end of a yo-yo string is held stationary. The yo-yo...

The top end of a yo-yo string is held stationary. The yo-yo itself is much more massive than the string. It starts from rest and moves down with constant acceleration 0.800 $\mathrm{m} / \mathrm{s}^{2}$ as it unwinds from the string. The rubbing of the string against the edge of the yo-yo excites transverse standing-wave vibrations in the string. Both ends of the string are nodes even as the length of the string increases. Consider the instant 1.20 s after the motion begins from rest. (a) Show that the rate of change with time of the wave-length of the fundamental mode of oscillation is 1.92 m/s. (b) What if? Is the rate of change of the wavelength of the second harmonic also 1.92 m/s at this moment? Explain your answer. (c) What if? The experiment is repeated after more mass has been added to the yo-yo body. The mass distribution is kept the same so that the yo-yo still moves with downward acceleration 0.800 $\mathrm{m} / \mathrm{s}^{2}.$ At the 1.20-s point in this case, is the rate of change of the fundamental wavelength of the string vibration still equal to 1.92 m/s? Explain. (d) Is the rate of change of the second harmonic wavelength the same as in part (b)? Explain.

Key Concepts

Harmonic Modes and Wavelength Relationships

For a string with fixed endpoints, the allowable wavelengths are quantized according to the harmonic number, typically given by the relation ?_n = 2L/n, where L is the length of the string and n is a positive integer. This relationship shows how different wave modes (fundamental, second harmonic, etc.) depend on the string length, which in turn affects how their wavelengths change when the length varies.

Uniform Acceleration and Kinematics

This concept involves objects moving with a constant acceleration, where relationships between displacement, velocity, and time are defined by equations of motion. In problems like this, knowing the instantaneous velocity of a moving object helps determine how fast certain related quantities (such as the unwinding string’s length) change over time.

Standing Waves on a String

Standing waves occur when waves traveling in opposite directions interfere, creating fixed nodes where there is no displacement. When a string is fixed at both ends, only certain wavelengths that satisfy this boundary condition are allowed, leading to the formation of distinct harmonic modes.

Time-Dependent Boundary Conditions

When the length of the vibrating medium changes with time—as with a string that is being unwound—the boundary conditions shift, causing the wavelengths of standing waves to change over time. Analyzing how these wavelengths vary involves applying the chain rule to relate the rate of change of the string length to the rate of change of the allowed wavelengths.

Transcript

00:01 Okay, so in this question, we are dealing with a yo -yo, and we're told that, you know, it is accelerating towards the ground.

00:10 Someone, you know, kind of starts playing with the yo -yo, and it starts accelerating towards the ground with an acceleration of 0 .80 meters per second squared.

00:23 And the time the moment in time that we are concerned with is t equals 1 .20 seconds i'm just going to rewrite that there we go and we this is you know a problem in the waves chapter because as the yo -yo is rolling off the string it rubs against the string and starts exciting transfers waves in the string as the length of the string increases.

00:57 And in part a, we are supposed to show that the rate of change in lambda over time in the wavelength is 1 .92 meters per second.

01:12 So what we were trying to show is that d lambda d t is equal to 1 .92 meters per second.

01:22 We're doing this specifically for the fundamental frequency of these transverse waves on the string.

01:30 So what we can do here, the first thing that we should do that will be helpful is to figure out the relationship between the length of the string and time.

01:42 So what is l as a function of time? and of course this comes back to a kinematics equation, and the one that's most pertinent here is one half a t squared plus vt plus l initial and so the thing is or and we can say v not as well but the thing is the initial length of the string and the initial speed of the string are both zero so the second and third terms drop out so our l of t becomes just one half a, t squared.

02:27 And we're given both the acceleration and the moment and time that we're concerned with.

02:35 So from there, we can take a look at how lambda is related to l.

02:43 So what is lambda as a function of l? well, the problem tells us that we can treat both the end of the string at the person's hand as a node and the end of the string at the person's hand as a node, and the end of the string at the yo -yo as a node, which makes sense.

02:59 Those are both being held in place.

03:01 So that means that our lambda is going to be equal to twice the length of the string.

03:06 So lambda as a function of l is 2l.

03:10 So now what we want to find is we want to find lambda d -t.

03:16 We have a function of lambda in terms of l, and we have a function of l in terms of t.

03:23 So we can write a function of lambda or sorry we have a function of lambda in terms of l and l in terms of t so we can write a function of lambda in terms of t as a t squared because 2l is 2l 2l 2l times 1 half a t is just a t squared so our formula for lambda over in terms of time is at squared.

03:56 So if we take d lambda d t and we just take the derivative of a t squared, we'll end up with 2a t.

04:09 And then we want to evaluate this at t equals 1 .2 seconds.

04:14 And so we can say that at the time that we care about, the rate of change over time of the wavelength of these transverse waves is equal to 2 times 0 .80 times 1 .20.

04:32 And if you go ahead and plug that into your calculator, you will get that d -lamda dt is in fact 1 .92 meters per second...

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