A wave on a string is described by y(x, t)=15.0 sin(πx / 8-4 πt) where x and y are in centimeter

A wave on a string is described by y(x, t)=15.0 sin(πx / 8-4...

A wave on a string is described by $$y(x, t)=15.0 \sin (\pi x / 8-4 \pi t)$$ where $x$ and $y$ are in centimeters and $t$ is in seconds. (a) What is the transverse speed for a point on the string at $x=6.00 \mathrm{cm}$ when $t=0.250$ s? (b) What is the maximum transverse speed of any point on the string? (c) What is the magnitude of the transverse acceleration for a point on the string at $x=6.00 \mathrm{cm}$ when $t=0.250 \mathrm{s} ?$ (d) What is the magnitude of the maximum transverse acceleration for any point on the string?

A wave on a string is described by
$$y(x, t)=15.0 \sin (\pi x / 8-4 \pi t)$$ where $x$ and $y$ are in centimeters and $t$ is in seconds. (a) What is
the transverse speed for a point on the string at $x=6.00 \mathrm{cm}$
when $t=0.250$ s? (b) What is the maximum transverse speed of any point on the string? (c) What is the magnitude of the
transverse acceleration for a point on the string at $x=6.00 \mathrm{cm}$
when $t=0.250 \mathrm{s} ?$ (d) What is the magnitude of the maximum
transverse acceleration for any point on the string?

Key Concepts

Wave Equation

A wave equation mathematically describes how a disturbance (or wave) propagates through space and time. It typically involves parameters such as amplitude, wavelength, frequency, and phase. In the context of a string, the wave equation provides information on how the displacement of points on the string varies with both position and time.

Transverse Velocity

Transverse velocity refers to the rate at which a point on the string moves perpendicular to the direction of wave propagation. It is obtained by taking the partial derivative of the displacement with respect to time. This derivative indicates how quickly the string’s displacement changes at a specific location and time.

Transverse Acceleration

Transverse acceleration is the rate of change of the transverse velocity with respect to time. It is found by taking the second partial derivative of the displacement function with respect to time. This measure provides insight into how the speed of the particle's displacement direction is changing, which is important in understanding the dynamic behavior of waves.

Angular Frequency and Wave Number

Angular frequency and wave number are fundamental parameters of a wave. The angular frequency, usually denoted by ?, describes how rapidly the wave oscillates in time. The wave number, denoted by k, describes how rapidly the wave oscillates in space. These parameters are essential for relating the time and spatial derivatives of the wave, and they play a key role in determining properties such as maximum transverse speed and acceleration.

Transcript

00:01 So in this question, we are given this equation, and we want to find various properties of this wave.

00:06 So in part a, we want to understand what is a transfer speed for a point, at a certain distance and at a certain time.

00:16 So in other words, we want to find what u is at x equals 6 centimeter and t equals 0 .250 second.

00:26 So we want to first find out what u is u equals the u is a transfer speed, which is a derivative of y with respect to t.

00:37 And when we find the derivative of y with respect to t, we can take the derivative of this equation.

00:44 It's negative 4 pi times 15, and let's double check.

00:49 Everything is in terms of centimeter times cosine pi x over 8 minus 4 pi t.

00:56 And then when t is 0 .25 this is just and when x is equal to 6 this equation becomes times 15 times cosine 6 over 8 is 4 over 3 pi minus all right so now we can put in the number it's equal to negative 70 pi times cosine negative 1 over 7.

01:32 That is 1 over square root 2.

01:37 So this is equal to negative 1 33 and we still remember the unit at centimeter per second.

01:46 So it's roughly 1 .3 meter per second and it's negative meaning it is going in the negative y direction.

02:00 So we have part b.

02:01 Part b asks us what is maximum transfer speed of any points.

02:05 This string.

02:09 So you, we have this equation over here, it's negative 60 pi cosine, pi x of 8 minus 4 pi t.

02:21 So this achieves maximum value when negative cosine something is just one.

02:27 So this achieves max value is pi.

02:32 All right, so 60 pi.

02:39 And we can also put into the decimal point.

02:46 In decimal number it is 188 centimeter per second...

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