As a Function of Position A sinusoidal transverse wave is...

As a Function of Position A sinusoidal transverse wave is traveling along a string in the negative direction of an $x$ axis. Figure $17-33$ shows a plot of the displacement as a function of position at time $t=0 \mathrm{~s}$; the $y$ intercept is $4.0 \mathrm{~cm}$. The string tension is $3.6 \mathrm{~N}$, and its linear density is $25 \mathrm{~g} / \mathrm{m}$. Find the (a) amplitude, (b) wavelength, (c) wave speed, and (d) period of the wave. (e) Find the maximum transverse speed of a particle in the string. (f) Write an equation describing the traveling wave.

As a Function of Position A sinusoidal transverse wave is traveling along a string in the negative direction of an $x$ axis. Figure $17-33$ shows a plot of the displacement as
a function of position at time $t=0 \mathrm{~s}$; the $y$ intercept is $4.0 \mathrm{~cm}$. The string tension is $3.6 \mathrm{~N}$, and its linear density is $25 \mathrm{~g} / \mathrm{m}$. Find the (a) amplitude,
(b) wavelength, (c) wave speed, and (d) period of the wave. (e) Find the maximum transverse speed of a particle in the string.
(f) Write an equation describing the traveling wave.

Key Concepts

Amplitude

Amplitude is the maximum displacement of points in the medium from the equilibrium position. It quantifies the wave's energy and is essential in defining the shape and intensity of the wave.

Wavelength

Wavelength is the spatial period of the wave, representing the distance over which the wave's shape repeats. It is fundamental in describing the size scale of the wave and is used together with the amplitude to characterize the wave.

Wave Speed

Wave speed is the rate at which a wave propagates through a medium. In the context of waves on a string, it is determined by the tension in the string and the linear mass density, often calculated using the formula v = sqrt(T/?), linking the mechanical properties of the medium to the propagation speed.

Period

The period is the time required for one complete oscillation or cycle of the wave to pass a fixed point. It is inversely related to the frequency and, along with the wavelength, it connects temporal and spatial descriptions of the wave movement.

Maximum Transverse Speed

The maximum transverse speed of a particle in the wave is the peak speed at which a point on the medium oscillates vertically, found using the product of the angular frequency and the amplitude. This speed provides insight into the dynamics of the oscillatory motion.

Traveling Wave Equation

A traveling wave is mathematically represented by an equation that includes the amplitude, wavelength, frequency (or period), and phase. The general form, y(x,t) = A sin(kx - ?t + ?), describes how the displacement varies with both position and time, encapsulating the wave’s spatial and temporal behaviors.

Transcript

00:03 Let's start from the part of this problem.

00:06 In this part of the problem, we have to find the amplitude of the oscillations that is written as ym.

00:14 It can be seen from the graph that the oscillation of the particles, which is the amplitude of the oscillations, which is the distance from mean position to the extreme position, is equals to ym, and this is equals to 5 .0 centimeter.

00:29 So this is the amplitude of the oscillations.

00:34 In part b, we have to describe the yf length, that is the lambda.

00:41 So from the graph we can see that the distance between the two conjunctive same phases is equals to say lambda and this is equal to 55 centimeter minus 15 centimeter so this is equals to 40 centimeter so the 40 centimeter is the waif length of this wave similarly in part c we have to describe the wave speed that is written as we we can write the speed of the v as v as v is equal to square root of tau divided by mu.

01:25 Here this tau is the tension on the string and this mu is the linear mass density of the string.

01:31 So let's set the values.

01:33 So square root of 3 .6 newton divided by 25 multiplied by turn 10.

01:40 Power minus 3 10 is power minus 3 kg per meter so here we will get the value for this speed as 12 meter per second this is the speed of the web let's move to the part uh uh d of this problem in this case we have to find the period of period of the web that is t so we can write the speed as v is equals to f lambda this is written as lambda uh divide by capital t so from here, we can write this equation for the t as t is equals to lambda divided by v.

02:28 So let's set the values into this equation.

02:30 So here we have 40 multiply by turn as far minus 2 meter divided by 2 meter divided by 12 meter per second.

02:44 So from here, we will get the value for this v as v is equals to 0 .033 meter per second.

02:55 So this is the speed of the web.

02:57 Sorry this is this is the period of the web so it will be t is equals to 0 .33 seconds or say this would be equals to 0 .03 seconds so this is the answer to the party of this problem now let's move to the party of this problem in party of this problem we have to find the maximum transfer speed of the particles in the string so let the speed is um which is the maximum particle speed we can write the equation for the maximum particle speed as um is equal to ym which is the amplitude of the oscillations multiply by the angular frequency which is omega so this can be written as ym into 2 pi f because this omega is equal to 2 pi f and here this is the frequency of the wave so this is written as a 2 pi m f now let's set the values into this quadrant so here we have u m is equals to 2 pi variance into here this ym which is equals to 5 .0 multiply by turn raise power minus two meter into the frequency which is equals to 30 hertz and this frequency can be obtained from this period by using this equation f is equals 2 1 divided by so from here we will get the value for this um as um is equal to 9 .4 meter per second.

04:44 So this is the maximum speed of the particles we take part in the oscillations...

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