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A transverse wave on a string has amplitude 0.300 cm,

Chapter 15, Problem 13E

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QUESTION:

A transverse wave on a string has amplitude 0.300 cm, wavelength 12.0 cm, and speed 6.00 cm/s. It is represented by y(x, t) as given in Exercise 15.12. (a) At time t = 0, compute y at 1.5-cm intervals of x (that is, at x = 0, x = 1.5 cm, x = 3.0 cm, and so on) from x = 0 to x = 12.0 cm. Graph the results. This is the shape of the string at time t = 0. (b) Repeat the calculations for the same values of x at times t = 0.400 s and t = 0.800 s. Graph the shape of the string at these instants. In what direction is the wave traveling? 15.12 .. ? ALC Speed of Propagation vs. Particle Speed. (a) Show that Eq. (15.3) may be written as (b) Use y(x, t) to find an expression for the transverse velocity vy of a particle in the string on which the wave travels. (c) Find the maximum speed of a particle of the string. Under what circumstances is this equal to the propagation speed v? Less than v? Greater than v?

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QUESTION:

A transverse wave on a string has amplitude 0.300 cm, wavelength 12.0 cm, and speed 6.00 cm/s. It is represented by y(x, t) as given in Exercise 15.12. (a) At time t = 0, compute y at 1.5-cm intervals of x (that is, at x = 0, x = 1.5 cm, x = 3.0 cm, and so on) from x = 0 to x = 12.0 cm. Graph the results. This is the shape of the string at time t = 0. (b) Repeat the calculations for the same values of x at times t = 0.400 s and t = 0.800 s. Graph the shape of the string at these instants. In what direction is the wave traveling? 15.12 .. ? ALC Speed of Propagation vs. Particle Speed. (a) Show that Eq. (15.3) may be written as (b) Use y(x, t) to find an expression for the transverse velocity vy of a particle in the string on which the wave travels. (c) Find the maximum speed of a particle of the string. Under what circumstances is this equal to the propagation speed v? Less than v? Greater than v?

ANSWER:

Solution 13E The wave function for a sinusoidal wave is given as, y(x,t) = A cos[ (x vt)] Given, amplitude A = 0.300 cm Wavelength = 12.0 cm At time t = 0 s, y(x,0) = Acos [ ()] …..(1) since , t = 0 Now, x intervals is given to be 1.5. Therefore, various x values starting from x = 0 will be x = 12.0 cm are, 0, 1.5, 3, 4.5, 6, 7.5, 9, 10.5 and 12 Substituting these values of x in equation (1), we get y(0) = 0.300 cm × cos[ 2 (0)] 12.0 y(0) = 0.300 cm Similarly, y(1.5) = 0.21 cm y(3) = 0 y(4.5) = 0.21 y(6) = 0.300 cm y(7.5) = 0.21 cm y(9) = 0 y(10.5) = 0.21 cm y(12) = 0.30 cm The graph of y(x) vs x is done as shown below. x y(x) cm

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