Chapter 16, Problem 050 For a certain transverse standing wave on a long string,

Question

Chapter 16, Problem 050 For a certain transverse standing wave on a long string, an antinode is at x 0 and an adjacent node is at x 0.40 m. The displacement y(t) of the string particle at x O is shown in the figure, where the scale of the y axis is set by ys 4.2 cm. When t0.50 -0.70 m ? What is the transverse velocity of the string particle at x-0.50 m at (c) t = 0.50 s and (d) s, what is the displacement of the string particle at (a)-0.50 m and (b) x 1.3 s? (a) Number (b) Number (c) Number (d) Number

Explanation / Answer

given

amplitude A (= ys) = 4.2 cm

time period T = 2.8 s

Angular frequency w = 2*pi / T = 2*pi/2.8 =0.71*pi rad/s

since node at x = 0.40 m

k*0.4 = pi/2

k = 5*pi/4

the standing wave funtion at x=0 ,t=0s is

y = A *cos(k*x)*sin(w*t)

y = (-0.042)*cos(5*pi/4 *x)*sin(0.71*pi*t) (taking A = -0.042 because at given time ys is negative)

(a)

the displacement for x = 0.50 m, t = 0.50 s

y = (-0.042)*cos(5*pi*0.50/4)*sin(0.71*pi*0.5)

y = 0.0134 m

(b)

the displacement for x = 0.70 m ,t= 0.50 s

y = (-0.042)*cos(5*pi*0.70/4)*sin(0.71*pi*0.50)

y = 0.034 m

(c)

the velocity of the wave is

v = dy/dt = (-0.042)*w*cos(k*x)*cos(w*t)

at x = 0.50m , t= 0.50 s

v = (-0.042)*0.71*pi*cos(5*pi*0.50/4)*cos(0.71*pi*0.50)

v = 0.0157 m/s

(d)

the velocity of wave at x = 0.5 m ,t = 1.3s

v = (-0.042)*0.71*pi*cos(5*pi*0.5/4)*cos(0.71*pi*1.3)

v = -0.034 m/s

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