In a demonstration, a 2.0 kg horizontal rope is fixed in place at its two ends (
ID: 1471327 • Letter: I
Question
In a demonstration, a 2.0 kg horizontal rope is fixed in place at its two ends (x = 0 and x = 3.0 m) and made to oscillate up and down in the fundamental mode, at frequency 5.1 Hz. At t = 0, the point at x = 1.5 m has zero displacement and is moving upward in the positive direction of a y axis with a transverse velocity of 5.0 m/s. What are (a) the amplitude of the motion of that point and (b) the tension in the rope? (c) Write the standing wave equation for the fundamental mode. Round numeric coefficients to three significant digits.
Explanation / Answer
a)
given
maximum speed = Vmax = A*w = 5
A*2*pi*f = 5
A*2*pi*5.1 = 5
A = 0.156 m <<--------answer
b)
f = (1/2L)*(sqrt(T/u))
linear density = u = m/L =2/3
5.1 = (1/(2*3))*sqrt(T*3/2)
T = 624.24 N
c)
wavelength = 2L = 6 m
k = 2pi/wavelength = pi/3
w = 2*pi*f = 2*pi*5.1 = 32 rad/s
y = A*sinkx*coswt
y = 0.156*sin(pix/3)*cos(32t)