A rope, under a tension of 114 N and fixed at both ends, oscillates in a second-

Question

A rope, under a tension of 114 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.27m)sin(pi times/5.0) sin(14 pi t). where x = 0 at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation? (a) Number 10 Units m (b) Number 70 Units m/s (c) Number 16.2857 Units kg (d) Number 0.30 units s

Explanation / Answer

a) given, y = 0.27sin(pi*x/5)sin(14pi*t)
a) length of rope = L
for second harmonic
lambda(wavelength) = L
but k = 2*pi/lambda
and from the equation, k (wavenumber) = pi/5
so, pi/5 = 2*pi/L
L = 10 m
b) speed of waves on rope = v
v = lambda/T
w = 2*pi/T
so, v = lambda*w/2*pi = L*w/2*pi
from equation, angular wave number, w = 14*pi
so v = 10*14*pi/2*pi = 70 m/s
c) we know that velocity on a string of a pulse is v = sqroot(T/mu)
where mu = m/L
m is mass
L is length of the string
70 = sqroot(114*10/m)
m = 0.2326 kg
d) for third harmonic osscilation
L = 3*lambda/2
and v = lambda/T
T = lambda/v = 2*10/3*70 = 0.0952 s

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