Question

A rope, under a tension of 311 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by   y = ( 0.393   m ) sin ( π ⁢ x / 5.00 ) sin ( 13.0 π ⁢ 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?

Answer 1

given

T = 311 N

compare the given equation with

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

A = 0.393 m
k = pi/5 rad/m
w = 13*pi rad/s
let L is the length of the rope.

a) we know, k = 2*pi/lamda

lamda = 2*pi/k

= 2*pi/(pi/5)

= 10 m

in second harmonic, L = lamda

= 10 m <<<<<<<<<<<------------------Answer

b) speed of the wave, v = w/k

= (13*pi)/(pi/5)

= 65 m/s <<<<<<<<<<<------------------Answer

c) we know, v = sqrt(T/(m/L))

v^2 = T/(m/L)

m/L = T/v^2

m = T*L/v^2

= 311*10/65^2

= 0.736 kg <<<<<<<<<<<------------------Answer

d) f2 = w/(2*pi)

= 13*pi/(2*pi)

= 6.5 Hz

fundamental frquency, f1 = f2/2

= 6.5/2

= 3.25 Hz

third hamrmonic, f3 = 3*f1

= 3*3.25

= 9.75 Hz

time period, T = 1/f3

= 1/9.75

= 0.102 s <<<<<<<<<<------------Answer