Question

A guitar string vibrates in its fundamental mode, with nodes at its ends. The length of the rope segment that vibrates freely is 0.386 m. The maximum transverse acceleration of a point at the midpoint of the segment is 8.40x103 m / s2, and the maximum transverse velocity is 3.80 m / s. a) Calculate the amplitude of this standing wave. b) How fast are transverse traveling waves in this rope?

a	a) 1.72 m b) 543 m/s
b	a) 1.72x10-3 m b) 272 m/s
c	a)6.45x10-3 m b) 234 m/s
d	a) 5.77x10-9 m b) 275 m/s

Answer 1

the Displacement of the string is

Y(x,t)= A sin kx sin wt

Distance between the adj nods = lambda/2

length of the string lamba = 2 L

Amplitude of the wave ASinkx

then sin kx = sin(2pi/2l) * L/2 = 1  

y(x,t) = A sin Wt

using L = 0.386 m

diff above eqn with respect to t,

Vy = dy/dy = W A Coswt

Max travservse velocity Vmax = W A = 3.8 m/s^2

Diff Vy w r t t to get accleration

a= dV/dt = -W^2 A sin wt

for max , a = -w^2 A

angular velocity W = a/V

W = (8.4 e 3)/3.8 = 2.21 e3 rad/s

Amplitude A =V/w = 3.8/2.21 e 3 = 1.72 *10^-3 m --(ANSWER)
----------------------

as V = Lambda*f

V = 2L * W/2pi

V = wL/pi

V = (0.386 * 2.21 e 3)/(3.14)

V = 272 m/s ------(ANSWER)

option B it is