Question

1)The equation of transverse wave on a string is y=0.12sin(20x-

Answer 1

a) wave speed = w/k = 5/20 = 0.25 m/s

b) wave speed = sqrt ( tension / linear density)

0.25 = sqrt (16.2/linear density)

linear density = 259.2 Kg/m

2)

y(15cm,t) = 10cm sin(2.5-8t)

(a) we have kx = 2.5

so, k*0.5 = 2.5

k = 5

What is the speed of the wave traveling along the string = w/k = 8/5 = 1.6 m/s is the answer

b) maximum speed of the paper

dy/dt = (0.1*-8) sin(2.5-8t) = -(0.8) sin(2.5-8t)

so, maximum speed of the paper = 0.8 m/s is the answer

Answer 2

1.a) Since angular velocity=20rad./s

so, frequency=angular velocity/2$\Pi$

so, f=3.184$s^{-1}$

and $2*\Pi /\lambda =k=5$

$\Rightarrow \lambda =1.256m$

so, velocity $velocity v=f*\lambda =4ms_{-1}$

1.b) since, $v=\sqrt{T/m/L}$ where T=Tension, m=mass, L=length

$mass per unit length=T/v^{^{2}}$

mass per unit length=1.0125kg/m

2.a) from 1.a velocity v=.3125m/s

2.b) differenciate y with repect to time and it's maximum value will be maximum velocity of paper=80cm/s.

Answer 3