Question

A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.10 mm cos[( 7.02 rad/m )x+( 749 rad/s )t].Being more practical-minded, you measure the rope to have a length of 1.34 m and a mass of 3.32 g .

Part A

Determine the amplitude.

Part B

Determine the frequency

Part C

Determine the wavelength.

Part D

Determine the wave speed.

Part E

Determine the direction the wave is traveling.

	−x−direction
	+x−direction

Part F

Determine the tension in the rope.

Part G

Determine the average power transmitted by the wave.

Answer 1

Given that

A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is

y(x,t)= 2.10 mm cos[( 7.02 rad/m )x+( 749 rad/s )t]

Mass of the rope (m) =3.32g =3.32*10^-3kg

Length of the rope is (L) =1.34m

a)

We know that

y(x.t) =Acos(wt+kx)

Comparing with the given equation we get

Amplitude is (A) =2.10mm =0.0021m

b)

The frequency of the wave is given by

w =2pif

f =w/2pi from the above equation w =749rad/s

=749/6.28 =119.2675HZ

c)

The wavelength of the wave is given by

k =2pi/lamda

lamda =2pi/k = 6.28/7.02 = 0.8945m (k=7.02 rad/m from the above equation ))

d)

The speed of the wave is given by

v =f*lamda =(119.2675HZ)( 0.8945m )=106.695m/s

e)

The direction of the string will be in the negative x-direction(-x)

f)

The tension in the rope is

v =Sqrt(T/u)

v² =T/u ===> T =v²*u =v²*(m/L) =(106.695m/s)²(3.32*10^-3kg/1.34) =28.2047N

g)

The  average power transmitted by the wave is

P =(1/2)uvw²A² =0.5*(3.32*10^-3kg/1.34)(106.695m/s)( 749 rad/s )²(0.0021m)² =0.327W