A transverse harmonic wave travels on a rope according to the following expressi

Question

A transverse harmonic wave travels on a rope according to the following expression:

y(x,t) = 0.17sin(2.6x + 17.3t)

The mass density of the rope is = 0.105 kg/m. x and y are measured in meters and t in seconds.

1)

What is the amplitude of the wave?

m

2)

What is the frequency of oscillation of the wave?

Hz

Your submissions:

17.3

Computed value:

17.3

Submitted:

Saturday, December 5 at 2:50 PM

Feedback:

This is the angular frequency (in rad/s) - we want the frequency (in Hz).

3)

What is the wavelength of the wave?

m

4)

What is the speed of the wave?

m/s

5)

What is the tension in the rope?

N

6)

At x = 3.4 m and t = 0.48 s, what is the velocity of the rope? (watch your sign)

m/s

7)

At x = 3.4 m and t = 0.48 s, what is the acceleration of the rope? (watch your sign)

m/s2

8)

What is the average speed of the rope during one complete oscillation of the rope?

m/s

Explanation / Answer

Given that

A transverse harmonic wave travels on a rope according to the following expression:

y(x,t) = 0.17sin(2.6x + 17.3t)

We know that

y(x,t) =Asin(kx+wt)

Comapring above two equation we get

a)

The amplitude of the wave is (A) =0.17m

b)

What is the frequency of oscillation of the wave is given by

w =2pif

Then frequency (f) =2pi/w from the given equation w =17.3rad/s

Now f =6.28/17.3 =0.3630Hz

3)

The wavelength of the wave is given by

k =2pi/lamda

Then the wavelength ( lamda) =2pi/k from the given equation k =2.6m-1

Then the wavelength ( lamda) =2pi/k = lamda =6.28/2.6 =2.415m

4)

The speed of the wave is given by

speed (v) =frequency(f) * wavelength (lamda)

v =(0.3630Hz)(2.415m) =0.8767m/s

5)

Given that

The mass density of the rope is ( ) = 0.105 kg/m

The tension in the rope is given by

v =Sqrt(T/)) ===>v2 =T/) ===>T =v2*

The the tension (T) =v2*=(0.8767m)2*(0.105 kg/m)=0.08070N

6)

At x = 3.4 m and t = 0.48 s, what is the velocity of the rope is given by substituting the values in the given equation y(x,t) = 0.17sin(2.6x + 17.3t)

v =dy/dt =(0.17)(17.3)cos(2.6x+17.3t)=(2.941)cos(2.6*3.4+17.3*0.48) =(2.941)cos(8.84+8.304)=2.8103m/s

7)

At x = 3.4 m and t = 0.48 s, what is the acceleration of the rope is given by substituting the values in the given equation y(x,t) = 0.17sin(2.6x + 17.3t)

a =dv/dt =d((0.17)(17.3)cos(2.6x+17.3t))/dt =-(0.17)(17.3)2sin(2.6x+17.3t) =-(50.8793)sin (2.6*3.4+17.3*0.48) =-(50.8793)sin(8.84+8.304)=-14.997m/s2 nearly -15m/s2

8)

The average speed of the wave is given by we know that one complete oscillation is to and from motion, during each trip there two amplitude dispalcements so the total of 4 amplitudes. Then the average is the total displacemnt over the time taken to the period

Average speed =4A/Period =4A/(1/f) =4*0.17m/(1/0.3630Hz) =0.68/2.7548=0.2468m/s

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