I and my colleagues were solving some physics problems relating to waves. We couldn't arrive at the solution of 2 problems, so we thought perhaps some parameters were omitted. So I suggested that I post a couple of them up here to see if anyone could arrive at a solution....


Questions

1. A steel wire having a diameter of 0.2mm is subject to a tension of 200N . Determine the velocity of propagation if transverse wave along the wire.


2 . A vibrating source at the end of a stretched strung has a displacement given by the equation y=0.1sin6t , where y is in meters and t is in seconds. The tension in the string is 4N and the mass per unit length is 0.01kgm/s .

a. What is the velocity in the strings
b. What is the frequency of the wave
c. what is the wave length.


cc: seunwen2 , privatetisa, simiolu1, STENON, dejt4u, Dominiquez

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KingErnesto:
I and my colleagues were solving some physics problems relating to waves. We couldn't arrive at the solution of 2 problems, so we thought perhaps some parameters were omitted. So I suggested that I post a couple of them up here to see if anyone could arrive at a solution....


Questions

1. A steel wire having a diameter of 0.2mm is subject to a tension of 200N . Determine the velocity of propagation if transverse wave along the wire.


2 . A vibrating source at the end of a stretched strung has a displacement given by the equation y=0.1sin6t , where y is in meters and t is in seconds. The tension in the string is 4N and the mass per unit length is 0.01kgm/s .

a. What is the velocity in the strings
b. What is the frequency of the wave
c. what is the wave length.


cc: seunwen2 , privatetisa, simiolu1, STENON, dejt4u, Dominiquez

Question 1 is 900 m/s.

I got this from knowing the density of steel to be 7.86*10^3 kg/m^3. Once you know this then calculate the circular area of the steel wire (which is pi*D^2/4) and then multiply by the density to give you the mass per unit length. It is pretty straight forward from there using the formula V= sqrt(T/M/L).


As for question 2, why is your unit of mass per unit length in kgm/s?. Once you let me know I will comment on question 2.

lordbeans:



Question 1 is 900 m/s.

I got this from knowing the density of steel to be 7.86*10^3 kg/m^3. Once you know this then calculate the circular area of the steel wire (which is pi*D^2/4) and then multiply by the density to give you the mass per unit length. It is pretty straight forward from there using the formula V= sqrt(T/M/L).


As for question 2, why is your unit of mass per unit length in kgm/s?. Once you let me know I will comment on question 2.

Thanks for the attempt, I appreciate. For the question 2, it's actually linear density, kg/m and not kg/s , my mistake.

KingErnesto:



Thanks for the attempt, I appreciate. For the question 2, it's actually linear density, kg/m and not kg/s , my mistake.

Velocity = sqrt(4/0.01)=20m/s;

since Sin is 2*pi periodic then the period of the sinusoidal function is 6t=2*pi=> t_p=2*pi/6;
but frequency= 1/t_p= 6/(2*pi)= 3/pi hz;

velocity=frequency*wavelength.

wavelength=velocity/frequency;

wavelength=20/(3/pi)= 20*pi/3 metres.

lordbeans:


Velocity = sqrt(4/0.01)=20m/s;

since Sin is 2*pi periodic then the period of the sinusoidal function is 6t=2*pi=> t_p=2*pi/6;
but frequency= 1/t_p= 6/(2*pi)= 3/pi hz;

velocity=frequency*wavelength.

wavelength=velocity/frequency;

wavelength=20/(3/pi)= 20*pi/3 metres.

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Thanks boss... thumbs up.