Professor's note : Everything is perfect, This is exactly what I wanted. Very good Job !! Keep it up. (you can use Widget -->insert symbol to get power raised to 2 and other symbols)
1a) Solution:
mass of object = 680g =.68kg
Spring Constant = 65 N/m
Pulled 12 cm from equilibrium = A = .12m
What we know:
we know the angular frequency can be calculated with the mass of the object and the spring constant.
We can also fine the period of the motion with the mass of the object and the spring constant.
The frequency of the object can be calculated with the period just calculated.
Necessary Equations:
Angular frequency.. w = (k/m)^1 /2
Period... T = 2(3.14) x (m/k)^1/2
Frequency... f = 1/T
1b)Solution:
Pulled 12 from x=0
What we know:
we know that the spring was pulled 12 cm from equilibrium and since we have to convert it to meters it would then be .12 m. The Amplitude is also the distance pulled from equilibrium therefore would equal .12m.
1c) solution:
Amplitude of object = .12m
angular frequency = w = (k/m)^1/2
What we know:
we know that in order to get the v max we need the amplitude of the object along with the angular frequency. We know that v max is located at x=0 .
Necessary equations:
Vmax = Aw
1d) Solution:
We know the amplitude and we know the angular frequency therefore we are able to solve for max acceleration.
What we know:
we know that the equation involves amplitude and angular frequency
Necessary equations:
A max = Aw^2
1e) Solution:
we need to find out how long it will take the object in simple harmonic motion to get from equilibrium (x=0) to x=6.
What we know:
We know in order to solve this equation with the proper equation, we need the distance changed with respect to time, the amplitude of the object, and also the angular frequency.
Necessary equations:
x(t) = Acos (wt) ... since we have one unknown in the equation we can solve for t (time) that we are trying to figure out. The equation when solved for t would then come out to be t = cos^-1 (x/A) / w.
2a) Solutions:
density = 7.2 g/m... change to kg/m .... .0072 kg/m
Force of tension = 150N
Length of cord = 90cm... change to meters... .9m
What we know:
We know in order to find the velocity of the standing wave we must know the force of tension and the density of the wire. Since we have both we are able to solve it.
Necessary Equations:
v = (Ft/u)^1/2
2b) Solution: length of wire = .9m
what we know:
3lambda = .9m.... because of three standing waves
necessary equations:
lamda =.9/3 m
2c) Solution: know the lambda and we know and the velocity of the wave.
what we know:
we know that we can use the equations v = lambda x frequency... we can solve for the frequency by dividing lambda by both sides.
necessary equations:
frequency = velocity/ lambda
Week 4:
What we know: Fundemental frequency = 420 Hz, speed of sound in air =340 m/s
a) Need to find the length of the pipe (L)
Necessary equations f = 1v/2L...
In order to find the length of the pipe at the given frequency you must solve the equation for L
L= (1 x 340)/ (2 x f)
L = .41 m
b) For b we have to figure out the frequency of the pipe at the second harmonic so we would change n to 2.
Necessary equations: f = (2 x 340)/ (2 x .41)
f = 840 Hz
c) For c the only thing we need to change in the equation is the speed of sounds, which was changed to 367 m/s.
Necessary equation: f = (2 x 367)/ (2 x .41)
f = 447 Hz
2) We know that pipe A has a frequency of 60 Hz at the first harmonic with both ends open. Pipe B has one end open so we know at its third harmonic, it has the same frequency of pipe A at the second harmonic.
We need to find the length of pipe A, so there for we use the equation f = 1v/2L... when plugged into the equation it looks like 60 = (1 x 340)/ 2L. After solving for L, we get 2.83 m.
Next we need to find the frequency of the second harmonic of pipe A in order to find the length of pipe B.
f = (2 x 340) / (2 x 2.83)
f = 120.1 Hz
Now we found the frequency of pipe A at the second harmonic, we can solve for the length of pipe B.
the equation changes slightly because the pipe has only one end open, f = (3v)/ (4L)..
When plugged in the the equation looks like, 120.1 = (3 x 340) / (4 x L)
Use algebra and solve for L.
L = 2.12
Professor's note : Everything is perfect, This is exactly what I wanted. Very good Job !! Keep it up.
(you can use Widget -->insert symbol to get power raised to 2 and other symbols)
1a) Solution:
mass of object = 680g =.68kg
Spring Constant = 65 N/m
Pulled 12 cm from equilibrium = A = .12m
What we know:
we know the angular frequency can be calculated with the mass of the object and the spring constant.
We can also fine the period of the motion with the mass of the object and the spring constant.
The frequency of the object can be calculated with the period just calculated.
Necessary Equations:
Angular frequency.. w = (k/m)^1 /2
Period... T = 2(3.14) x (m/k)^1/2
Frequency... f = 1/T
1b)Solution:
Pulled 12 from x=0
What we know:
we know that the spring was pulled 12 cm from equilibrium and since we have to convert it to meters it would then be .12 m. The Amplitude is also the distance pulled from equilibrium therefore would equal .12m.
1c) solution:
Amplitude of object = .12m
angular frequency = w = (k/m)^1/2
What we know:
we know that in order to get the v max we need the amplitude of the object along with the angular frequency. We know that v max is located at x=0 .
Necessary equations:
Vmax = Aw
1d) Solution:
We know the amplitude and we know the angular frequency therefore we are able to solve for max acceleration.
What we know:
we know that the equation involves amplitude and angular frequency
Necessary equations:
A max = Aw^2
1e) Solution:
we need to find out how long it will take the object in simple harmonic motion to get from equilibrium (x=0) to x=6.
What we know:
We know in order to solve this equation with the proper equation, we need the distance changed with respect to time, the amplitude of the object, and also the angular frequency.
Necessary equations:
x(t) = Acos (wt) ... since we have one unknown in the equation we can solve for t (time) that we are trying to figure out. The equation when solved for t would then come out to be t = cos^-1 (x/A) / w.
2a) Solutions:
density = 7.2 g/m... change to kg/m .... .0072 kg/m
Force of tension = 150N
Length of cord = 90cm... change to meters... .9m
What we know:
We know in order to find the velocity of the standing wave we must know the force of tension and the density of the wire. Since we have both we are able to solve it.
Necessary Equations:
v = (Ft/u)^1/2
2b) Solution: length of wire = .9m
what we know:
3lambda = .9m.... because of three standing waves
necessary equations:
lamda =.9/3 m
2c) Solution: know the lambda and we know and the velocity of the wave.
what we know:
we know that we can use the equations v = lambda x frequency... we can solve for the frequency by dividing lambda by both sides.
necessary equations:
frequency = velocity/ lambda
Week 4:
What we know: Fundemental frequency = 420 Hz, speed of sound in air =340 m/s
a) Need to find the length of the pipe (L)
Necessary equations f = 1v/2L...
In order to find the length of the pipe at the given frequency you must solve the equation for L
L= (1 x 340)/ (2 x f)
L = .41 m
b) For b we have to figure out the frequency of the pipe at the second harmonic so we would change n to 2.
Necessary equations: f = (2 x 340)/ (2 x .41)
f = 840 Hz
c) For c the only thing we need to change in the equation is the speed of sounds, which was changed to 367 m/s.
Necessary equation: f = (2 x 367)/ (2 x .41)
f = 447 Hz
2) We know that pipe A has a frequency of 60 Hz at the first harmonic with both ends open. Pipe B has one end open so we know at its third harmonic, it has the same frequency of pipe A at the second harmonic.
We need to find the length of pipe A, so there for we use the equation f = 1v/2L... when plugged into the equation it looks like 60 = (1 x 340)/ 2L. After solving for L, we get 2.83 m.
Next we need to find the frequency of the second harmonic of pipe A in order to find the length of pipe B.
f = (2 x 340) / (2 x 2.83)
f = 120.1 Hz
Now we found the frequency of pipe A at the second harmonic, we can solve for the length of pipe B.
the equation changes slightly because the pipe has only one end open, f = (3v)/ (4L)..
When plugged in the the equation looks like, 120.1 = (3 x 340) / (4 x L)
Use algebra and solve for L.
L = 2.12