Professor's note: All is very good, to save your time and serve the purpose of this exercise, let's avoid the numbers. Give me the steps to solve a given problem in words - explain as much as you want ! --- keep up the good work !! --- Solutions : Week 2
1.
m=680 9 = 0.68 kg
k=65N/m
x=12cm=0.12m
angular frequency= (k/m)^(1/2) = (65/0.68)^(1/2) = 9.78
angular frequency = 2 pi f ------> f= angular frequency/2pi = 9.78/2pi = 15.4 Hz
Period=1/f = 1/15.4 = 0.065 sec
A= 12cm =0.12m
Vmax=A (omega) = 0.12X9.78 =1.2 m/s
Vmax is at the origin at x=0
Amax= A (omega^2) = 0.12(9.78^2) = 11.5m/s^2
t@6cm = pi/(3(omega)) = pi/ (3(9.78))=0.107 sec
t@0cm= x(t)=Acos(omega(t))
0=0.12cos(9.78t)
divide both sides by 0.12
0/0.12=0 so
0=cos(9.78t)
cos^-1(0)=9.78t
t=(cos^-1(0))/9.78 = 0.16
t@0-t@6= 0.16-0.107=0.053 sec
2.
mu=7.2g/m
T=150N
x=90cm=.90m
mu=7.2g/m(1kg/1000g)=0.0072kg
v=(t/mu)^(1/2)= (150/0.0072)^(1/2)=144m/s
lambda=2/3x= (2/3)(90)= 60 cm = 0.6 m
v=lambda(f) ------> f=v/lambda = 144/0.6=240Hz
Solutions: Week 3:
1. We know:
m(ball)=5kg
m(string)=0.6g=0.0006 kg
g=9.8 m/s^2
T=2s
The final equation we will use in v= (tension/ mu)^(1/2)
But first we need to find the tension, length of the string and mu in order to use that equation.
The tension is the upward force of the ball on the string. This means it is equal to the downward weight of the string.
So: Tension=mg = (5kg)(9.8m/s^2)
Since we are given the period, we can find the length of the string using the equation T= 2(pi) ((l/g)^(1/2))
2=2(pi) ((l/9.8)^(1/2))
doing some algebra we can reach: l= 9.8(pi)^(1/2)
Now that we have length we can find mu:
mu=m/l
The mass is 5kg and the length we just found
Now we have tension and mu so we can plug into the original equation
v= (tension/ mu)^(1/2) and get our answer.
2.
We know:
Vo=80m/s
vs=54 m/s
v=340m/s
fs=440hz
This problem is where both the observer and the source are moving so the following equation can be used:
fo=fs ((v+vo)/(v-vs))
We have all the values needed so a simple plug in will give you an answer.
Solutions: Week 4:
1. We know:
Pipe is open on both ends
fundamental frequency
420 Hz
v=340m/s
a)Since the pipe is open at both ends the equation f= nV/2L can be used.
The f= 420(given), the n = 1 (b/c fundamental or first harmonic), the v= 340(given). Since there are only 4 variables in the equation and we have three, we can plug in and solve for L.
420=(1x340)/(2xL)
Do the algebra and get an answer for L.
b) Here we want to find the second harmonic so we have to to change our n to 2 and use the length found in part a to solve for f. The same equation f=nV/2L will be used. The f is what we are now solving for. The n=2 (2nd harmonic), v=340(given), l =0.41m(answer from part a).
f=(2x340)/(2x0.41)
c) The only thing that is changing here is the speed of sound. We use everything from part a and just have v=367 instead of 340.
f=(1x367)/(2x0.41)
2.We know:
Pipe A= open both ends, 1st harmonic, 60 Hz
Pipe B= one end open, 3rd harmonic, same frequency of pipe A in 2nd harmonic
We know the equation for both ends open is f=nV/2L and we have enough information to find the length of pipe A.
f=60 Hz, n=1, v=340, and l is what we're looking for.
60= (1X340)/(2L)
Do the algebra and solve for L
L=2.83m
Next we need to find the frequency of the second harmonic for A in order to proceed with finding the length of B
We are looking for f, n=2 (2nd harmonic), v= 340(speed of sound in air), L=2.83m (found in the previous step)
So, f= (2X340)/(2X2.83)
f=120.1 Hz
So since the frequency of B is the same as the frequency of A in the 2nd harmonic f=120.1 Hz for pipe B
B is a one end open pipe so the equation changes to f=nV/4L
F=120.1 Hz, n=3(3rd harmonic), v=340, l is what we are looking for
120.1=(3X340)/(4XL)
Do the algebra and solve for L
L=2.12 m
Solutions: Week 7:
1. I will call q1 the negative charge in which we are trying to find the force acted on it. Q2,q3,q4 are all positive. This means they all have attractive forces to the q1. I set the charges up so that q1 is in the bottom left corner, q2 is the upper left corner, q3 is the upper right corner, and q4 is the lower right corner. We know:
q=2 microC
k=8.9 e9
r= the 10 cm distance between charges which is 0.1 m
Based on this information we can use the equation f=k ((q1)(q2))/ r^2, to find the force acting on q1 due to q2 and q4.
F12= k(2microC)(2microC)/(0.1^2), this also equals f14 because all of the numbers are the same.
To find F13 we need to find the difference between q1 and q3 since they are diagonal from each other.
We use the equation a^2=b^2+c^2, since b and c are both 10 cm we can plug in: a=(square root)(0.1m^2)+(0.1m^2)
Now we can plug in for F13=k(2microC)(2microC)/(0.14^2)
Next we have to find the sum of the forces in the x and y directions.
The sum of forces in the x direction = f13 cos60+f14cos0
The sum of the forces in the y direction = f13sin60+f12cos0
Plug in and get two answers. In order to find the net force we have to square those two numbers and then square root them.
Fnet= (square root)(sum of fx^2+sum of fy^2)
2. q1 is placed on the bottom left, q2 is the top of the triangle, and q3 is the bottom right corner (just what i chose for my diagram). Since all charges are positive, they are repulsive to each other. We know:
q=3 microC
r=4cm=0.04m
k=8.9 e 9
Choosing q1 to be the test charge:
F12=k q1 q2/r^2
F13=k q1 q2/r^2
Plug in and both answers will be the same since all of the givens are the same.
Find the sum of the forces in the x direction = F13
Find the sum of the forces in the y direction = F12
F net= (square root) (sum of x^2)+(sum of y ^2)
--- keep up the good work !! ---
Solutions : Week 2
1.
m=680 9 = 0.68 kg
k=65N/m
x=12cm=0.12m
angular frequency= (k/m)^(1/2) = (65/0.68)^(1/2) = 9.78
angular frequency = 2 pi f ------> f= angular frequency/2pi = 9.78/2pi = 15.4 Hz
Period=1/f = 1/15.4 = 0.065 sec
A= 12cm =0.12m
Vmax=A (omega) = 0.12X9.78 =1.2 m/s
Vmax is at the origin at x=0
Amax= A (omega^2) = 0.12(9.78^2) = 11.5m/s^2
t@6cm = pi/(3(omega)) = pi/ (3(9.78))=0.107 sec
t@0cm= x(t)=Acos(omega(t))
0=0.12cos(9.78t)
divide both sides by 0.12
0/0.12=0 so
0=cos(9.78t)
cos^-1(0)=9.78t
t=(cos^-1(0))/9.78 = 0.16
t@0-t@6= 0.16-0.107=0.053 sec
2.
mu=7.2g/m
T=150N
x=90cm=.90m
mu=7.2g/m(1kg/1000g)=0.0072kg
v=(t/mu)^(1/2)= (150/0.0072)^(1/2)=144m/s
lambda=2/3x= (2/3)(90)= 60 cm = 0.6 m
v=lambda(f) ------> f=v/lambda = 144/0.6=240Hz
Solutions: Week 3:
1. We know:
m(ball)=5kg
m(string)=0.6g=0.0006 kg
g=9.8 m/s^2
T=2s
The final equation we will use in v= (tension/ mu)^(1/2)
But first we need to find the tension, length of the string and mu in order to use that equation.
The tension is the upward force of the ball on the string. This means it is equal to the downward weight of the string.
So: Tension=mg = (5kg)(9.8m/s^2)
Since we are given the period, we can find the length of the string using the equation T= 2(pi) ((l/g)^(1/2))
2=2(pi) ((l/9.8)^(1/2))
doing some algebra we can reach: l= 9.8(pi)^(1/2)
Now that we have length we can find mu:
mu=m/l
The mass is 5kg and the length we just found
Now we have tension and mu so we can plug into the original equation
v= (tension/ mu)^(1/2) and get our answer.
2.
We know:
Vo=80m/s
vs=54 m/s
v=340m/s
fs=440hz
This problem is where both the observer and the source are moving so the following equation can be used:
fo=fs ((v+vo)/(v-vs))
We have all the values needed so a simple plug in will give you an answer.
Solutions: Week 4:
1. We know:
Pipe is open on both ends
fundamental frequency
420 Hz
v=340m/s
a)Since the pipe is open at both ends the equation f= nV/2L can be used.
The f= 420(given), the n = 1 (b/c fundamental or first harmonic), the v= 340(given). Since there are only 4 variables in the equation and we have three, we can plug in and solve for L.
420=(1x340)/(2xL)
Do the algebra and get an answer for L.
b) Here we want to find the second harmonic so we have to to change our n to 2 and use the length found in part a to solve for f. The same equation f=nV/2L will be used. The f is what we are now solving for. The n=2 (2nd harmonic), v=340(given), l =0.41m(answer from part a).
f=(2x340)/(2x0.41)
c) The only thing that is changing here is the speed of sound. We use everything from part a and just have v=367 instead of 340.
f=(1x367)/(2x0.41)
2.We know:
Pipe A= open both ends, 1st harmonic, 60 Hz
Pipe B= one end open, 3rd harmonic, same frequency of pipe A in 2nd harmonic
We know the equation for both ends open is f=nV/2L and we have enough information to find the length of pipe A.
f=60 Hz, n=1, v=340, and l is what we're looking for.
60= (1X340)/(2L)
Do the algebra and solve for L
L=2.83m
Next we need to find the frequency of the second harmonic for A in order to proceed with finding the length of B
We are looking for f, n=2 (2nd harmonic), v= 340(speed of sound in air), L=2.83m (found in the previous step)
So, f= (2X340)/(2X2.83)
f=120.1 Hz
So since the frequency of B is the same as the frequency of A in the 2nd harmonic f=120.1 Hz for pipe B
B is a one end open pipe so the equation changes to f=nV/4L
F=120.1 Hz, n=3(3rd harmonic), v=340, l is what we are looking for
120.1=(3X340)/(4XL)
Do the algebra and solve for L
L=2.12 m
Solutions: Week 7:
1. I will call q1 the negative charge in which we are trying to find the force acted on it. Q2,q3,q4 are all positive. This means they all have attractive forces to the q1. I set the charges up so that q1 is in the bottom left corner, q2 is the upper left corner, q3 is the upper right corner, and q4 is the lower right corner. We know:
q=2 microC
k=8.9 e9
r= the 10 cm distance between charges which is 0.1 m
Based on this information we can use the equation f=k ((q1)(q2))/ r^2, to find the force acting on q1 due to q2 and q4.
F12= k(2microC)(2microC)/(0.1^2), this also equals f14 because all of the numbers are the same.
To find F13 we need to find the difference between q1 and q3 since they are diagonal from each other.
We use the equation a^2=b^2+c^2, since b and c are both 10 cm we can plug in: a=(square root)(0.1m^2)+(0.1m^2)
Now we can plug in for F13=k(2microC)(2microC)/(0.14^2)
Next we have to find the sum of the forces in the x and y directions.
The sum of forces in the x direction = f13 cos60+f14cos0
The sum of the forces in the y direction = f13sin60+f12cos0
Plug in and get two answers. In order to find the net force we have to square those two numbers and then square root them.
Fnet= (square root)(sum of fx^2+sum of fy^2)
2. q1 is placed on the bottom left, q2 is the top of the triangle, and q3 is the bottom right corner (just what i chose for my diagram). Since all charges are positive, they are repulsive to each other. We know:
q=3 microC
r=4cm=0.04m
k=8.9 e 9
Choosing q1 to be the test charge:
F12=k q1 q2/r^2
F13=k q1 q2/r^2
Plug in and both answers will be the same since all of the givens are the same.
Find the sum of the forces in the x direction = F13
Find the sum of the forces in the y direction = F12
F net= (square root) (sum of x^2)+(sum of y ^2)