1. A 1.5 kg object is suspended from a vertical spring whose spring constant is 120 N/m (a). Find the amount that the spring is stretched from its unstrained length (b). The object is pulled straight down by an additional distance of 0.25 m and released from rest. Find the speed with which the object passes through its original position on the way up.
Solution: (a) Givens : mass of the object, m spring constant of the spring, k
What we know: When an object is vertically hanging the force acting on the object, F1 Force acting on the spring, F2 Since the system is at equilibrium, the magnitude of the two forces are equal
Necessary equations: For the gravitational force, F1 = mg Force on the spring F2 = - kx The system reaches a stable state (equilibrium) --> F1 = F2 mg = kx (we are taking only the magnitude and the negative sign is ignored) x = mg/k
(b) Given : the spring is stretched an additional amount = 0. 25 m Thus the new x = (mg/k) + 0.25 Initial speed of the bob = 0 m/s (as it is released from rest)
What we know: at the equilibrium, the object has a maximum speed and x = 0 at the beginning, just before the release, it has a maximum x (we call it A) and v = 0
Necessary equations: Conservation of Energy ½ mv² = ½ k A² (NOTE: v in this equation is the maximum value of v)
2. A simple pendulum is made from a 0.78 m long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released how much time elapses before it attains its greatest speed?
Givens: Length of the string, L = 0.78 m
initial speed of the bob (when pulled to a side), v =0
What we know: Period is the time taken to complete a full cycle, T
When the pendulum bob passes the lowest point (when the string is verticle), it attains the maximum speed
--> thus the maximum speed is attained in quater way of the full cycle.
--> Let's take the time required to reach this position as t1
Necessary equations:
Picture2.png
Here is the "Haunted Swing"
Week two (September 17th , 2012) 1). A block whose mass m is 680 g is fastened to a spring whose spring constant k is 65 N/m. The block is pulled a distance x = 12 cm from it’s equilibrium position at x = 0 on a frictionless surface and released from rest at t = 0. 1.What are the angular frequency, the frequency and the period of the resulting motion. 2.What is the amplitude of the oscillation? 3.What is the maximum speed v and where is the block when it has this speed? 4.What is the magnitude a of the maximum acceleration of the block? 5. How long would the block take to go from x = 0 position to x = 6 cm
2) A guitar string has a linear density of 7.2 g/m and it is under a Tension of 150 N. The fixed supports are 90 cm apart. The string is oscillating forming a standing wave pattern with three full loops.
I. Find the speed of the wave
II. Find the wavelength
III. Find the frequency
Solutions are posted after each part of the problem in a different color. Given parameters: mass = 680 g, convert it into kg spring constant, k = 65 N/m pulled distance, x or A = 12 cm , convert to meters (in an ideal case, the maximum distance that the block would travel is 12 cm, other cases due to damping A decreases with time) 1.What are the angular frequency, the frequency and the period of the resulting motion. Use the equations given below,
Picture2.gif
Where, T is the period, f is the frequency and w is the angular frequency
2.What is the amplitude of the oscillation? As mentioned above, A is given = 12 cm 3.What is the maximum speed v and where is the block when it has this speed? v (max) = A * angular frequency 4.What is the magnitude a of the maximum acceleration of the block? a (max) = A * (ang. frequency)² 5. How long would the block take to go from x = 0 position to x = 6 cm Refer to your first HW assignment problem # 10 to # 12 and go through the solutions x (t) = A cos (wt) ----------------> (1) x = 0 , say the time = t1 Use eq. (1) and substitute x =0 0 = A cos (wt1) divide both sides by A 0 = cos (wt1) take the inverse of cosine, cos-¹ (0) = wt1 (pi/2) radians = wt1 We get the answer in radians, to satisfy the units of w, which is rad/s t1 = 1/w (pi/2) ----------------- (2) Follow the same steps for x = 6 cm , for time = t2 0.06 = 0.12 cos (wt2) 1/2 = cos (wt2) thus t2 = 1/w (pi/3) ---------------------> (3)
To find the time required to go from x = 0 to x = 6 cm t1 -t2 = (1/w) [ pi/2 - pi/3] answer will be in seconds
2) A guitar string has a linear density of 7.2 g/m and it is under a Tension of 150 N. The fixed supports are 90 cm apart. The string is oscillating forming a standing wave pattern with three full loops. It always helps to make a sketch of the given problem Given information :
Picture5.png
linear density, µ =7.2 g/m ---> change to kg/ m Tensional force, F = 150 N distance between three loops = 90 cm
Things we should know length between the two nodes in a loop = lambda/2 3l = 0. 9 m I. Find the speed of the wave
v = sqrt(F/µ)
II. Find the wavelength
3l= 0. 9 m
III. Find the frequency Use the other equation for v v = f * l
Week 03 (September 24th, 2012)
1). A simple pendulum consists of a ball of mass M = 5 kg hanging from a uniform string of mass m = 0.6 g and unknown length L. The pendulum is located in gravitational field g = 9.8 m/s2 and has period T = 2 s. Determine the speed of a transverse wave in the string when the pendulum is stationary and hangs vertically. For simplicity, neglect the string’s weight compared to the ball’s weight. Likewise, neglect the ball’s radius compared to the string’s length. Answer in units of m/s.
(I) Givens :
mass of ball (kg)
mass of string (.6g)
period (s)
gravitational field (m/s2)
What we know:
The velocity is the sqrt of force divided by mu (units are m/s)
The force is the mass of the ball times gravitational field µ is mass divided by the unknown length Necessary equations:
V = √(F/µ)
T=2π√(L/g) F=mg µ= mass of the string/length of the string
Note : Some values are given in CGS units (cm and g). Remember to convert them to SI units We discussed this problem in the class as well.
2). A police car is chasing a speeding Porsche. Assume that the Porsche's maximum speed is 80.0 m/s and the Police car's is 54.0 m/s. At the moment both cars reach their maximum speed, what frequency will the Porsche driver hear if the frequency of the police car 's siren is 440 Hz? Take the speed of sound in air to be 340 m/s
Hint :
Carefully decide on the right equation. Here both the observer (Porsche) and the source (police car) are moving.
Source is moving toward the observer
Observer is moving away from the source
Carefully analyze the equation to determine the signs (+ or -) in the numerator and the denominator
equation, fo = fs (1 +/- vo/v) (1 +/- vs/v)
Since the source is moving toward, frequency must go up, but the term related to vs (1 +/- vs/v) is in the denominator and to make the frequency go up --> denominator should go down --> andwe pick the negative sign.
Observer is moving away from the source and the frequency should go down. The term (1 +/- vo/v) is in the numerator and to make frequency go down we pick the negative sign.
Week 04 (October 1st, 2012)
Professor's Note : I will be posting the solutions shortly. Jade and Shelby are doing wonderful job -- please visit their pages as well.
1. A pipe that is open at both ends has a fundamental frequency of 420 Hz when the speed of sound in air is 340 m/s.
a) What is the length of this pipe?
Steps to solve the problem: f = nv/2L , where n = 1, 2, 3, 4 ..... f = 420 Hz v = 340 m/s n =1 plug the values to get L
b) What is the second harmonic?
f = nv/2L n =2 L = found in the first part v = 340 m/s find f for the second harmonic
c) What is the fundamental frequency of this pipe when the speed of sound in air is increased to 367 m/s due to a rise in the temperature of the air?
f = nv/2L
n = 1
L = found in the first part
v = 367 m/s
find f for the fundamental
2. An organ pipe A, with both ends open, has a frequency of 60 Hz at the first harmonic. The third harmonic of organ pipe B, with one end open, has the same frequency as the second harmonic of pipe A. How long are the pipes?
For the organ pipe A, f = nv/2L n =1 f = 60 Hz v = 340 m/s Using the given information and the equation, find the length of the pipe A
f (organ pipe A) = 2V/2LA
f (organ pipe B)
3V/ 4LB
2V/2LA = 3V/ 4LB
Since LA is known can find LB
Week 05 ----- TEST WEEK -----===
- No Quiz week
Week 06 (October 15th, 2012 )
01. Find the energy in MeV, released when alpha decay converts radium (Ra , A = 226 and Z = 88) with atomic mass 226.02540 u, into Radon (Rn, with A= 222 and Z= 86) with atomic mass 222.01757 u. The atomic mass of an alpha particle is 4.002603 u
First write the equation:
Then find the mass difference (DM) in u
u = atomic mass units
convert u to MeV
MeV is the unit of energy
convert MeV to Joules (not necessary to do -- but you can)
Follow the example did in the class
02. The number of radioactive nuclei present at the start of an experiment is 4.6 x 10^15. The number present 20 days later is 8.14 x 10^14. What is the half life of the nuclei - in days?
Steps to solve the problem:
Givens: N0 = original amount N = remaining amount after 20 days t = time = 20 days
What can we find using the information? we can find lambda (the decay constant)
Use N = N0 exp (-lambda)t
find lambda
t½ = half life t½ = ln 2 lambda
03.
Picture2_-121.png
Find the net resistance between the two open points
If the open ends are connected to a battery of 6 V, find the net current passing through the circuit.
(Hint : recall the characteristics of the resistors that are in parallel and in series, to identify such)
Steps to solve : (discussed in the class)
Start from the far end of the circuit. 4 and 6 are in series. find the equivalent of those two resistors --- (1)
8 and 9 (middle loop) are in parallel - find the equivalent resistor --- (2)
Equivalent R in (1) and that of (2) are in parallel
Below find the math:
This problem is very similar to what we discussed in the class, except there is one additional branch.
(For each step sketch a diagram. That will help you see the parallel or series arrangement of resistors)
1. Start from the right hand-side of the circuit (not from the open end)
2. The two resistors, 6 and 4 are in series, there is a wire connected between, (still the same current passes through both)
--> Rnet = 6 + 4 = 10 Ohm
3. Pay attention to the parallel pair now, 9 Ohm and 8 Ohm resistors.
-->1/Rnet = 1/9 + 1/8 Rnet = 72/17 Ohm = 4.24 Ohm
4. Now, consider the 10 Ohm and 4.24 Ohm resistors, you should realize that they are in parallel.
5. Find the resistor to replace those two resistors, since they are in parallel,
1/Rnet = 1/10 + 1/4.24 Rnet = 2.98 Ohm
6. Now that we have the resistor to replace all resistors in the right-handside of the circuit, we have 6 Ohm, 2.98 Ohm and 3 Ohm in series.
Rnet = 6 + 2.98 + 3 = 11.98 Ohm
7) Find the current, use V = I R 6 = I (11.98) Find I
Week 07 (October 22nd, 2012)
A nice video for you to go through the Coulomb's Law
--- Focus on exam ---
01. Four point charges, each of magnitude 2 μC, are placed at the corners of a square 10 cm on a side. The value of Coulomb’s constant is 8.98755 × 10^9 N m²/C². If three of the charges are positive and one is negative, find the magnitude of the force experienced by the negative charge.
Your knowledge in vectors, taking x and y components, finding net forces are extremely important in this section. You all must have done this in PHY 1 (it is a compulsory part in the course).
Please note, I can't re-teach this during the class time, and I am requesting you to refresh your knowledge and come to the class.
I am willing to help you outside the class, if necessary.
Steps to solve the problem
Draw a diagram
Mark the forces, considering the sign (+ or -) on the charges
Figure out (or calculate if necessary ) the distance between any two charges.
.
02. Three point charges, each of magnitude 3 μC, are placed at the corners of an equi-lateral triangle 4 cm on a side. The value of Coulomb’s constant is 8.98755 × 10^9 N m²/C². If three of the charges are positive, find the magnitude of the force experienced by one charge at a corner. Answer in units of N
-- Please refer to the class notes. We discussed one similar to this !
Week 8 (26th, November 2012)
A concave mirror has a focal length of 30.0 cm. The distance between an object and its image is 45.0 cm. Find the object and image distances, assuming that, a) the object lies beyond the center of curvature b) the object lies within the focal point
Given parameters
concave mirror
f = + 30 cm
a) If the object lies beyond the center of curvature, the image forms on the same side of the object, in front of the object.
Thus p - i = 45
i = p- 45
use the thin lens formula, 1/p + 1/i = 1/f
1/p + 1/(p-45) = 1/30
solve for p, you can either feed the information into the calculator or use the common denominator method - the basic math tricks
b) If the object lies within the focal length , the image forms behind the mirror (opposite to the object) - or forms a virtual image
Week one (03rd of September -2012)
Enjoy the video ....
Problems:
1. A 1.5 kg object is suspended from a vertical spring whose spring constant is 120 N/m
(a). Find the amount that the spring is stretched from its unstrained length
(b). The object is pulled straight down by an additional distance of 0.25 m and released from rest. Find the speed with which the object passes through its original position on the way up.
Solution:
(a) Givens : mass of the object, m
spring constant of the spring, k
What we know:
When an object is vertically hanging the force acting on the object, F1
Force acting on the spring, F2
Since the system is at equilibrium, the magnitude of the two forces are equal
Necessary equations:
For the gravitational force, F1 = mg
Force on the spring F2 = - kx
The system reaches a stable state (equilibrium) --> F1 = F2
mg = kx (we are taking only the magnitude and the negative sign is ignored)
x = mg/k
(b) Given : the spring is stretched an additional amount = 0. 25 m
Thus the new x = (mg/k) + 0.25
Initial speed of the bob = 0 m/s (as it is released from rest)
What we know: at the equilibrium, the object has a maximum speed and x = 0
at the beginning, just before the release, it has a maximum x (we call it A) and v = 0
Necessary equations:
Conservation of Energy
½ mv² = ½ k A² (NOTE: v in this equation is the maximum value of v)
2. A simple pendulum is made from a 0.78 m long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released how much time elapses before it attains its greatest speed?
Givens: Length of the string, L = 0.78 m
initial speed of the bob (when pulled to a side), v =0
What we know: Period is the time taken to complete a full cycle, T
When the pendulum bob passes the lowest point (when the string is verticle), it attains the maximum speed
--> thus the maximum speed is attained in quater way of the full cycle.
--> Let's take the time required to reach this position as t1
Necessary equations:
Here is the "Haunted Swing"
Week two (September 17th , 2012)
1). A block whose mass m is 680 g is fastened to a spring whose spring constant k is 65 N/m. The block is pulled a distance x = 12 cm from it’s equilibrium position at x = 0 on a frictionless surface and released from rest at t = 0.
1.What are the angular frequency, the frequency and the period of the resulting motion.
2.What is the amplitude of the oscillation?
3.What is the maximum speed v and where is the block when it has this speed?
4.What is the magnitude a of the maximum acceleration of the block?
5. How long would the block take to go from x = 0 position to x = 6 cm
2) A guitar string has a linear density of 7.2 g/m and it is under a Tension of 150 N. The fixed supports are 90 cm apart. The string is oscillating forming a standing wave pattern with three full loops.
I. Find the speed of the wave
II. Find the wavelength
III. Find the frequency
Solutions are posted after each part of the problem in a different color.
Given parameters:
mass = 680 g, convert it into kg
spring constant, k = 65 N/m
pulled distance, x or A = 12 cm , convert to meters (in an ideal case, the maximum distance that the block would travel is 12 cm, other cases due to damping A decreases with time)
1.What are the angular frequency, the frequency and the period of the resulting motion.
Use the equations given below,
2.What is the amplitude of the oscillation?
As mentioned above, A is given = 12 cm
3.What is the maximum speed v and where is the block when it has this speed?
v (max) = A * angular frequency
4.What is the magnitude a of the maximum acceleration of the block?
a (max) = A * (ang. frequency)²
5. How long would the block take to go from x = 0 position to x = 6 cm
Refer to your first HW assignment problem # 10 to # 12 and go through the solutions
x (t) = A cos (wt) ----------------> (1)
x = 0 , say the time = t1
Use eq. (1) and substitute x =0
0 = A cos (wt1)
divide both sides by A
0 = cos (wt1)
take the inverse of cosine, cos-¹ (0) = wt1
(pi/2) radians = wt1 We get the answer in radians, to satisfy the units of w, which is rad/s
t1 = 1/w (pi/2) ----------------- (2)
Follow the same steps for x = 6 cm , for time = t2
0.06 = 0.12 cos (wt2)
1/2 = cos (wt2)
thus t2 = 1/w (pi/3) ---------------------> (3)
To find the time required to go from x = 0 to x = 6 cm
t1 -t2 = (1/w) [ pi/2 - pi/3]
answer will be in seconds
2) A guitar string has a linear density of 7.2 g/m and it is under a Tension of 150 N. The fixed supports are 90 cm apart. The string is oscillating forming a standing wave pattern with three full loops.
It always helps to make a sketch of the given problem
Given information :
linear density, µ =7.2 g/m ---> change to kg/ m
Tensional force, F = 150 N
distance between three loops = 90 cm
Things we should know
length between the two nodes in a loop = lambda/2
3l = 0. 9 m
I. Find the speed of the wave
v = sqrt(F/µ)
II. Find the wavelength
3l= 0. 9 m
III. Find the frequency
Use the other equation for v
v = f * l
Week 03 (September 24th, 2012)
1). A simple pendulum consists of a ball of mass M = 5 kg hanging from a uniform string of mass m = 0.6 g and unknown length L. The pendulum is located in gravitational field g = 9.8 m/s2 and has period T = 2 s. Determine the speed of a transverse wave in the string when the pendulum is stationary and hangs vertically. For simplicity, neglect the string’s weight compared to the ball’s weight. Likewise, neglect the ball’s radius compared to the string’s length. Answer in units of m/s.
(I) Givens :
mass of ball (kg)
mass of string (.6g)
period (s)
gravitational field (m/s2)
What we know:
The velocity is the sqrt of force divided by mu (units are m/s)
The force is the mass of the ball times gravitational fieldµ is mass divided by the unknown length
Necessary equations:
V = √(F/µ)
T=2π√(L/g)F=mg
µ= mass of the string/length of the string
Note : Some values are given in CGS units (cm and g). Remember to convert them to SI units
We discussed this problem in the class as well.
2). A police car is chasing a speeding Porsche. Assume that the Porsche's maximum speed is 80.0 m/s and the Police car's is 54.0 m/s. At the moment both cars reach their maximum speed, what frequency will the Porsche driver hear if the frequency of the police car 's siren is 440 Hz? Take the speed of sound in air to be 340 m/s
Hint :
Carefully decide on the right equation.
Here both the observer (Porsche) and the source (police car) are moving.
- Carefully analyze the equation to determine the signs (+ or -) in the numerator and the denominator
equation,fo = fs (1 +/- vo/v)
(1 +/- vs/v)
Since the source is moving toward, frequency must go up, but the term related to vs (1 +/- vs/v) is in the denominator and to make the frequency go up --> denominator should go down --> andwe pick the negative sign.
Observer is moving away from the source and the frequency should go down. The term (1 +/- vo/v) is in the numerator and to make frequency go down we pick the negative sign.
Week 04 (October 1st, 2012)
Professor's Note : I will be posting the solutions shortly. Jade and Shelby are doing wonderful job -- please visit their pages as well.
1. A pipe that is open at both ends has a fundamental frequency of 420 Hz when the speed of sound in air is 340 m/s.
a) What is the length of this pipe?
Steps to solve the problem:
f = nv/2L , where n = 1, 2, 3, 4 .....
f = 420 Hz
v = 340 m/s
n =1
plug the values to get L
b) What is the second harmonic?
f = nv/2L
n =2
L = found in the first part
v = 340 m/s
find f for the second harmonic
c) What is the fundamental frequency of this pipe when the speed of sound in air is increased to 367 m/s due to a rise in the temperature of the air?
f = nv/2L
n = 1
L = found in the first part
v = 367 m/s
find f for the fundamental
2. An organ pipe A, with both ends open, has a frequency of 60 Hz at the first harmonic. The third harmonic of organ pipe B, with one end open, has the same frequency as the second harmonic of pipe A. How long are the pipes?
For the organ pipe A,
f = nv/2L
n =1
f = 60 Hz
v = 340 m/s
Using the given information and the equation, find the length of the pipe A
f (organ pipe A) = 2V/2LA
f (organ pipe B)
3V/ 4LB2V/2LA = 3V/ 4LB
Since LA is known can find LB
Week 05 ----- TEST WEEK -----===
- No Quiz week
Week 06 (October 15th, 2012 )
01. Find the energy in MeV, released when alpha decay converts radium (Ra , A = 226 and Z = 88) with atomic mass 226.02540 u, into Radon (Rn, with A= 222 and Z= 86) with atomic mass 222.01757 u. The atomic mass of an alpha particle is 4.002603 u
First write the equation:
- Then find the mass difference (DM) in u
u = atomic mass units02. The number of radioactive nuclei present at the start of an experiment is 4.6 x 10^15. The number present 20 days later is 8.14 x 10^14. What is the half life of the nuclei - in days?
Steps to solve the problem:
Givens:
N0 = original amount
N = remaining amount after 20 days
t = time = 20 days
What can we find using the information? we can find lambda (the decay constant)
Use N = N0 exp (-lambda)t
find lambda
t½ = half life
t½ = ln 2
lambda
03.
Find the net resistance between the two open points
If the open ends are connected to a battery of 6 V, find the net current passing through the circuit.
(Hint : recall the characteristics of the resistors that are in parallel and in series, to identify such)
Steps to solve : (discussed in the class)
- Start from the far end of the circuit. 4 and 6 are in series. find the equivalent of those two resistors --- (1)
- 8 and 9 (middle loop) are in parallel - find the equivalent resistor --- (2)
- Equivalent R in (1) and that of (2) are in parallel
Below find the math:This problem is very similar to what we discussed in the class, except there is one additional branch.
(For each step sketch a diagram. That will help you see the parallel or series arrangement of resistors)
1. Start from the right hand-side of the circuit (not from the open end)
2. The two resistors, 6 and 4 are in series, there is a wire connected between, (still the same current passes through both)
--> Rnet = 6 + 4 = 10 Ohm
3. Pay attention to the parallel pair now, 9 Ohm and 8 Ohm resistors.
-->1/Rnet = 1/9 + 1/8
Rnet = 72/17 Ohm = 4.24 Ohm
4. Now, consider the 10 Ohm and 4.24 Ohm resistors, you should realize that they are in parallel.
5. Find the resistor to replace those two resistors, since they are in parallel,
1/Rnet = 1/10 + 1/4.24
Rnet = 2.98 Ohm
6. Now that we have the resistor to replace all resistors in the right-handside of the circuit, we have 6 Ohm, 2.98 Ohm and 3 Ohm in series.
Rnet = 6 + 2.98 + 3 = 11.98 Ohm
7) Find the current, use V = I R
6 = I (11.98)
Find I
Week 07 (October 22nd, 2012)
A nice video for you to go through the Coulomb's Law
--- Focus on exam ---01. Four point charges, each of magnitude 2 μC, are placed at the corners of a square 10 cm on a side. The value of Coulomb’s constant is 8.98755 × 10^9 N m²/C². If three of the charges are positive and one is negative, find the magnitude of the force experienced by the negative charge.
Steps to solve the problem
.
02. Three point charges, each of magnitude 3 μC, are placed at the corners of an equi-lateral triangle 4 cm on a side. The value of Coulomb’s constant is 8.98755 × 10^9 N m²/C². If three of the charges are positive, find the magnitude of the force experienced by one charge at a corner. Answer in units of N
-- Please refer to the class notes. We discussed one similar to this !
Week 8 (26th, November 2012)
A concave mirror has a focal length of 30.0 cm. The distance between an object and its image is 45.0 cm. Find the object and image distances, assuming that,
a) the object lies beyond the center of curvature
b) the object lies within the focal point
Given parameters
concave mirror
f = + 30 cm
a) If the object lies beyond the center of curvature, the image forms on the same side of the object, in front of the object.
Thus p - i = 45
i = p- 45
use the thin lens formula, 1/p + 1/i = 1/f
1/p + 1/(p-45) = 1/30
solve for p, you can either feed the information into the calculator or use the common denominator method - the basic math tricks
b) If the object lies within the focal length , the image forms behind the mirror (opposite to the object) - or forms a virtual image
Thus p + i = 45
i = 45 -p
use the thin lens formula, 1/p + 1/i = 1/f
1/p + 1/-(45-p) = 1/30
solve for p