So I'm dong a lab about simpkle harmonic motion and I have to answer questions at the end of the lab were I'm provided this question to answer
here's a step that's not the actual question but the question makes reference to it
Step 3
using the point and the scale, note the equilibrium position of your mass on the srping, pull the mass down a given distance, record it, and release the mass. Record the position that the mass rises to. Do this for each mass.
Here's a question that asked me about step 3 which is not the actual question I have but my answer helps to answer my question
How did the distance above and below the equilibrium positions compare for each of the masses in step 3?
here's the answer I wrote
When we performed step three in the procedure that when we pulled the mass spring system down 4.0 cm from the equilibrim position it went past this equilibrium position 4.0 cm. For every single different mass that was used to conduc this experiment the same thing also occuered.
HERE'S MY QUESTION
2. Using your experimetnally determined spring constant, your largest mass, and your amplitude for that mass from step 3 calculate,
a. the total energy of you srping-mass system;
b. the velocity of the mass when it passes through its equilibrim position (E total = E kinetic)
c. The potential energy and kinetic energy of your simple Harmonic Oscillator when its displacement is half the amplitude. (x = A/2 E potential = 1/2 kx^2 & E total = E potential + E kinetic)
d. the velocity of the mass when its displacement is half the amplitude.
here's my data
largest mass = .117 kg
Amplitude = 4.0 cm
100 cm = m
(4.0 cm)/(100 cm) = .04 m
force of gravity = -.49 N
X (this distance is the x i pulled down the srping to find the observed period that i found by taking the time and dividing it by oscillations. I also had to calculate the time.) = -.01 m
spring constant k = 49 N/m
T observed = .33 s
T calculated = .20 s
Ok I don't know how to answer these questions. If you could please show me how to them step by step showing the formulas used and eveyrhting including the correct amount of sig figs that would really be great!
Also the next question asks me this
Write the equation for the motion of your simple Harmonic Pscillator form problem 2 if at t=0 it is at the top of its motion?
I'm completley lost
thanks for your time!
I had to type this whole thing in again =_=
You have a long post. You must have spent quite a while typing it, probably twice.
The question deals with a lab experiment concerning simple harmonic motion. It would be hard for you to follow if you do not have much notion about it. So I suggest you try to read up on the subject, while I look at your data.
You can read your text-book on the subject (best choice), or you can read a Wiki article such as :
http://en.wikipedia.org/wiki/Simple_harmonic_motion
When we mention amplitude, we usually the distance between the extreme positions, which means that if the mass is pulled down 4 cm, and it bounces up above the equilibrium position 4 cm, then the amplitude would be 8 cm. Could you confirm that?
It is not clear how you determined the spring constant. Was it given to you or you did some observations?
In your experiment, how did you determine the period? It is the time taken to go through a complete cycle, i.e. when the mass returns to the same position AND the movement is in the same direction.
The Wiki article might help you clarify a lot of these questions.
YES this is whay happened i gues I should fix this
I pullled it down 4 cm it went up past equilibrium by 4 cm amplitutde 8 cm sorry for being unclear
I observed it with a stop watch to complete 10 oscilations then divided by 10 then I calculated it by this formula
T = 2 pi sqrt (m/k)
thanks again
Kelly,
Don't worry, I didn't forget about you.
I will get back to you, hoping you would be reading up on simple harmonic motion in the mean time. It is a lab, and you will have to write it up. The better you understand the subject matter, the faster you will be able to finish the lab.
MathMate.
Kelly,
"X (this distance is the x i pulled down the srping to find the observed period that i found by taking the time and dividing it by oscillations. I also had to calculate the time.) = -.01 m"
Can you explain a little more about what you did here, and what does the -0.01m represent?
In a lab, it would be easier to present the observations as is. Leave the calculations for later.
Can I assume that you have used only one spring? If that's not the case, you would need to identify the springs just like what you do for the masses.
It is clearer if you can make a table that have the following headings:
Step 3
Mass
Distance pulled below equilibrium point
Distance above equilibrium point
Step X : period
mass m =
Number of complete oscillations (n)=
Time required (t)= seconds.
Period (T) = t/n seconds
The governing simple harmonic motion (SHM) equation is
T=2 pi sqrt(m/k)
Spring constant k
=
Physics in the Spring
Introduction
The concept of this lab that is being investigated is simple harmonic motion. Our textbook provides the following definition, ¡°vibration about an equilibrium position in which a resorting force is proportional to the displacement from equilibrium¡±. The purpose of this lab is to find several different spring constants and then to solve for the period of the spring. In this experiment I will be using different masses attached to a spring and then recording its displacement and finding the spring constant. I will then set the spring into oscillation and record the period. I will then calculate the period and compare these two values.
Sketch
Materials
Spring Stand (attached onto it is a ruler that was also used)
Spring
Mass stand
Several different masses
Stop watch
Procedure
Using at least three different masses, generate a graph of displacement (x- axis) vs. Force (y-axis) for your spring. Use at least three data points.
Using the slope of this graph, determine the spring constant, k, of your spring.
Using the pointer and the scale, note the equilibrium position of your mass on the spring, pull the mass down a given distance, record it, and release the mass. Record the position that the mass rises to. Do this for each mass.
Using three different masses, set the system in oscillation and measure the time required for ten oscillations for each mass. Using this data, determine the period, T, of each mass on the spring.
Compare the actual period, with the predicted period determined from your spring constant determined in step 2.
Data
gravity = 9.81 m/s^2
Equilibrium Position with no Mass= .90 cm ¡À .5 cm
mass of stand = 17.4 g ¡À 5.0 g
Mass (g) Equilibrium Position with Mass (cm) Amplitude (cm) ¦¤t (for 10 oscillations) (s)
30 1.8 ¡À .5 4.0 ¡À .5 3.3 ¡À .5
50 2.5 ¡À .5 4.0 ¡À .5 4.7 ¡À .5
100 4.5 ¡À .5 4.0 ¡À .5 5.2 ¡À .5
Observations
I observed that when I performed step three in the procedure that I pulled the mass spring system down 4.0 cm from the equilibrium position with mass that it went past this equilibrium position 4.0 cm. I observed this for every single different mass that was used to conduct this experiment.
Analysis
¡ÆF=0
We know this to be so because when it¡¯s in equilibrium it¡¯s in equal barium and the net force is zero when all of the present forces acting on it are added together and it¡¯s not moving anywhere. The two forces that are present are elastic and gravity. We can derive the next formula with this reasoning.
¡ÆF= F_elastic+ F_g=0
We can now simplify this equation using the following formulas:
Hooke¡¯s Law
F_elastic=-kx
Force of Gravity
F_g=-mg
¡ÆF= -kx-mg=0
We can then rearrange this formula for the spring constant, k.
¡ÆF= -kx-mg=0
+mg +mg
(-kx=mg) x^(-1)
(-k=mgx^(-1) )-
k=(mgx^(-1) )-
Using this graph below the spring constant, k, can be found using the formula for slope.
Slope
m=¦¤y¦¤x^(-1)
Plugging our variables in, the force of gravity and displacement, for our graph we get this formula.
m=F_g x^(-1)
Using what we have already established above we can simplify the formula.
m=-mgx^(-1)
This formula gives us a modified formula from the one we already have established to find the spring constant. So we set the slope equal to the spring constant, k.
m=-mgx^(-1)=k
The work below is a sample of the first mass.
mass=30g mass of stand=17.4 g Equlibrium Position with no Mass=.90cm Equilibrium Position with Mass=1.8 cm
m=mass+mass of stand=30 g+17.4 g=47 g 1000 g=kg (47 g)/(1000 g)=.05 kg g=9.81m/s^2 N=kg m/s^2 x=Equilbrium Position with no Mass-Equilbrium Position with Mass=.90 cm-1.8 cm=-.90 cm 100 cm=m (-.90 cm)/(100 cm)=-.01 m
F_gravity= -mg=-.05 kg(9.81 m/s^2 )¡Ö -.49 N
m=-mgx^(-1)=k=-.49 N(-.01 m)^(-1)=49N/m
This graph shows the force of gravity with respect to their displacements. This graph shows that as the force of gravity decreases the displacement also decreases along with the slope, the spring constant.
In order to find period observed I need to take my recorded time for ten oscillations and divide it by ten to get the time of one period. The first period will be used as a sample.
∆T=3.3s # osciliations=10
T (observed)=(∆T )/(# oscillations)=(3.3 s)/10=.33 s
In order to calculate the period I¡¯ll use the formula provided in our textbook.
Period of a Mass-Spring System in Simple Harmonic Motion
T (calculated)=2¦Ð¡¼(mk^(-1) )^2¡½^(-1)
I¡¯ll calculate the period using the data collected for the first mass and the value I calculated for the spring constant in my analysis, k, as a sample for this calculation.
mass=30g mass of stand=17.4 g k=49N/m
m=mass+mass of stand=30 g+17.4 g=47 g 1000 g=kg (47 g)/(1000 g)=.05 kg
T (calculated)=2¦Ð(.05kg(49 N/m)^(-1) )^(2^(-1) )=.20 s
Below is proof that the units do actually cancel out.
N=kg/m^2
(kg/((N/m) ))^((1/2) )=((kg m)/(kg m/s^2 ))^((1/2))=((kg m s^2)/(kg m))^((1/2))=(s^2 )^((1/2) )=s
In order to find the percent error between the two times the following formula will be used:
Percent Error
percent error=|experimental-accepted|/accepted 100
In this lab the experimental value is the period, T calculated, sense it was found throughout performing this lab, and the accepted value is the period, T observed, sense it was the actual period that occurred. The percent error for the first mass will be used as a sample for this calculation.
experinemntal= .20 s accepted=.33 s
percent error=|experimental-accepted|/accepted 100=|.20 s- .33 s|/(.33 s) 100=40 %
Mass (kg) Fg (N) x (m) k (N/m) T (observed)(s) T (Calculated) (s) Percent error (T observed Vs. T calculated) (%)
0.05 -0.49 -0.01 49 0.33 0.20 40
0.07 -0.69 -0.02 35 0.47 0.28 40
0.117 -1.15 -0.04 29 0.52 0.40 23
Conclusion
The purpose of this lab was to find several spring constants, observe the periods of a spring, and calculate the periods of a spring set in harmonic motion. I observed a period to be .33 s which I later calculated to be .20 s which gave me a percent error of 40. I observed a period to be .47 s which I later calculated to be .28 s which gave me a percent error of 40 also. I observed a period of .52 s and later calculated it to be .40 s which gave me a percent error of 23. The concept, simple harmonic motion, was demonstrated in this lab because throughout it I had to set a spring into simple harmonic motion numerous times while collect data and later performed an analysis. In my analysis I had to find several spring constants, for example 49 N/m for the first mass, which I was only able to find because of the fact that the spring was in simple harmonic motion. My force of gravity vs. displacement graph supports this conclusion because the spring constant can be found, which can only be found if the spring is indeed in simple harmonic motion, by observation of the graph¡¯s slope. By also calculating the spring constant by using Hooke¡¯s Law, which only works when in simple harmonic motion, I can also find the spring constant by calculation and get the same answer. This would allow me to conclude that my graphs demonstrate the concept, simple harmonic motion. My results are consistent with my classmates. I observed in my analysis that when I increased my mass by a lot that my percent error went down. So I can conclude form this observation that this experiment can be improved by using larger masses, sense according to my analysis, produce smaller percent errors.
Questions
When I performed step three in the procedure I found that when I pulled the mass spring system down 4.0 cm from the equilibrium position with mass that it went past this equilibrium position 4.0 cm. For every single different mass that was used to conduct this experiment the same thing also occurred.
Largest mass=.117 kg Amplitude=8.0 cm 100 cm=m (8.0 cm)/(100 cm)=.08 m k=29N/M
E=(kA^2 ) 2^(-1)
∅=¦Ð2^(-1)
Of course the mass of the stand matters. The mass of everything attached to the end of the spring maters sense it¡¯s connected to the end of the spring which makes it a variable in its simple harmonic motion. The mass of the stand must be added to the mass of the mass. When you find the spring constant, k, you must also factor in the force of gravity on the mass. When you change the mass this value changes which changes you spring constant which also changes the period if you use the formula I used to calculate it. This allows me to conclude that the mass does actually matter sense changing it would change everything I calculated in my analysis.
Stiffer springs that only compress a couple of centimeters under thousands of Newton¡¯s of force have higher spring constants because they are ¡°stiffer¡± and have high spring constants. If the springs in my care were replaced with very stiff springs my car wouldn¡¯t vibrate as freely as softer springs with lower constants. Sense I would have high spring constants the amplitude and period of motion would also decrease and my car would bounce less freely because of this increase in the spring constant.
oh come on sigh
No problem, I can help you with these questions step by step.
(a) To calculate the total energy of the spring-mass system, you need to consider both the kinetic energy and potential energy. The formula for total energy (E_total) is:
E_total = E_kinetic + E_potential
1. Calculate the potential energy (E_potential):
E_potential = (1/2) * k * x^2
Substitute the values:
E_potential = (1/2) * 49 N/m * (0.04 m)^2
E_potential = 0.049 J (rounded to 3 significant figures)
2. Calculate the kinetic energy (E_kinetic):
E_kinetic = E_total - E_potential
Since E_total is equal to E_kinetic, we can rewrite the equation as:
E_kinetic = E_total - E_potential
E_kinetic = 0.049 J
Therefore, the total energy of the spring-mass system is 0.098 J (0.049 J for both potential and kinetic energy).
(b) The velocity of the mass when it passes through its equilibrium position is equal to the maximum velocity of the simple harmonic oscillator. To calculate it, you can use the formula:
v = sqrt((2 * E_kinetic) / m)
Substitute the values:
v = sqrt((2 * 0.049 J) / 0.117 kg)
v = 0.641 m/s (rounded to 3 significant figures)
So, the velocity of the mass when it passes through its equilibrium position is 0.641 m/s.
(c) To find the potential energy (E_potential) and kinetic energy (E_kinetic) when the displacement is half the amplitude (x = A/2 = 0.02 m), use the formulas:
E_potential = (1/2) * k * x^2
E_kinetic = E_total - E_potential
Substitute the values:
E_potential = (1/2) * 49 N/m * (0.02 m)^2
E_potential = 0.0196 J (rounded to 3 significant figures)
E_kinetic = 0.098 J - 0.0196 J
E_kinetic = 0.0784 J (rounded to 3 significant figures)
So, when the displacement is half the amplitude, the potential energy is 0.0196 J and the kinetic energy is 0.0784 J.
(d) The velocity of the mass when its displacement is half the amplitude can be calculated using the formula:
v = sqrt((2 * E_kinetic) / m)
Substitute the values:
v = sqrt((2 * 0.0784 J) / 0.117 kg)
v = 0.858 m/s (rounded to 3 significant figures)
Therefore, the velocity of the mass when its displacement is half the amplitude is 0.858 m/s.
Now, moving on to the next question:
To write the equation for the motion of your simple harmonic oscillator from problem 2, if at t=0 it is at the top of its motion, you need to consider the equation of simple harmonic motion:
x(t) = A * cos(ωt + φ)
Since the motion starts at the top, the initial phase angle φ is 0° (or π/2 radians). Therefore, the equation becomes:
x(t) = A * cos(ωt)
where A is the amplitude and ω is the angular frequency (ω = 2πf, where f is the frequency).
I hope this helps you answer the questions! Let me know if you have any further questions.