Mass vs. Average Period Squared

In summary: The graph makes sense, the higher the mass the less time it takes to do an oscillation. However, you may not have correctly calculated the slope of the graph.
  • #1
noname1
134
0
I have to plot the average period squared vs the mass, now my question is which goes to the x-axis and which goes to the y?
 
Physics news on Phys.org
  • #2
X: Independent variable
Y: Dependent variable.

So, did you change mass to find different periods or did you change the periods of find different masses?
 
  • #3
i added mass to see the variation in oscillation periods
 
  • #4
That means that the period DEPENDS on the mass, making mass the independent variable and period the dependent variable.

Remember:
X is what you change
Y is what you measure
 
  • #5
for example to calculate the k constant i believe the formula is

k = (4pi²)/slope

but the slope of the graph has to be t² and kilograms correct? i have the values in t² and the mass in grams
 
  • #6
If you want the spring constant to come out as Newtons per meter then yes the mass has to be in kilograms.
 
  • #7
Is it supposed to give a huge difference from the elongation method?

My assignment says i have to do 2 tables where one measures force against displacement and than this one above

for force vs displacement i get k as 34.6 while when i do it with the formula i got above i get k as-8.77, is this normal?
 
  • #8
You can't calculate the slope the same way for force vs. displacement. Remember, [tex]F = kx[/tex], so when you graph that equation you are going to end up with a line anyways. What do you notice about the slope?
 
  • #9
The instructions say that the slope of the graph force vs displacement which i got a slope of 34.6 N/m and when i did the slope calculation of the oscillation graph i got -4.5 for the slope where i imputed in the formula

k = (4pi²)/slope

which gave me -8.77 N/m
 
  • #10
If you plot displacement against weight, yes. If you plot it against mass, the k = m * 9.8
 
  • #11
The first graph i have
y axis - Force (N)
x axis - displacement (m)

Where i got the slope of 34.6 which the instructions say its the k constant

the second graph
y axis - t²
x axis - mass (kg)

where i got slope of -4.5

which i imputed here

k = (4pi²)/slope

which gave me -8.77 N/m

Now i was supposed to calculate the difference between the two

with these values i get a difference of -125% which i think is a huge difference
 
  • #12
You swapped X and Y on the first graph. You change force and measure displacement.

I'm not quite sure how you ended up with a negative slope, there should be a strong positive correlation between mass and period squared. Besides, the spring constant can't be negative. Can you list some data points?
 
  • #13
First graph

x f(N)
0.98
1.96
2.94
3.92

y displacement (m)
0.03
0.06
0.088
0.115Second graph

x mass(kg)
.3
.4
.5
.6

y t²(s)
2.72
2.13
1.66
1.41

i used this website to help input data to get info
http://www.shodor.com/unchem/math/lls/leastsq.html
 
Last edited:
  • #14
Ok the first graph seems fine but I can't understand why there's a negative slope on the second one. Can you make sure you're using periods and not frequencies? That's the only thing I can think of that would decrease like that.
 
  • #15
we had to count how many secs it took to do 25 oscillations

.3kg - took 15.19 sec
.4kg - 17.16 sec
.5kg - 19.38 sec
.6kg- 21.09 sec

than i took the average period T (sec)

divided 25 (oscillations) by the secs it took and squared them

25/15.19 = 1.65 = t² = 2.72
25/17.16 = 1.46 = t² = 2.13
25/19.38 = 1.29 = t² = 1.66
25/21.09 = 1.19 = t² = 1.41and that's how i got my values.

in a way the graph makes sense, the higher the mass the less time it takes to do an oscillation but i know o have something wrong on this graph
 
Last edited:
  • #16
Ah, that's it. An oscillation divided by time is a frequency. You want to use the reciprocal of that (period). I did that for the excel chart I made of your data and got both measurements of k almost identical.
 
  • #17
So i am supposed to divide these values by one and than square them, is that what i did wrong?

25/15.19 = 1.65 = t² = 2.72
25/17.16 = 1.46 = t² = 2.13
25/19.38 = 1.29 = t² = 1.66
25/21.09 = 1.19 = t² = 1.41
 
Last edited:
  • #18
Or just divide the values you already have by one since you already squared them. If you don't understand why, frequency is how many oscillations, revolutions, etc. the object makes per second, thus having the unit of hertz. Period is how many seconds it takes the object to make one revolution.

The frequency of the Earth is 1/24 hours = 0.04 per hour (meaning that it makes 1/24 of a rotation per hour) and its period is 24 hours (meaning that it takes 24 hours to make one revolution.
 
  • #19
Your a life saver, i have been around for so long that i was getting mad lol I just have 1 more question i had to calculate the energy stored in one of the spring, the units of the energy stored is in Joules correct?
 
  • #20
Yes, so long as the spring constant is N/m and the displacement is m.
 

What is mass vs. average period squared?

Mass vs. average period squared is a concept in physics that describes the relationship between the mass of an object and the average period of its oscillation or vibration. It is also known as the "mass-spring system" and is commonly used in experiments to study the motion of objects.

How is mass vs. average period squared calculated?

The formula for mass vs. average period squared is T^2 = 4π^2(m/k), where T is the average period, m is the mass of the object, and k is the spring constant. This formula is derived from Hooke's Law and can be used to calculate the mass of an object if the average period and spring constant are known.

What is the significance of mass vs. average period squared?

The relationship between mass and average period squared is important because it allows us to understand the behavior of objects that are in motion, such as pendulums or springs. It also helps us to determine the mass of objects that are difficult to measure directly, by using the principles of oscillation and vibration.

How is mass vs. average period squared used in scientific research?

Mass vs. average period squared is used in many scientific experiments and research studies, particularly in the fields of physics and engineering. It is often used to study the behavior of systems that involve oscillation or vibration, such as bridges, buildings, and mechanical devices. It can also be used to determine the mass of celestial objects, such as planets and stars.

What are some real-life applications of mass vs. average period squared?

Mass vs. average period squared has many practical applications in our daily lives. For example, it is used in the design and construction of buildings and bridges to ensure their stability and safety. It is also used in the development of musical instruments, as well as in medical devices that use vibrations, such as ultrasound machines. Additionally, the principles of mass vs. average period squared are used in sports, such as in the design of golf clubs and tennis rackets.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
1K
Replies
44
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
979
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
2
Replies
43
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
867
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
546
Back
Top