How do significant digits work in physics calculations?

AI Thread Summary
In physics calculations, significant digits are crucial for accuracy. When calculating the period and frequency of a guitar string vibrating 750 times in 3.00 seconds, the period is calculated as 3.00/750, resulting in 0.0040 seconds, which correctly reflects the significant figures. The frequency is found by dividing 750 by 3.00, yielding 250 Hz, also adhering to significant figure rules. The general guideline is to express answers with the least number of significant figures from the given data, which in this case is two. Understanding and applying significant digits ensures precision in scientific calculations.
Jayak
Messages
2
Reaction score
0
A guitar string vibrates 750 times in 3.00s. Calculate period and Frequency.

I am in grade 11 and it is my first week in physics 11 so I'm new at this! My question is on how the significant digits work. I don't know where I should leave my answer..Here is my attempt:

Period: 3.00/ 750 = 0.0040s.. is that right with the about of sig fig?

Frequency 750/3.00 = 250hz

Help me! Thanks
 
Last edited:
Physics news on Phys.org
A
 
Last edited:
Jayak said:
A guitar string vibrates 750 times in 3.00s. Calculate period and Frequency.

I am in grade 11 and it is my first week in physics 11 so I'm new at this! My question is on how the significant digits work. I don't know where I should leave my answer..Here is my attempt:

Period: 3.00/ 750 = 0.0040s.. is that right with the about of sig fig?

Frequency 750/3.00 = 250hz

Help me! Thanks

As a rule of thumb, take your answer out to the number of significant digits given in the problem. For example, if you are given that some quantity of the problem equals 0.751, you were given three significant figures. Thus, if you put something in your calculator and get an answer out to 10 decimal places, you should only take three significant figures. Also, it is generally a good idea to give an answer out to the least number of significant figures given in the question. In your case above, the quantity 3.00 has three significant figures, while 750 has only two. Therefore, your answer should most reasonably have two significant figures, which you did! Hope this helps!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top