David R.

asked • 08/03/15

T=2p vL/g Solve for g

Please help me
 
p is Pi 
 
v is the square root symbol

2 Answers By Expert Tutors

By:

Mark M. answered • 08/03/15

Tutor
5.0 (278)

Mathematics Teacher - NCLB Highly Qualified

See tutors like this
See tutors like this
T = 2π √(L/)g                 Divide both sides by 2π    
T / (2π) = √(L/g)            Square both sides
T2 / (2π)2 = L/g              Reciprocal of equals are equal
(2n)2 / T2 = g/L              Multiply both sides by L
(2n)2L / T2 = g
Upvote • 0 Downvote
More
Report

Son N.

I think you copied the problem incorrectly, otherwise it looks good.
Report

08/03/15

Mark M.

You asked if "Lg" were both under the radical. The student answered yes. So what did I copy incorrectly?
Report

08/03/15

Son N. answered • 08/03/15

Tutor
4.9 (86)

Math and Physics Tutor

See tutors like this
See tutors like this
Does the square root include L and g or only L?
Upvote • 0 Downvote
More
Report

David R.

Yes, vL/g
Report

08/03/15

Son N.

The problem gives you the equation T = π√(L/g) and asks you to solve for g.
 
When a problem asks you to solve for a variable, in this case g, all it is asking you to do is get the variable on one side of the equals sign be itself. 
 
The way Mark did it is just fine but I think he missed a division sign so I'll do it again here. 
 
1st) Divide both sides by √(L/g): T/√(L/g) = π√(L/g)/√(L/g)
 
We now have: T√(g/L) = π
Remember, dividing by √(L/g) is the same as multiplying by √(g/L).
 
2nd) Square both sides:  (T√(g/L))2 = π2
 
This gets rid of the square root, so we have: T2g/L = π2
 
3rd) Multiply both sides by L, then divide by T2: (T2g/L)L = π2L
 
This cancels out the L  on the left side: T2g = n2L
 
Now divide by T2: T2g/T2 = n2L/T2
 
The T2 on the left side cancels out and we're left with: g = n2L/T2
 
This is exactly the answer we're looking for!
 
I hope this helps you with any future problems like this.
 
 
 
Report

08/03/15

Son N.

And of course I forgot the 2...
Report

08/03/15

Mark M.

"Remember, dividing by √(L/g) is the same as multiplying by √(g/L)."
No! That is applicable only to fractions. Here the fraction is under a radical. The radical must be removed before the fraction can be used as a divisor.
Report

08/03/15

Son N.

I think this is actually ok for this radical since the radical here can be seen as an exponent and exponents of the same power can be combined or deconstructed.  
 
Meaning, √(L/g) = √(L)/√(g).
 
This makes it completely valid for me to divide by √(L/g) before squaring it. 
 
Report

08/03/15

Patrick R.

Both Mark and Son are correct. You both derived the same answer. Son forgot the 2 in 2pi and mark changed 2pi to 2n for simplicity reasons. If you look at both of your answers, they are identical if adding the 2 back into the equation Son derived that was mistakenly dropped. You ABSOLUTELY can divide by the sqrtL/g before squaring it to remove radical EVEN WITH L/G BEING UNDERNEATH A SQRT or THE RADICAL can be undone first. Both ways WORK! Just pointing out the obvious here, but there are more than one way to derive an equation and arrive at the same answer while staying within the bounds of the many rules of Mathematics. Only thing I see incorrect is Marks reasoning for squaring both sides before making the sqroot of L/g divisible. That is only a matter of preference IN THIS CASE and not a RULE. One can clearly see from both examples, that by rearranging the equation for the period of pendulum in SHM, gravity or g=(2pi)^2*L(Length of pendulum)/Tsqrd, which is simply the period of the pendulum^2 
Report

05/13/17

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.