word problem

Question

The period T(in seconds) of a simple pendulum as a function of its length L (in feet) is given by

 

T(L)=2pisqrtL/32

a) Find the lenght of the pendulum as a function of T given by L(T). That is find the inverse of T(L)

b) Use the results in (a) to find the lenght of a pendulum whose period is 3 seconds

Answer 1

Hi Logan,

 

a)

 

We know that

 

T = 2*π*sqrt(L/32)

 

To solve for L, we must first move 2*n over:

 

T/(2*π) = sqrt(L/32)

 

Next, we eliminate the square root by squaring both sidse:

 

(T/(2*π))² = L/32

 

or

 

T²/(4π²) = L/32

 

Lastly, multiply both sides by 32 to yield:

 

32T²/(4π²) = L

 

and simplify:

 

8T²/π² = L

 

Hence, L(T) = 8T²/π²

 

b)

 

If the period is 3, i.e. T = 3, we simply calculate

 

L(3) = 8*3²/π²

 

which is approximately 7.3 ft.

Answer 2

This question is really just about functions - that is, the formula given for T(L) allows you to plug in L to get a value for T. An inverse function for this would allow you to find a function L(T) that does the opposite.

 

The easiest way to find an inverse function in this case is to solve the original function for L:

 

 

T = 2pi * sqrt (L/32)

 

*note: next time, you may want to put L/32 in parenthesis; the way it's written makes it unclear whether it's L/32 or only L inside the radical. I knew this solely because I've seen the formula before.*

 

 

First, divide both sides by 2pi:

 

T / 2pi = sqrt (L/32)

 

 

To get rid of the radical, square both sides:

 

 

T² / (2pi)² = L/32

 

Next, multiply both sides by 32 (note that I am also squaring the denominator now):

 

32T² / 4pi² = L

 

Since 2pi was in parenthesis, both parts had to be squared.

 

32 divides evenly by 4, so this just becomes:

 

L = 8T² / pi²

 

 

To solve (b), you obviously need to have the correct answer for (a). Once you have this, just plug T = 3 into that function:

 

L = 8 (3)² / pi²

 

L = 72 / pi²

 

 

The period of a pendulum is the amount of time it takes to swing once in both directions and return to its starting point.

 

A good rule of thumb for solving algebraic equations for a different variable is to "use the order of operations backwards." For instance, if you wanted to solve the equation 3 (x+4)² - 5 = y for x, you would do addition (of 5) first, then division (of 3), then square root (exponent), then subtract 4 (parenthesis). A common technique I use in tutoring is to show that the closer people are, the less they'll want to part from each other. In the above equation, x is very closely connected with 4 and with the exponent, but only distantly acquainted with the 5 and is separated from the 3 by the parenthesis and the exponent itself.