Problem 5E

The period T of a simple pendulum is given by where L is the length of the pendulum and g is the acceleration due to gravity.

a. Assume g = 9.80 m/s2 exactly, and that L= 0.742 ± 0.005 m. Estimate T, and find the uncertainty in the estimate.

b. Assume L = 0.742 m exactly, and that T = 1.73 ± 0.01 s. Estimate g, and find the uncertainty in the estimate.

Solution 5E

Step1 of 3:

Let T be the period of of a simple pendulum and it is given by

Where,

L = length of the pendulum

g = acceleration due to gravity

Here our goal is:

a).We need to Estimate T, and find the uncertainty in the estimate by Assuming g = 9.80 exactly, and that L = 0.7420.005m

b).We need to Estimate g, and find the uncertainty in the estimate by Assuming L = 0.742 m exactly, and that T = 1.730.01 s.

Step2 of 3:

a).

From the given information we have g = 9.80, L = 0.742,

Here

mathematical constant and its value is 3.14

=

= 6.28

= 6.28(0.2751)

= 1.7280

Hence, 1.7280.

Consider,

Differentiate above equation with respect to “L” we get

Let u = L/g

du = 1/g

Apply chain rule to above equation to get a differential equation and it is given by

=

=

Substitute “u” value in above equation we get

=

=

=

=

=

= 1.1645

Hence, = 1.1645.

Now,

The estimate of uncertainty is given by

= (1.1645)(0.005)

= 0.005822

Hence, = 0.005822

Therefore, The estimate of uncertainty is T = 1.72800.0058 s.

Step3 of 3:

b).

From the given information we have T = 1.73, L = 0.742,

Here

=

T=

Square above equation on both side we get

=

g =

Now,

g =

= 4(9.8596)(0.742)(0.3341)

= 9.7775

Hence, g = 9.7775.

Consider,

g =

Differentiate above equation with respect to “T” we get

=

=

=

= -8

= -8(9.8596)(0.742)(0.1931)

= - 11.3035

Hence, = -11.3035.

Now,

The estimate of uncertainty is given by

=

= (11.3035)(0.01)

= 0.1130

Hence, = 0.1130

Therefore, The estimate of uncertainty is g = 9.77750.1130