Question

Example 2.12 The period of oscillation of
a simple pendulum is T =
T = 27/L/g
Measured value of Lis 20.0 cm known to 1
mm accuracy and time for 100 oscillations
of the pendulum is found to be 90 s using
a wrist watch of 1 s resolution. What is the
accuracy in the determination of g ?
Answer g= 41L/T2
and AT=
t
At
AT At
Here, T=
Therefore,
n
n
T
The errors in both L and t are the least count
errors. Therefore,
(Ag/g) = (AL/L) + 2(AT/T)
0.1
1
+2 =0.027
20.0 90
Thus, the percentage error in g is
100 (Ag/g) = 100(AL/L) + 2 x 100 (AT/T)
= 3%
*****CAN ANYONE TELL ME WHY WE HAVEN'T TAKEN THE ABOVE TIME VALUE FOR 1 OSCILLATION ???WHY 100??​

Answer 1

Answer:

Example 2.12 The period of oscillation of

a simple pendulum is T =

T = 27/L/g

Measured value of Lis 20.0 cm known to 1

mm accuracy and time for 100 oscillations

of the pendulum is found to be 90 s using

a wrist watch of 1 s resolution. What is the

accuracy in the determination of g ?

Answer g= 41L/T2

and AT=

t

At

AT At

Here, T=

Therefore,

n

n

T

The errors in both L and t are the least count

errors. Therefore,

(Ag/g) = (AL/L) + 2(AT/T)

0.1

1

+2 =0.027

20.0 90

Thus, the percentage error in g is

100 (Ag/g) = 100(AL/L) + 2 x 100 (AT/T)

= 3%

*****CAN ANYONE TELL ME WHY WE HAVEN'T TAKEN THE ABOVE TIME VALUE FOR 1 OSCILLATION ???WHY 100??