## Question

the time period of oscillatn of a simple pendulum is 2π (l/g)1/2.  l is about 10cm & is known to 1mm accuracy. the time period of oscillatn is about 0.5s . time of 100 oscillatn with a wrist watch of 1s revolutn . what is the accuracy in determinatn of g?

ans 5%

Solution by Joshi sir.

T = 2π (l/g)1/2

so g α  l/T2

so % error in g = % error in l + 2 % error in T

=  0.1*100/10  + 2* 1*100/50 = 1+4 = 5

here you shoult remember that if you count 100 osccilations then time will be counted for 100 osccilation.
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}