If light rigid rod with a length of 0.53 m extends straight out from one end of meter stick then
(a) the time period is 1.4604 sec and
(b) the period differ from the period of a simple pendulum 1 m long is 27.19 % .
In the question ,
it is given that ,
the total length of the rod(L) is = 0.53 m
Part(a)
we know that , Time period (T) is = 2π√(L/g)
T² = 4π²*(L/g)
T² = (4 * 3.14 * 3.14 * 0.53)/9.8
On simplifying further ,
we get ,
T² = 20.902352/9.8 = 2.13289306
So , T = 1.4604 sec
Part(b)
given that the length of simple pendulum is = 1 m .
So , the new time period (T')² = 4π²*(L/g)
(T')² = (4*(3.14)²*1)/9.8
(T')² = 39.4384/9.8 = 4.02432
T' = 2.006 sec
percent difference of the period of stick with simple pendulum is
= (T' - T)/T' × 100
= (2.006 - 1.4604)/2.006 × 100
= 0.5456/2.006 × 100
= 0.2719 × 100
= 27.19 %
the period differ by 27.19% .
Therefore , (a) the time period is 1.4604 sec
(b) the time period differ by 27.19% .
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