(i) What is the change in the time period of simple pendulum if its length changes by 1%? (ii) When the length of a simple pendulum is increased by 21% what is the change in its time period?

(i) What is the change in the time period of simple pendulum if its le

1.	(i) What is the change in the time period of simple pendulum if its length changes by 1%? (ii) When the length of a simple pendulum is increased by 21% what is the change in its time period?
Answer» Solution :Expression of the time period of a simple PENDULUM is `T=2pisqrt(l/g)`………1 Differentiating w.r.t `l""(dT)/(dl)=((2pi)/(sqrt(g)))1/(2l^(1//2))` `dT=((2pi)/(sqrt(g)))1/2(dl)/(l^(1/2))`……….2 Dividing 2 with 1 `(dT)/T=(((2pi)/(sqrt(g)))(d/l)/(2l^(1//2)))/(((2pi)/(sqrt(g)))l^(1//2))=(dT)/T=1/2((dl)/l)` `((dT)/T)%=1/2((dl)/l)%` In this PROBLEM, % CHANGE in l=1% `:.` % change in `T=1/2(1)=0.5%` (ii) Expression for the time period of a simple pendulum `T=2pisqrt(l/g)` When length CHANGES from `l_(1)` to `l_(2)` time period changes from `T_(1)` to `T_(2)` KEEPING g constant `(T_(2))/(T_(1))=sqrt((l_(2))/(l_(1)))"" :. (T_(2))/(T_(1))=sqrt((121l)/(100l)0=11/10` `=((T_(2))/(T_(1))-1)=(11/10-1)=1/10implies(DeltaT)/T% =1/10xx100=10%` `:.` Increase in time period =10%