The time t of a complete oscillation of a simple pendulum of length l is given by the equation `T=2pisqrt(1/g)` where g is constant. What is the percentage error in T when l is increased by 1%? Comment the above result.

The time period T of oscillation of a simple pendulum of length l is given by T=2pi. sqrt((l)/(g)) . Find the percentage error in T corresponding to 2% in the value of l.

Check the correctness of the relation T=2pisqrt((l)/(g)) , where T is the time period, l is the length of pendulum and g is acceleration due to gravity.

The length of a metallic wire is L_1 : when tension on the wie is T_1 , and length is L_2 , when the tension in the wire is T_2 , Find original length of the wire.

Time for 20 oscillations of a pendulum is measured as t_1 = 39.6 s ,t_2 = 39.9s , t_3 = 39.5s , What is the precision in the measurements ? What is the accuracy of the measurement ?

Answer the following questions : Time period of a particle in SHM depends on the force constant k and mass m of the particle: T= 2pisqrt(m/k) .A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

A physical quantity X is related to four measurable quantities a, b, c and d as follows. X = a^2 b^3 c^(5//2) d^-2 The percentage error in the measurement of a,b,c and d are 1%,2%, 3% and 4%, respectively. What is the perecentage error in quantity X? If the value of X calculated on the bais of the above relation is 2.7 63, to what value should you round off the result?