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%?

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 (i) on increase of 2% in the value of l. (ii) decrease of 2% in the value of l.

The time period 'T' of a simple pendulum of length 'l' is given by T = 2pisqrt(l/g) .Find the percentage error in the value of 'T' if the percentage error in the value of 'l' is 2%.

Show that the expression of the time period T of a simple pendulum of length l given by T = 2pi sqrt((l)/(g)) is dimensionally currect

Show that the expression of the time period T of a simple pendulum of length l given by T = 2pi sqrt((l)/(g)) is dimensionally correct

The period of oscillation of a simple pendulum is given by T = 2pisqrt((L)/(g)) , where L is the length of the pendulum and g is the acceleration due to gravity. The length is measured using a meter scale which has 500 divisions. If the measured value L is 50 cm, the accuracy in the determination of g is 1.1% and the time taken for 100 oscillations is 100 seconds, what should be the possible error in measurement of the clock in one minute (in milliseconds) ?

A simple pendulaum consists of a small mass suspended from a string that is fixed at its upper end and has negligible mass. The length of time, t second, for complete swing of a simple pendulum can be modeled by the equation t = 2pisqrt(L/(32)) , where L is the length, in feet, of the string. If the time required for a complete swing of Pendulum 1 is triple the time required for a complete swing of Pendulum 2, the length of Pendulum 1's string is how many times the length of Pendulum 2's string?

The period of oscillation of a simple pendulum is given by T=2pisqrt((l)/(g)) where l is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measrued by a stop watch of least count 0.1 s. The percentage error is g is

The period of oscillation of a simple pendulum is given by T=2pisqrt((l)/(g)) where l is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measrued by a stop watch of least count 0.1 s. The percentage error is g is

The graph between time period ( T ) and length ( l ) of a simple pendulum is

The time period of oscillation of a simple pendulum is given by T=2pisqrt(l//g) The length of the pendulum is measured as 1=10+-0.1 cm and the time period as T=0.5+-0.02s . Determine percentage error in te value of g.