The length of a pendulum is measured as 20.0 cm. The time interval for 100 oscillations is measured as 90 s with a stop watch of 1 s resolution. Find the approximate percentage change in g.

The period of oscillation of a simple pendulum is given by T=2pi sqrt((l)/(g)) . The length l of the pendulum is about 0.5s. The time of 100 oscillations is measured with a watch of 1 s resolution. Calcualte percentage error in measurment of g.

The time period of a pendulum is given by T = 2 pi sqrt((L)/(g)) . The length of pendulum is 20 cm and is measured up to 1 mm accuracy. The time period is about 0.6 s . The time of 100 oscillations is measured with a watch of 1//10 s resolution. What is the accuracy in the determination of g ?

The time period of a pendulum is given by T = 2 pi sqrt((L)/(g)) . The length of pendulum is 20 cm and is measured up to 1 mm accuracy. The time period is about 0.6 s . The time of 100 oscillations is measured with a watch of 1//10 s resolution. What is the accuracy in the determination of g ?

The measurement value of length of a simple pendulum is 20 cm known with 2 mm accuracy. The time for 50 oscillations was measured to be 40 s within 1s resolution. Calculate the percentage accuracy in the determination of acceleration due to gravity g from the above measurement.

The period of oscillation of a simple pendulum is T=2pisqrt(L/(g)) . L is about 10 cm and is known to 1 mm accurancy. The period of oscillation is about 0.5 second. The time of 100 oscillation is measured with a wrist watch of 1 s resolution. What is the accurancy in the determination of g ?

The period of oscillation of a simple pendulum is T = 2 pi sqrt((L)/(g)) .L is about 10 cm and is known to 1mm accuracy . The period of oscillation is about 0.5 s . The time of 100 oscillation is measured with a wrist watch of 1 s resolution . What is the accuracy in the determination of g ?