The period of oscillation of a box suspended by a string of length 1 is `2pisqrt((l)/(g))` This expression is valid only for

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 periodic time (T) of a simple pendulum of length (L) is given by T=2pisqrt((L)/(g)) . What is the dimensional formula of Tsqrt((g)/(L)) ?

The time period of an oscillating body is given by T=2pisqrt((m)/(adg)) . What is the force equation for this body?

The period of oscillation of a freely suspended bar magnet is 4 second. If it is cut into two equal parts length wise then the time period of each part will be

The time period of oscillation of a body is given by T=2pisqrt((mgA)/(K)) K: Represents the kinetic energy, m mass, g acceleration due to gravity and A is unknown If [A]=M^(x)L^(y)T^(z) , then what is the value of x+y+z?

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.

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 ?