Period of simple pendulam is given as `T = 2pi sqrtfrac(1)(g)`.Verify this formula using dimensional method.

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 measured by a stop watch of least count 0.1 s. The percentage error is g is

If the time period of a simple pendulum is T = 2pi sqrt(l//g) , then the fractional error in acceleration due to gravity is

The period of oscillation of a simple pendulum is T = 2pisqrt((L)/(g)) . Meaured value of L is 20.0 cm know to 1mm accuracy and time for 100 oscillation of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accracy in the determinetion of g is :

The time period of a simple pendulum inside a stationary lift is T. If it starts descending with acceleration g/5. the new period of the simple pendulum will be

What is conical pendulum? Show that its time period is given by 2pi sqrtfrac(lcostheta)(g) , where l is the length of the string, theta is the angle that the string makes with the vertical and g is the acceleration due to gravity.

A student measures the value of g with the help of a simple pendulum using the formula g = (4pi^(2)L)/(T^(2)) . He measures length L with a meter scale having least count 1 mm and finds it 98.0 cm . The time period is measured with the help of a watch of least count 0.1s . The time of 20 oscillations is found to be 40.4 s. The error Deltag in the measurment of g is (" in" m//s^(2)) .

The acceleration due to gravity on the surface of moon is 1.7 ms ^-2 . What is the time period of a simple pendulum on the surface of moon if its time period of the surface of earth is 3.5 s ? (g on the surface of the earth is 9.8 ms^-2

Answer the following . The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angle of oscillation, a more involved analysis shows that T is greater than 2 (pi) sqrt ((l/g)) . Think of a qualitative argument to appreciate this result.

If a simple pendulum is taken to place where g decreases by 2%, then the time period

Time period of a pendulum on earth surface is T_1 . It is arranged on earth surface at a height R and thus its time period is T_2 . What is the ratio of T_1 and T_2 ?