The angle made by the string of a simple pendulum with the vertical depends on time as `theta=pi/90sin[(pis^-1)t]`. Find the length of the pendulum if `g=pi^2ms^-2`
Instantaneous angle (in radian) between string of a simple pendulum and verical is given by theta = (pi)/(180) sin2pit . Find the length of the pendulum if g = pi^(2) m//s^(2) .
Time period of a simple pendulum only depends on the _______________ of the string.
Find the length of seconds pendulum at a place where g = pi^(2) m//s^(2) .
Find the length of seconds pendulum at a place where g = pi^(2) m//s^(2) .
The given figure shows the variation of the kinetic energy of a simple pendulum with its angular displacement (theta) from the vertical. Mass of the pendulum bob is m = 0.2 kg. Find the time period of the pendulum. Take g = 10 ms^(-2) .
Assertion : If the length of the simple pendulum is equal to the radius of the earth, the period of the pendulum is T=2pisqrt(R//g) Reason : Length of this pendulum is equal to radius of earth.
Show that the angle made by the string with the vertical in a conical pendulum is given by costheta=(g)/(Lomega^(2)) , where L is the string and omega is the angular speed.
Calculate the tiem period of a simple pendulum of length one meter. The acceleration due ot gravity at the place is pi^2ms^-2 .
While measuring the acceleration due to gravity by a simple pendulum , a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period . His percentage error in the measurement of g by the relation g = 4 pi^(2) ( l // T^(2)) will be
When the length of a simple pendulum is increased by 36 cm, its time period of oscillation is found to increase by 25 %. The initial length of the simple pendulum 1s (Acceleration due to gravity= 9.8ms^(-2) )
