A simple pendulum executing S.H.M. is falling freely along with the support. Then

The length of simple pendulum executing SHM is increased by 21% . By what % time period of pendulum increase ?

A simple pendulum executes S.H.M. in water with a time period T_(w) . When the bob executes S.H.M. in air, time period is T_(a) , IF the density of bob of simple pendulum is 5000/3 kg/ m^(3) and frictional force of water is neglected, then find the relation between T_(w) and T_(a) .

The length of simple pendulum executing SHM is increased by 69% The percentage increase in the time period of the pendulum is

The length of simple pendulum executing SHM is increased by 69% The percentage increase in the time period of the pendulum is

Answer the following questions: (a) Time period of particle is S.H.M. depends on the force constant k and mass m of the particle : T=2pisqrt(m//k). A simple pendulum executes S.H.M. approximately. Why then is the time-period of a pendulum independent of the mass of the pendulum? (b) The motino of simple pendulum is approximatley simple harmonic for small angles of oscillation. For large angle of oscillation, a more involved analysis (beyond the scope of this book) shows that T is greater that 2pisqrt(l//g) . Think of a quanlitative argument to appreciate this result. (c) A man with a wrist watch on his hand falls from the top of tower. Does the watch give correct time during the free fall? (d) What is teh frequency of oscillation of a simple pendulum mounted iin a cabin that is freely falling under gravity?

Time period of a particle in SHM is T=2pisqrt(m/k) . A simple pendulum executes SHM approximately. Why then is the time period of pendulum is independent of mass?

A simple pendulum is executing simple harmonic motion with a time period T. If the length of the pendulum is increased by 21%, the percentage increase in the time period of the pendulum of is

Show that a simple pendulum executes S.H.M. when the angular displacement is small.

Answer the following . Time period of a particle in S.H.M. depends on the force constant k and mass m of the particle : T = 2 (pi) (( sqrt m /k)). A simple pendulum executes SHM approximately . Why then is the time period of a pendulum independent of the mass of the pendulum ?