A body of mass 'm' is suspended to an ideal spring of force constant 'k'. The expected change in the position of the body due to an additional force 'F' acting vertically downwards is

If a body of mass 0.98 kg is made to oscillate on a spring of force constant 4.84 N/m the angular frequency of the body is

A body of mass 1 kg oscillates on a spring of force constant 16 N/m. Calculate the angular frequency.

A body of mass 1 kg suspended by a light vertical spring of spring constant 100 N/m executes SHM. Its angular frequency and frequency are

A particle of mass m is hanging vertically by an ideal spring of force constant K . If the mass is made to oscillate vertically, its total energy is

A body of mass 0.5 kg is suspended by a weightless spring of force constant 500 Nm^(-1) . Another body of mass 0.1 kg moving vertically upwards with velocity of 5 ms^(-1) hits the suspended body and gets embeded into the suspended body. What will be the frequency of oscillation ?

A block is suspended by an ideal spring of force constant force F and if maximum displacement of block from its initial mean position of rest is x_(0) then

A number of forces acting on a body changes velocity of the body. The forces are

A mass M is suspended by two springs of force constants K_(1) and K_(2) respectively as shown in the diagram. The total elongation (stretch) of the two springs is

Two blocks of masses m_(1) and m_(2) are connected by an ideal sprit, of force constant k . The blocks are placed on smooth horizontal surface. A horizontal force F acts on the block m_(1) . Initially spring is relaxed, both the blocks are at rest. What is acceleration of centre of mass of system at the instant of maximum elongation of spring