A spring of a negligible mass is stretched by a force of 20 g wt through a distance of 5 cms. A total mass of 50 g. is attached to the spring and set into oscillations. Find the periodic time of oscillation. Given `g = 9.8 frac(m)(s^2)`

The period of oscillation of a mass M, hanging from a spring of force constant k is T. When additional mass m is attached to the spring, the period of oscillation becomes 5T/4. m/M =

A weightless spring of length 60 cm and force constant 200 N/m is kept straight and unstretched on a smooth horizontal table and its ends are rigidly fixed. A mass of 0.25 kg is attached at the middle of the spring and is slightly displaced along the length. The time period of the oscillation of the mass is

A stone of mass 50 g is tied to the end of a string 2 m long and is set into rotation in a horizontal circle with a uniformspeed of 2 m//s . Then tension in the string is

A mass M is suspended from a massless spring. An additional mass m stretches the spring further by a distance x . The combined mass will oscillate with a period

The time period of vertical oscillations of a light spring with a load of mass M is T. If another load of mass m is attaches to the load M, the time period of oscillation is doubled. The value of m is

A mass is suspended from a vertica spring which is executing SHM of frequency 5 Hz. The spring is unstretched at the highest point of oscillation. Maximum speed of the mass is (take, acceleration due to gravity, g=10m//s^(2) )

The period of a simple pendulum is found to increase by 50% , when the length of the pendulum is increased by 0.6 m. Calculate the initial length and initial period of oscillation at a place where g = 9.8 frac(m)(s^2)

A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from the balance , when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body ?

A string breaks under a load of 4 kg. A mass weighing 200 g is attached to the end of this string which is one metre long and rotation when the string breaks, Is nearly(g=10 m//s^2 )

A load of mass 100 gm increases the length of wire by 10 cm. If the system is kept in oscillation, its time period is (g=10m//s^(2))