H is known that a load with a mass of 200 g will stretch a spring 100 cm. The spring is then
stretched an additional 5.00 cm and released. Find:
a the spring constant, period of vibration and frequency
b. the maximum acceleration
e the velocity through equilibrium positions
d. the equation of motion

Explanation:

The spring is stretched by a force = 200 x 980 dynes through a length 100 cm . By Hooks law  The force F = - k x

here k is spring constant  and x is displacement of weight .

Thus  200 x 980 = -  k x 100

or k = 1960 dynes/cm

The time period of spring can be found by relation

T = 2π

= 2π =  2 sec

The frequency of vibration is taken as the reciprocal of time period

Thus frequency ν = = = 0.5 revolution / sec

b. The maximum acceleration is at the end points of vibration , and is equal to acceleration due to gravity .

c. The velocity at mean position can be calculated from the kinetic energy relation of spring .

The kinetic energy of spring = k x²

and it is converted into kinetic energy of mass at mean position

Thus   k x² =   m v²

or v = x

= x 5 =  50 cm/sec

double the distance = divide strength by 4