Question

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​

Answer 1

The velocity at mean position is 50 cm/sec

Step-by-step 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
`\sqrt{(m)/(k) }`

= 2π
`\sqrt{(200)/(1960) }` = 2 sec

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

Thus frequency ν =
`(1)/(T)` =
`(1)/(2)` = 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 =
`(1)/(2)` k x²

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

Thus
`(1)/(2)` k x² =
`(1)/(2)` m v²

or v =
`\sqrt{(k)/(m) }` x

=
`\sqrt{(1960)/(200) }` x 5 = 50 cm/sec