Question

a mass of 0.30kg is attached to a spring and is set in motion with a period of 0.24 s. what is the spring constantof the spring ​

Answer 1

Answer:

The spring constant of the spring is 205.42 N/m.

Explanation:

Springs have their own natural "spring constants" that define how stiff they are. The letter k is used for the spring constant, and it has the units N/m.

k = -F/x

The period of a spring-mass system is proportional to the square root of the mass and inversely proportional to the square root of the spring constant.

Given:

mass of object in SHM = m = 0.30 kg

Time period of the spring mass system = 0.24s

Spring constant = k ?

Finding 'k' using Time period 'T':

We know that

Step-by-step explanation:

Answer 2

The spring constant of the spring is 205.42 N/m.

Step-by-step explanation:

Springs have their own natural "spring constants" that define how stiff they are. The letter k is used for the spring constant, and it has the units N/m.

k = -F/x

The period of a spring-mass system is proportional to the square root of the mass and inversely proportional to the square root of the spring constant.

Given:

mass of object in SHM = m = 0.30 kg

Time period of the spring mass system = 0.24s

Spring constant = k ?

Finding 'k' using Time period 'T':

We know that
`T = 2pi\sqrt{(m)/(k) }`

`(T)/(2pi) = \sqrt{(m)/(k) }\\ (T^(2) )/(4(3.14)^(2) ) = (m)/(k)\\ k = (m*4*9.86)/(T^(2) ) \\k = (0.3*4*9.86)/(0.24^(2))\\k = 205.42 N/m}`