A mass of 0.250 kg is attached to a spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the force constant of the spring?

The force constant of the spring is approximately 24.038 newtons per meter.

As we are talking about Simple Harmonic Motion. In this exercise we need to determine the Spring Constant (), in newtons per meter, from the equation of the Period (), in seconds, which is described below:

(1)

Where is the mass of the moving element, in kilograms.

If we know that and , then the spring constant of the spring is:

The force constant of the spring is approximately 24.038 newtons per meter.

Please see this question related to Simple Harmonic Motion for further details: link

Force constant of the spring (k) = 24.07 N/m

Concept/Theory:

The period of a spring-mass system is proportional to the square root of the mass (m) and inversely proportional to the square root of the force constant of the spring (k).

Equation of period:

Explanation:

Mass = 0.250 kg

Period = 0.640 s

By substituting values in the equation, we get:

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer