Simple Harmonic Motion of an ideal spring

[Jtappan]

Homework Statement

An ideal spring has a spring constant k = 29 N/m. The spring is suspended vertically. A 1.4 kg body is attached to the unstretched spring and released. It then performs oscillations.
(a) What is the magnitude of the acceleration of the body when the extension of the spring is a maximum?
____ m/s2
(b) What is the maximum extension of the spring?
____ m

Homework Equations

a$$_{m}$$=$$\frac{k}{m}$$A

The Attempt at a Solution

I don't know how to start this problem. Any help?

 

[Kurdt]

What are the forces acting on the mass. To work out what the acceleration is you need to find the net force at the maximum extension.

 

[ogb p]

To start, we need to understand the concept of simple harmonic motion. This is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium. In this case, the ideal spring follows this type of motion since it has a constant spring constant k.

Now, to answer the first question, we can use the equation for acceleration in simple harmonic motion: a_m = (k/m)A, where k is the spring constant and m is the mass of the body attached to the spring. A is the maximum amplitude, which in this case is the maximum extension of the spring. Therefore, when the extension of the spring is a maximum, the body experiences the maximum acceleration, which is equal to (k/m)A. Plugging in the given values, we get a_m = (29 N/m)/(1.4 kg) * A = 20.7 A m/s^2.

For the second question, we can use the equation for spring potential energy: U_s = (1/2)kx^2, where x is the displacement from equilibrium. At maximum extension, the spring is stretched to its maximum displacement x_max. Therefore, we can rearrange the equation to solve for x_max: x_max = √(2U_s/k). Plugging in the given values, we get x_max = √(2 * (29 N/m) * (1.4 kg) * 9.8 m/s^2) = 0.47 m. This is the maximum extension of the spring when the 1.4 kg body is attached to it and released.