SIMPLE HARMONIC MOTION

Question

A 2.0 kg object attached to a spring, oscillates with simple harmonic motion. The spring constant is 200 N/m, and the maximum force acting on the object is measured to be 3.5 N

A.) What is the amplitude of the oscillation?

B) what is the total energy of the mass-spring system as it oscillates?

C) What is the maximum speed of the mass?

Answer 1

Here we have a nice spring problem, for which we need to know a thing or two about springs. Nice and well known equations for springs are as follows:

F = -k*x (1)

Where F is the force the spring is exerting, "k" is the spring constant, and "x" is the displacement from the equilibrium point.

U = 0.5 * k * x^2 (2)

Where "U" is the potential energy in the spring, "k" and "x" are the same as above.

Part A.) We want to know the max amplitude. To find this, we must ask ourselves what we know about the point of max amplitude, and see if the problem gave us any clues. At max amplitude, we have the largest value of "x", since we are as far away from the equilibrium as we will get. Also, if we have the largest displacement, by equation (1) above we will also have the largest Force "F". This makes intuitive sense; if you pull of a spring, it resists you more the further you stretch it.

Hey guess what, they told us the Max force, which means we can use equation (1) to find the max displacement "x".

F_max = -k * x_max

(F_max) / k = x_max (ignore the negative right now, all I want is the value, not direction)

3.5 / 200 = x_max

0.0175 = x_max

x_max is the max displacement, also known as the amplitude. Value will be in meters.

Part B.) For this part, we need to know the total energy of the spring-mass system. By equation (2), we know how to calculate the total energy in the spring given a certain amount of stretching (x). You might wonder about the mass, which isnt in equation (2). This is a valid concern, here is what is happening: When the spring is fully extended, the mass is not moving, therefore there is NO kinetic energy (kinetic energy has the mass term). When the spring rebounds and is passing through the equilibrium position, there is no potential energy in the spring (its not stretched at all), and all the energy that was in the spring before is now in the kinetic energy of the mass.

There is a constant trade-off between spring energy and kinetic energy of the mass. The total energy of the system, however, is constant. So we can just calculate the energy stored in the spring at max displacement (x_max) and that will be the total energy of the mass-spring system, lets use equation (2):

U = 0.5*k*(x_max)^2 = 0.5 * 200 * (0.0175)^2

U = 100 * 0.00030625 = 0.030625 Joules

U = 0.030625 Joules

Part C.) As mentioned in Part B, all of the energy in the spring at max amplitude will transfer into the mass at the moment the mass passes through the equilibrium position of the spring. So, we just need to set this total energy we calculated "U" equal to the equation for kinetic energy

KE = 0.5 * m * v^2

Where "m" is the mass, and "v" is the velocity (speed)

And we know that, in the middle of the spring's travel, that the "U" from Part B will be equal to the KE. Lets take the equation for KE and solve for v

KE = 0.5 * m * v^2

KE / (0.5 * m) = v^2

SQRT( KE / (0.5 * m) ) = v (sqrt means square root)

SQRT( 0.030625 / (0.5 * 2) ) = v

SQRT( 0.030625 ) = v

0.175 = v

This value will be in meters per second.

Hope this Helps!