Spring constant of object in simple harmonic motion

In summary, the spring constant of a vertical spring in simple harmonic motion with a 15.0-N object oscillating at the end can be found by using the frequency of oscillation. The formula for the frequency is related to the spring constant and mass by the equation omega = sqrt(k/m). Using this information, the spring constant can be calculated to be approximately 166.67 N/m.
  • #1
qlzlahs
13
0

Homework Statement


A 15.0-N object is oscillating in simple harmonic motion at the end of an ideal vertical spring. Its vertical position y as a function of time t is given by y(t)=4.50 cos[(19.5s−1)t−π/8] in centimeters.
What is the spring constant of the spring?

Homework Equations


y component of spring force = -k*displacement = mg
k = (mg)/displacement

The Attempt at a Solution


I thought the spring constant of the spring was basically (mg)/displacement. mg = 15 N, and the total displacement of the spring is 4.50*2 = 9.00 centimeters. I thought the spring constant was 15N/0.09m = 166.67 N/m.
 
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  • #2
Hello, and welcome to PF!
qlzlahs said:
I thought the spring constant of the spring was basically (mg)/displacement.
This formula for the spring constant applies to the situation where the mass is hanging at rest. The displacement here would be the amount the spring is stretched while hanging at rest. This displacement is not related at all to the 4.50 cm amplitude of oscillation. Since you are not given the amount that the spring is stretched if the mass is hanging at rest, you will not be able to get the spring constant this way.

Do you know any relation between the spring constant and the period or frequency of oscillation?
 
  • #3
Hi qlzlahs, Welcome to Physics Forums.

qlzlahs said:

Homework Statement


A 15.0-N object is oscillating in simple harmonic motion at the end of an ideal vertical spring. Its vertical position y as a function of time t is given by y(t)=4.50 cos[(19.5s−1)t−π/8] in centimeters.
What is the spring constant of the spring?

Homework Equations


y component of spring force = -k*displacement = mg
k = (mg)/displacement

The Attempt at a Solution


I thought the spring constant of the spring was basically (mg)/displacement. mg = 15 N, and the total displacement of the spring is 4.50*2 = 9.00 centimeters. I thought the spring constant was 15N/0.09m = 166.67 N/m.
I presume that your equation for the displacement as a function of time, y(t)=4.50 cos[(19.5s−1)t−π/8] was meant to be

y(t)=4.50 cos[(19.5s−1)t−π/8]

so that the units of the argument of the cosine would make sense?

Note that the relationship between displacement and weight of the mass on a spring applies to the static case when there's no motion -- the mass is hanging motionless. In that case the mass is stationary at an equilibrium position, where the weight is equal to the force that the spring provides.

Unfortunately you aren't given information about this equilibrium scenario. Instead you are given information for oscillations about the equilibrium.

So, look into your text and class notes (or on the web) for information on the period or frequency of mass-spring oscillation. You should find that it is related to the spring constant. Can you pick out the oscillation frequency from your displacement formula?

Edit: Ah. I see that TSny got there before me. Oh well, carry on...
 
  • #4
gneill said:
Edit: Ah. I see that TSny got there before me. Oh well, carry on...
Please feel free to continue contributing to this thread!
 
  • #5
I found that omega equals the square root of (k/m), and used it to find the answer. Thank you both!
 

What is the spring constant of an object in simple harmonic motion?

The spring constant of an object in simple harmonic motion is a measure of the stiffness of the spring. It is represented by the symbol k and is measured in units of force per unit length, such as N/m or lb/in.

How is the spring constant related to the frequency of simple harmonic motion?

The spring constant is directly proportional to the square of the frequency of simple harmonic motion. This means that as the spring constant increases, the frequency of the motion also increases.

How does the mass of the object affect the spring constant in simple harmonic motion?

The mass of the object does not affect the spring constant in simple harmonic motion. The spring constant only depends on the stiffness of the spring and the distance it is stretched or compressed.

What happens to the spring constant when the spring is stretched or compressed?

The spring constant remains constant as long as the material and dimensions of the spring do not change. However, if the spring is stretched or compressed beyond its elastic limit, the spring constant may change.

How can the spring constant be measured experimentally?

The spring constant can be measured experimentally by using Hooke's law, which states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed. By measuring the force and displacement of the spring, the spring constant can be calculated using the equation k = F/x.

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