Question abou Simple Harmonic Motion

In summary, a spring mass system with a mass of 1 kg and spring constant of 100 N/m is initially displaced by 0.5 cm and released from rest. The speed of the mass when it crosses the equilibrium position can be determined using the equations x(t)=A sin(wt+phi) and v(t)=Aw cos(wt+phi). The value of A is 0.5 cm, and the speed will be maximum when cos(wt+phi)=1. Additionally, there is an equation that links the spring constant to the angular frequency of the system. The amplitude of the system will be equal to the initial displacement given to the mass.
  • #1
jcsolis
38
1

Homework Statement



A spring mass system has the following parameters: Mass m=1 kg, spring constant k=100 N/m. The oscillations are started by initially displacing the mass so that the spring is compressed by 0.5 cm and releasing it from rest. What is the speed of the mass (in m/s) when it crosses the equilibrium position?



Homework Equations



x(t)=A sin(wt+phi)
v(t)=Aw^2 cos(wt+phi)

The Attempt at a Solution



Ok, so the spring is compressed by 0.5 cm from its equilibrium position. So when x(t)=0.5 cm, v(t)=0 at t=0

after solving the equation for motion x(t) I got that A=0.5 cm

I don´t know what is next, can somebody help me?
 
Physics news on Phys.org
  • #2
jcsolis said:
v(t)=Aw^2 cos(wt+phi)

I think that your omega shouldn't be squared here but you should check it also remember that as the spring passes through its equilibrium position it would have maximum speed and this can only happen when cos(wt+phi) = 1. (well + or - 1)

Also remember that there's an equation which links K to omega. I'm not sure if you'd have to derive this formula the question doesn't ask you to but it probably depends on how many marks its worth, i'd guess you can just use it though.
 
Last edited:
  • #3
I'm not sure whether the question gave you those equations or not, but if it didn't then Sanitykey is right; the omega shouldn't be squared, there should be a negative sign (to keep direction), and the equation of x(t) should correspond to cosine and v(t) corresponds to sine.
If you apply a certain amount of force to a system (do work to it), then whatever the initial displacement you gave it is going to be the amplitude. So if you pulled the mass .5 cm from the equilibrium point, then the spring isn't going to displace further than + or - .5cm (a bound system).
 

Related to Question abou Simple Harmonic Motion

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion where a system, such as a mass attached to a spring, oscillates back and forth around a central equilibrium point with a constant amplitude and period.

2. What factors affect the period of simple harmonic motion?

The period of simple harmonic motion is affected by the mass of the object, the spring constant, and the amplitude of the motion. The period is also independent of the initial velocity and the displacement of the object.

3. How is simple harmonic motion related to Hooke's law?

Hooke's law states that the force exerted by a spring is directly proportional to the displacement of the object from its equilibrium position. Since simple harmonic motion involves a spring, it follows Hooke's law and the motion can be described using the equation F = -kx, where k is the spring constant and x is the displacement.

4. Can simple harmonic motion occur in different directions?

Yes, simple harmonic motion can occur in any direction as long as there is a restoring force that is proportional to the displacement from the equilibrium position. This means that the motion can be horizontal, vertical, or even circular.

5. What are some real-life applications of simple harmonic motion?

Simple harmonic motion can be observed in many everyday objects, such as a swinging pendulum, a spring on a pogo stick, or a mass on a spring. It is also used in engineering and technology, such as in the suspension systems of cars and in the tuning of musical instruments.

Similar threads

  • Introductory Physics Homework Help
Replies
16
Views
494
  • Introductory Physics Homework Help
2
Replies
51
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
949
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
915
  • Introductory Physics Homework Help
Replies
5
Views
919
  • Introductory Physics Homework Help
Replies
1
Views
1K
Replies
13
Views
406
  • Introductory Physics Homework Help
Replies
10
Views
2K
Back
Top