SHM - mass-spring system on frictionless horizontal surface

In summary, a 250 g block attached to a spring on a frictionless surface is compressed by a 2.5 N force and released from rest, resulting in an oscillation with a 1.0 s period. To find the spring constant, use the equation T=2Pi*sqroot(m/k). The amplitude of the oscillation can be found using the equation E=1/2mv^2 + 1/2kx^2=1/2kA^2, with A representing the amplitude. The block's velocity at 30.0 cm from its starting point can be calculated using v=omega*sqroot(A^2-x^2), where x represents the distance from equilibrium. The block
  • #1
trusean
9
0

Homework Statement


A 250 g block is resting on a frictionless horizontal surface. The block is attached to a
spring. The mass-spring system is compressed by a 2.5 N force and then released from
rest.
(a) The resulting oscillation has a 1.0 s period. What is the spring constant?
(b) What is the amplitude of the oscillation?
(c) How fast is the block moving when it is 30.0 cm from its starting point?
(d) How long does the block take to travel this distance?


Homework Equations


T=2Pi/omega
T=2Pi*sqroot(m/k)
E=1/2mv^2 + 1/2kx^2=1/2kA^2
v=omega*sqroot(A^2-x^2)
m=mass
k=spring constant
A=amplitude

The Attempt at a Solution



I have solved for a) the spring constant, 9.87N/m. Then I tryed to use Fs=2.5N to solve for the A by applying the equation F=-kx, where x would repersent the amplitude. This is not working out because my A=0.25m and when inputed into equation for velocity, I get a negitive value under the square root, not allowed. So I was wondering if anyone had some input on this problem. Thank you.
 
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  • #2
Realize that the starting point is not the equilibrium point. 30 cm from the starting point will bring you how far from equilibrium?
 
  • #3
Sorry it took so long to answer. Having the amplitude equal 25cm, I am thinking it is located 25cm from the equilibrium, and thus 30cm from the starting point would be 5cm from the equilibrium?
 
  • #4
You got it!
 
  • #5
cool, thank you very much for your help, have a good one. And by the way, I just found this forum and I think this is a great tool for students, keep it up. peace
 

What is SHM in the context of a mass-spring system on a frictionless horizontal surface?

SHM stands for Simple Harmonic Motion, which refers to the back-and-forth oscillatory motion of a mass attached to a spring on a frictionless horizontal surface. This motion is periodic, with a constant amplitude and frequency, and can be described by a sinusoidal function.

What determines the period of a mass-spring system in SHM?

The period of a mass-spring system in SHM is determined by the mass of the object (m) and the spring constant (k). It can be calculated using the formula T = 2π√(m/k), where T is the period in seconds.

What happens to the frequency of SHM if the mass is increased?

The frequency of SHM decreases as the mass is increased. This is because a larger mass requires a stronger force to move it, which leads to a decrease in the frequency of oscillation. However, the period remains constant as long as the spring constant remains the same.

How does the amplitude affect the motion of a mass-spring system in SHM?

The amplitude of SHM refers to the maximum displacement of the mass from its equilibrium position. A larger amplitude leads to a larger distance covered by the mass during each oscillation, resulting in a longer period and a lower frequency. However, the amplitude does not affect the period or frequency of SHM.

What is the difference between a simple pendulum and a mass-spring system in SHM?

A simple pendulum consists of a mass attached to a string or rod, while a mass-spring system in SHM has a mass attached to a spring on a frictionless horizontal surface. The motion of both systems is periodic, but the period and frequency of a mass-spring system can be adjusted by changing the mass or spring constant, while the period of a simple pendulum is only affected by its length.

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