SHM horizontal spring mass system

In summary: A/4?i don't have any idea how to do this, but i think once i get part a i will understandYou can use the same approach as in part a, but with a different value for x. Remember to use the appropriate equation for the total energy of a spring. In summary, the problem involves a mass attached to a spring with a spring constant k, and pulled aside a distance A from its equilibrium position. The speed at x=0 is equal to the maximum speed, which can be found using the conservation of energy. The net force on the mass at position x=A is equal to -kA. To find the speed at x=A/4, the
  • #1
bionerdette
1
0
This problem states that a mass m is attatched to a spring with spring constant k and is pulled aside a distance A from its equilibrium position x=0.

a) what is the speed as it passes the position x=0?
I know that at A and -A v=0 and at x=0 v=max, but I can't remember how to get an equation for what that max is!

b)what is the net force on the mass at position x=A?
would this be F=-kA?

c)what is the speed when x=A/4?
i don't have any idea how to do this, but i think once i get part a i will understand

any help would be greatly appreciated!
 
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  • #2
Hi bionerdette and welcome to PF,

For future reference, we have Homework forums for such questions, but don't worry you thread will be moved there soon. With regards to your question,
bionerdette said:
a) what is the speed as it passes the position x=0?
I know that at A and -A v=0 and at x=0 v=max, but I can't remember how to get an equation for what that max is!
I would consider conservation of energy.
bionerdette said:
b)what is the net force on the mass at position x=A?
would this be F=-kA?
Indeed it would :smile:
 
  • #3


a) The speed of the mass at position x=0 can be determined by using the equation for the velocity of a mass in SHM, which is v = ω√(A^2-x^2), where ω is the angular frequency of the system. In this case, ω = √(k/m), where k is the spring constant and m is the mass. Therefore, the speed at x=0 can be calculated as v = √(k/m)√(A^2-0) = A√(k/m). This means that the maximum speed of the mass at x=0 is directly proportional to the amplitude A and the square root of the ratio of the spring constant and mass.

b) The net force on the mass at position x=A can be determined by using Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration. In this case, the acceleration is given by a = -(k/m)x, where x is the displacement from equilibrium. Therefore, the net force can be calculated as F = ma = -(k/m)x = -kA. This means that the net force on the mass at position x=A is equal to the spring constant multiplied by the displacement from equilibrium.

c) The speed of the mass when x=A/4 can be determined by using the same equation as part a), but with x=A/4. This means that the speed can be calculated as v = √(k/m)√(A^2-(A/4)^2) = √(3/4)A√(k/m). This shows that the speed at x=A/4 is equal to a fraction of the maximum speed at x=0, and the fraction is dependent on the amplitude A and the ratio of the spring constant and mass.
 

What is SHM (Simple Harmonic Motion) in a horizontal spring mass system?

SHM refers to the repetitive back and forth motion of a mass attached to a spring when it is displaced from its equilibrium position. This motion follows a sinusoidal pattern and is characterized by a constant amplitude and frequency.

What factors affect the period of a SHM horizontal spring mass system?

The period of a SHM horizontal spring mass system is affected by the mass of the object attached to the spring, the spring constant, and the amplitude of the motion. The period is also independent of the object's initial displacement or velocity.

How is the displacement of a mass in SHM related to the force exerted by the spring?

The displacement of the mass is directly proportional to the force exerted by the spring. This means that as the displacement increases, the force also increases, resulting in a stronger restoring force that brings the mass back to its equilibrium position.

What is the equation for calculating the frequency of a SHM horizontal spring mass system?

The frequency of a SHM horizontal spring mass system can be calculated using the equation f = 1/2π√(k/m), where f is the frequency, k is the spring constant, and m is the mass of the object.

How does the amplitude affect the maximum velocity and maximum acceleration in SHM?

The maximum velocity and maximum acceleration in SHM are directly proportional to the amplitude of the motion. This means that as the amplitude increases, the maximum velocity and acceleration also increase, resulting in a faster and more forceful motion.

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