What is the potential energy at 1/2A in simple harmonic motion?

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In summary, the potential energy of a particle undergoing simple harmonic motion with an amplitude of A and connected to a spring with force constant K is 1/4 of the maximum potential energy, which can be calculated using the formula U=1/2kx^2.
  • #1
Eagles342
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A particle of mass is connected to a spring with a force constant K. The particle undergoes simple harmonic with an amp A. What is the potential energy of the partic when the position is (1/2A)?



Homework Equations


E=1/2kA^2
1/2kdelta^2=1/2mv^2+1/2kx^2


The Attempt at a Solution



U=1/2kx^2=1/2k(1/2A)^2=1/4(1/2KA^2)=1/4E= K=E-U= 3/4? Not really sure at all can somebody help i don't think i did it correctly
 
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  • #2
Welcome to PF;
You started strong - the formula for PE is correct.
You substituted x=A/2 fine.

You have been asked to find U - so stop when you have.
 
  • #3
Besides, E-U = 3/4 makes no sense.
 
  • #4
1/4(1/2KA^2) doesn't equal 1/4E, it equals 1/4 Umax.
gotta be careful.
 
  • #5


I can provide a more detailed and accurate explanation of the potential energy at 1/2A in simple harmonic motion.

In simple harmonic motion, the potential energy of a particle is directly related to its displacement from equilibrium. This can be represented by the equation U = 1/2kx^2, where U is the potential energy, k is the force constant of the spring, and x is the displacement from equilibrium.

In this scenario, the particle has an amplitude of A, meaning its maximum displacement from equilibrium is A. At the position of 1/2A, the particle's displacement from equilibrium is 1/2A. Plugging this value into the equation, we get:

U = 1/2k(1/2A)^2
U = 1/4kA^2

Therefore, the potential energy at 1/2A in simple harmonic motion is 1/4kA^2.

It is also important to note that potential energy is a relative quantity and is typically measured in comparison to a reference point. In this case, the reference point is the equilibrium position, where the potential energy is 0. This means that the actual potential energy of the particle at 1/2A is 1/4kA^2 greater than the potential energy at equilibrium.

I hope this explanation helps clarify the concept of potential energy in simple harmonic motion. Remember to always carefully consider the variables and equations involved in a problem to arrive at the correct solution.
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction to the displacement. This results in a smooth, repetitive motion, like that of a pendulum or a mass on a spring.

2. What is the equation for simple harmonic motion?

The equation for simple harmonic motion is x = A sin(ωt + φ), where x is the displacement from equilibrium, A is the amplitude (maximum displacement), ω is the angular frequency, and φ is the phase angle.

3. What is the difference between simple harmonic motion and uniform circular motion?

The main difference between simple harmonic motion and uniform circular motion is the type of force that causes the motion. In simple harmonic motion, the force is linear and directed towards the equilibrium point, while in uniform circular motion, the force is centripetal and directed towards the center of the circle.

4. How does the mass affect simple harmonic motion?

The mass does not affect the frequency or period of simple harmonic motion, but it does affect the amplitude. A larger mass will result in a smaller amplitude, and vice versa. However, the period (time for one complete cycle) and frequency (number of cycles per unit time) will remain the same.

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

Some common examples of simple harmonic motion include a pendulum clock, a bouncing spring, a guitar string, and a swinging child on a playground swing. Simple harmonic motion can also be seen in the vibrations of atoms and molecules in a solid or the vibrations of air molecules in a sound wave.

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