Solving Simple Harmonic Motion Problem with Energy

In summary, the conversation involved finding the velocity of a 500g block on a spring after it has been pulled 20cm and released, with a motion period of 0.8s. The process involved finding the spring constant (k) using the formula w = sqrt(k/m), finding the total energy using the initial values, and then using this total energy to find the velocity at a specific point (15.4cm from equilibrium). The final answer should be around 1.
  • #1
Glorzifen
25
0

Homework Statement


A 500g block on a spring is pulled 20cm and released. The motion has a period 0.8s.

What is the velocity when the block is 15.4cm from the equilibrium?


Homework Equations


Etotal = 0.5kx2 + 0.5mv2
w = [tex]\sqrt{k/m}[/tex]
w = 2pi*f
f = 1/T

The Attempt at a Solution



[FINDING K]
f = 1/0.8 = 1.25
w = 2pi*(1.25) = 7.85
7.85 = [tex]\sqrt{k/0.5}[/tex]
k = 30.8

[FINDING TOTAL ENERGY BY USING THE INITIAL VALUES]
Etotal = 0.5(30.8)(0.2)2
= 0.616

[USING THE TOTAL ENERGY WITH THE SECOND SET OF VALUES TO FIND V]
0.616 = 0.5(30.8)(0.154)2 + 0.5(0.5)(v)2
x = 1.7

Make sense to me...just hope its right. Any comments? Thanks as always - I appreciate the effort.
 
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  • #2
Glorzifen said:


[USING THE TOTAL ENERGY WITH THE SECOND SET OF VALUES TO FIND V]
0.616 = 0.5(30.8)(0.154)2 + 0.5(0.5)(v)2
x = 1.7


It is all right, up to the last line. What is x? and what is 1.7? And how much is v?

ehild
 
  • #3
yeah its all good except you missed some calculations. My answer is around 1 I think.
 

What is simple harmonic motion and how is it related to energy?

Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium. This means that as an object moves away from its equilibrium position, a force is applied to return it to its original position. This motion can be described using energy, specifically kinetic and potential energy.

How do you calculate the potential energy in a simple harmonic motion problem?

The potential energy in a simple harmonic motion problem is equal to 1/2 times the spring constant (k) times the square of the displacement (x). This can be represented by the equation U = 1/2kx^2.

What is the equation for the total energy in a simple harmonic motion problem?

The total energy in a simple harmonic motion problem is the sum of the kinetic and potential energy. This can be represented by the equation E = 1/2mv^2 + 1/2kx^2, where m is the mass of the object and v is its velocity.

How can the conservation of energy be used to solve simple harmonic motion problems?

The conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. In simple harmonic motion problems, this means that the total energy at any point in the motion will remain constant. This can be used to solve for unknown variables such as displacement, velocity, or spring constant.

What are some real-life examples of simple harmonic motion and how is energy involved?

Examples of simple harmonic motion in everyday life include the motion of a pendulum, a swinging door, or a mass on a spring. In each of these cases, energy is involved as the object moves back and forth between its equilibrium position and its maximum displacement. This energy is constantly changing between kinetic and potential energy, but the total energy remains constant due to the conservation of energy principle.

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