(Not so) Simple Harmonic Motion Problem

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SUMMARY

The discussion focuses on a simple harmonic motion problem involving a block of mass 0.0750 kg attached to a spring with a spring constant of 82.0 N/m. The force exerted by the spring before release is calculated to be +9.84 N. The angular frequency of the oscillatory motion is determined to be 10.5 rad/s, leading to a maximum speed of 1.25 m/s and a maximum acceleration of 13.2 m/s². The participant successfully solved the problem using established equations for harmonic motion.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Knowledge of angular frequency calculation (w = (k/m)^(1/2))
  • Familiarity with maximum speed and acceleration formulas (v(max) = Aw, a(max) = Aw²)
  • Basic grasp of oscillatory motion concepts
NEXT STEPS
  • Study the derivation of the equations of motion for simple harmonic oscillators
  • Learn about energy conservation in harmonic motion
  • Explore the effects of damping on oscillatory systems
  • Investigate real-world applications of simple harmonic motion in engineering
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for examples of simple harmonic motion problems.

shurleec
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Homework Statement


A block of mass m= 0.0750 kg is fastened to an unstrained horizontal spring whose spring constant is k= 82.0 N/m. The block is given a displacement of +0.120m, where the + sign indicates that the displacement is along the +x axis, and then released from rest.

a. What is the force (magnitude and direction) that the spring exerts on the block just before the block is released?

b. Find the angular frequency w of the resulting oscillatory motion.

c. What is the maximum speed of the block?

d. Determine the magnitude of the maximum acceleration of the block.



Homework Equations


F= -kx
x= Acos(wt)
w= 2pif
w = (k/m)^.5
v(max)= Aw
v(sho)= -Asin(wt)
a(max)= Aw^2


The Attempt at a Solution



a. I plugged the numbers into the equation, and got the answer + 9.84 N.
b. I plugged numbers into the equation w = (k/m)^.5 to get 10.5.
c. Since f = 1/t, I used w=2pif and set it equal to the answer in part b to solve for f, then got t. I use x = Acos(wt) to get amplitude. Then, I plugged everything into V(max) = Aw. The answer I got was 1.25m/s.
d. a(max)= Aw^2. I plugged numbers in and got 13.2m/s^2.

I'm sorry. I initially needed help, but figured it out as I was typing out my inquiries, so why waste all my work. I couldn't find this problem online at all so, yeah. [=
 
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Welcome to PF!

Hi shurleec! Welcome to PF! :smile:

Sometimes just typing it out for someone else to see makes it all clearer! :wink:
 

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