Simple Harmonic Motion and angular frequency

Click For Summary
SUMMARY

The discussion focuses on calculating the angular frequency of a horizontal plank undergoing simple harmonic motion when displaced from its equilibrium position. The key equations involved are the angular frequency formula, ##\omega = \frac{\tau}{I}##, and the spring force equation, F = kx. The user incorrectly equated the spring force to torque, leading to mismatched units in their calculations. The correct approach requires proper application of torque and moment of inertia to derive the angular frequency accurately.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Familiarity with torque and moment of inertia concepts
  • Knowledge of spring force and Hooke's Law
  • Basic proficiency in angular motion equations
NEXT STEPS
  • Review the derivation of angular frequency in simple harmonic motion
  • Study the relationship between torque and angular motion
  • Learn about the moment of inertia for different shapes
  • Explore applications of Hooke's Law in oscillatory systems
USEFUL FOR

Physics students, educators, and anyone studying mechanics, particularly those interested in oscillatory motion and its mathematical modeling.

Abid Rizvi
Messages
20
Reaction score
0

Homework Statement


A horizontal plank of mass m and length L is pivoted at one end. The plank's other end is supported by a spring of force constant k (see the figure below). The plank is displaced by a small angle θ from its horizontal equilibrium position and released. Find the angular frequency with which the plank moves with simple harmonic motion. (Use any variable or symbol stated above as necessary.)

The picture is uploaded

Homework Equations


##\omega## = ##\frac{\tau}{I}## (where omega is angular frequency)
F = kx

The Attempt at a Solution


So I said the force that puts the plank back to equilibrium is kx. Using the definition of arc length, I said kx = k##\theta##L. A force of Mg is also acting on the plank, so I had the total torque = K##\theta##L-Mg*##\frac{L}{2}## I know that the moment of inertia is ##\frac{1}{3}##M##L^2##. Using the formula for ##\omega##, I had ##\sqrt{\frac{k\theta L-Mg\frac{L}{2}}{3ML}}## But this is incorrect. What am I doing wrong?
 

Attachments

  • 15-p-057.gif
    15-p-057.gif
    5 KB · Views: 711
Last edited:
Physics news on Phys.org
Abid Rizvi said:
kx = kθ\thetaL
This is a force, not a torque.

The units don't match afterwards due to this mistake.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 15 ·
Replies
15
Views
6K
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
7
Views
1K
  • · Replies 51 ·
2
Replies
51
Views
4K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
13
Views
2K