Pendulum Motion (Simple Harmonic Motion)

Click For Summary
SUMMARY

The discussion focuses on calculating the time it takes for a pendulum with a mass of 180 g on a 2.5 m string to swing from an initial angle of 6.6 degrees to an angle of 2.7 degrees on the opposite side. The angular frequency (omega) is calculated as 1.98 rad/s using the formula omega = sqrt(g/L). The time to swing to 6.6 degrees is determined to be 0.59 seconds, but the calculation for 2.7 degrees is incorrect due to the misapplication of the phase constant. The correct approach involves recognizing the continuous sinusoidal motion of the pendulum.

PREREQUISITES
  • Understanding of Simple Harmonic Motion (SHM)
  • Familiarity with angular frequency and its calculation
  • Knowledge of trigonometric functions in relation to SHM
  • Ability to manipulate equations involving angular displacement
NEXT STEPS
  • Learn about the period of pendulum motion and its relationship to angular displacement
  • Study the concept of phase constants in harmonic motion
  • Explore the mathematical representation of sinusoidal functions in SHM
  • Investigate the impact of varying amplitudes on pendulum motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for examples of pendulum dynamics in Simple Harmonic Motion.

Gotejjeken
Messages
27
Reaction score
0

Homework Statement


A 180 g mass on a 2.5m long string is pulled 6.6 degrees to one side and released. How long does it take for the pendulum to reach 2.7 degrees on the opposite side?

Homework Equations



omega = sqrt(g/L)
theta = S/L
theta(t) = A * cos(omega * t + phi)

The Attempt at a Solution



I am looking at this problem in two parts, one is the time it takes the pendulum to swing 6.6 degrees (0.12 rad) and the other is how long it takes for the pendulum to go the remaining 2.7 degrees (0.047 rad). I have:

omega = sqrt(g/L) = 1.98 rad/s

For time it takes to swing 0.12 rad:

theta = S/L => s = L * theta => s = 0.3m = A

theta(t) = A * cos(omega * t + phi) => 0.12rad = 0.3 * cos(1.98t) => t = 0.59s

For time it takes to swing 0.047 rad:

theta = S/L => s = L * theta => s = 0.118m = A

theta(t) = A * cos(omega * t + phi) => 0.047rad = 0.118 * cos(1.98t) => t = 0.59s

Now, I know for a fact the second part is wrong as the pendulum doesn't go nearly as far. For the first part I left the phase constant as 0 since the pendulum is being held to the far right of its motion, however for the second part I'm not sure what to do with the phase constant. I know that the phase constant should probably be negative as it is pendulum motion to the left, however I'm not sure how to find it. I think if I can find the phase constant for the second part, use it to find the time, and add the times together I should get total time.

Any help is appreciated.
 
Physics news on Phys.org
It would be best to work with 3 significant figures everywhere (0.12 rad isn't particularly accurate).

You write:
theta(t) = A * cos(omega * t + phi)
But theta is an angle and A is a distance in metres. . . .
It's the angular displacement that behaves simple-harmonically (i.e. varies sinusoidally)--there's usually no need to convert to a linear distance. In fact you could leave those angles in degrees; mathematically, degrees make as much sense as metres in this context, and they save work.

(Perhaps a simpler way to think about this part: What is the period of the motion? What fraction of that period is the time taken for the mass to swing from the extreme to the centre? --Think about what a sine or cosine curve looks like and what sorts of symmetry it has.)

For the second part, I think you're making it a bit more complicated than it needs to be. Remember the mass is in continuous sinusoidal motion from the time it's released.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
Replies
1
Views
1K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
7
Views
2K
  • · Replies 51 ·
2
Replies
51
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K