Frequency and angular frequency of sinusoidal vibrations in s.h.m.

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SUMMARY

The discussion focuses on calculating the frequency and angular frequency of sinusoidal vibrations in simple harmonic motion (SHM) using specific equations. The user calculated a frequency of 11 Hz based on an acceleration amplitude of 10g (98.1 m/s²) and a vibration amplitude of 2 cm. The angular frequency was derived as ω = 2πf, resulting in an angular frequency of 69 s⁻¹, which raised concerns due to its high value compared to typical examples. A suggestion was made to use the equation a = -ω² * A for a more accurate calculation, yielding an angular frequency of 70 s⁻¹.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with the equations for frequency and angular frequency
  • Knowledge of acceleration and displacement in SHM
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the frequency formula f = (√(a/A))/2π
  • Learn about the relationship between angular frequency and acceleration in SHM
  • Explore the implications of large acceleration amplitudes on frequency calculations
  • Investigate the use of the equation a = -ω² * A in different SHM scenarios
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and simple harmonic motion, as well as educators looking to clarify concepts related to frequency and angular frequency calculations.

Bugsy23
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Homework Statement


I need to calculate the frequency, angular frequency and period of sinusoidal vibrations in simple harmonic motion. The acceleration amplitude is 10g (98.1m s-2) and the vibration amplitude is 2cm

Homework Equations


The equation I have for frequency is
f=(√(a/A))/2*pi
where a is the acceleration amplitude and A is the vibration amplitude
And the angular frequency equation is
[tex]\omega[/tex]=2*pi*f

The Attempt at a Solution



for the frequency I got
f=√(98.1ms-2/2x10-2m)=70/2*pi=11Hz
which seems quite plausible, but when I then try to calculate angular frequency I get
[tex]\omega[/tex]=2pi*11=69s-1
which seems too large a number, since all the examples I have seen have angular frequencies of between about 0 and 3s-1. Is this actually right or have I gone wrong somewhere?
 
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I think you got the answer right. if you think its large, look at acceleration, it's quite large too, and displacement amplitude is small...
 
Just in addition, I think you can also use the equation a=-ang freq^2 * max displacement. Its much faster, however you get an answer of 70s^-1 for the angular frequency. Which is more accurate i do not know. Somebody'll probably tell me I am wrong now. :)
 

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