Simple Harmonic motion (non-calculus)

In summary, Atoms in a solid are not stationary, but vibrate about their equilibrium positions. The frequency of vibration is about f = 2 E12 Hz and the amplitude is about 1.1 E-11m. The maximum speed of a typical atom can be found using the relation \omega ^2 = \frac{k}{m} and equating mechanical energy to kinetic energy. The maximum speed is approximately 140 m/s. The maximum acceleration can be found using the relation a = \omega ^2 A, where A is the amplitude of the atom's vibration. The maximum acceleration is approximately 1.7 E15 m/s^2.
  • #1
suwarna07
33
0

Homework Statement


Atoms in a solid are not stationary, but vibrate about their equilibrium positions. Typically, the frequency of vibration is about f = 2 E12 Hz and the amplitude is about 1.1 E-11m. For a typical atom, what is its (a) maximum speed?

Homework Equations


T= 1/f.
v = change in displacement/ change in time

The Attempt at a Solution



I found the period which is 1/f. For T i got 5 E -13.
We know the maximum speed is at its equilibrium position.
So, to find the time taken from the highest point (A) to equilibrium point, I divided T/4 which is 1.25 E -13.
Finally, to find velocity I divided 1.25 E-13 (change in time) by 1.1 E-11 (change in displacement or Amplitude) and i got 88 m/s.
BUT the correct answer is 140 m/s.

Can someone tell me what did i do wrong in here? AND this is non-calculus based physics course.
 
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  • #2
it's more simple if you use energy relation.

for maximum velocity, you can equate ME = KE or [tex]\frac{1}{2}k A^2 = \frac{1}{2}m v^2[/tex].

use relation [tex]\omega ^2 = \frac{k}{m}[/tex] and you will get your velocity to about 140m/s
 
  • #3
lepton5 said:
it's more simple if you use energy relation.

for maximum velocity, you can equate ME = KE or [tex]\frac{1}{2}k A^2 = \frac{1}{2}m v^2[/tex].

use relation [tex]\omega ^2 = \frac{k}{m}[/tex] and you will get your velocity to about 140m/s

ohh yea thank you so much and is there any way we can find the max acceleration too?
 
  • #4
suwarna07 said:

Homework Statement


Finally, to find velocity I divided 1.25 E-13 (change in time) by 1.1 E-11 (change in displacement or Amplitude) and i got 88 m/s.
BUT the correct answer is 140 m/s.

Can someone tell me what did i do wrong in here? AND this is non-calculus based physics course.

It is worth noting (so that you can avoid doing this again in future problems) that your method gives you the average velocity. If acceleration were a constant, the maximum speed would be twice the average. But, as you subsequent post males clear, acceleration is changing.
 
  • #5
Fewmet said:
It is worth noting (so that you can avoid doing this again in future problems) that your method gives you the average velocity. If acceleration were a constant, the maximum speed would be twice the average. But, as you subsequent post males clear, acceleration is changing.

ohh i see! anyway i can find the max acceleration too?
 
  • #6
you can find acceleration just using simple kinematic eqn, that relate Vmax, Vmin, a, t.
 
  • #7
lepton5 said:
you can find acceleration just using simple kinematic eqn, that relate Vmax, Vmin, a, t.

whats the time tho?
 
  • #8
time t (period / 4) to reach max velocity from min velocity.
 
  • #9
lepton5 said:
time t (period / 4) to reach max velocity from min velocity.

so the period is 1/f = 5 E-13
5E-13/ 4 = 1.25E-13

v1 = v0 + at ( i thought we were suppose to use this only when the a is constant but anyway

138 = a(1.25 E-13)
a = 1.1 E 15

but the answer key says that the answer should be 1.7 E 15... so i am wondering is the answer close enough or is something wrong?
 
  • #10
suwarna07 said:
so the period is 1/f = 5 E-13
5E-13/ 4 = 1.25E-13

v1 = v0 + at ( i thought we were suppose to use this only when the a is constant but anyway

138 = a(1.25 E-13)
a = 1.1 E 15

but the answer key says that the answer should be 1.7 E 15... so i am wondering is the answer close enough or is something wrong?

sory my bad, mix it up with kinematics.

you should you use a (acceleration) for SHM, that is [tex] a = \omega ^2 A[/tex]
 
  • #11
lepton5 said:
sory my bad, mix it up with kinematics.

you should you use a (acceleration) for SHM, that is [tex] a = \omega ^2 A[/tex]

cool now i got it.
tx a lot for helping me man
 

Related to Simple Harmonic motion (non-calculus)

What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium and always acts towards the equilibrium point. This results in a repetitive back-and-forth motion that follows a sinusoidal pattern.

What are some examples of simple harmonic motion?

Some common examples of simple harmonic motion include a mass on a spring, a pendulum, and a vibrating guitar string. Any object that oscillates or vibrates with a constant period can be considered to be exhibiting simple harmonic motion.

How is simple harmonic motion different from other types of periodic motion?

The main difference between simple harmonic motion and other types of periodic motion is that the restoring force in simple harmonic motion is directly proportional to the displacement, while in other types of periodic motion, the restoring force may vary with the displacement or may not always act towards the equilibrium point.

What is the equation for simple harmonic motion?

The equation for simple harmonic motion is x = A*sin(ωt + φ), where x is the displacement from equilibrium, A is the amplitude, ω is the angular frequency (equal to divided by the period), and φ is the phase angle.

How is simple harmonic motion related to energy?

In simple harmonic motion, the total energy of the system remains constant. The kinetic energy and potential energy of the oscillating object are constantly being exchanged, but the total energy remains the same. This is known as the law of conservation of energy.

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