Simple Harmonic motion (non-calculus)

[suwarna07]

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.

 

[lepton5]

it's more simple if you use energy relation.

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

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

 

[suwarna07]

it's more simple if you use energy relation.

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

use relation \omega ^2 = \frac{k}{m} 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?

 

[Fewmet]

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.

 

[suwarna07]

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?

 

[lepton5]

you can find acceleration just using simple kinematic eqn, that relate V_max, V_min, a, t.

 

[suwarna07]

you can find acceleration just using simple kinematic eqn, that relate V_max, V_min, a, t.

whats the time tho?

 

[lepton5]

time t (period / 4) to reach max velocity from min velocity.

 

[suwarna07]

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?

 

[lepton5]

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 a = \omega ^2 A

 

[suwarna07]

sory my bad, mix it up with kinematics.

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

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