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The potential energy of a particle executing SHM alon x-axis is U=U0(1-cosax).What is the frequency of oscillation? The potential energy of a particle executing SHM alon x-axis is U=U0(1-cosax).What is the frequency of oscillation?
The potential energy of a particle executing SHM alon x-axis is U=U0(1-cosax).What is the frequency of oscillation?
Dear student U=U0(1-cosax) Force=-dU/dx=-aU0sinax=-w^2 (x) from here you can easliy calculate frequency on comparing the two.. Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation. All the best. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar.. Askiitians Expert Sagar Singh B.Tech, IIT Delhi
Dear student
U=U0(1-cosax)
Force=-dU/dx=-aU0sinax=-w^2 (x)
from here you can easliy calculate frequency on comparing the two..
Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation. All the best. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar.. Askiitians Expert Sagar Singh B.Tech, IIT Delhi
Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.
All the best.
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Askiitians Expert
Sagar Singh
B.Tech, IIT Delhi
U = U0(1-cosax) F = -dU/dx = -Uoasinax for small x , sinax is small or sinax = ax F = -U0a2x a = -Uoa2x/m (a = -w2x) -w2x = -Uoa2x/m w = (Uoa2/m)1/2
U = U0(1-cosax)
F = -dU/dx = -Uoasinax
for small x , sinax is small or sinax = ax
F = -U0a2x
a = -Uoa2x/m (a = -w2x)
-w2x = -Uoa2x/m
w = (Uoa2/m)1/2
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