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Old Dec 14th 2016, 03:23 AM   #1
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calculation frequency of oscillation.

A particle of mass 10g is placed in potential field given by V= (50x^2+100)erg/g. what will be the frequency of oscillation?

2. Relevant equations
n(frequency)=2pi(K/m)^1/2

3. The attempt at a solution
F= -dU/dx . given is potential field. so dU/dx= (2*50x +0)=-100x. equating it with -Kx(since F= -Kx) , K=100. so omega(angular frequency)=(K/m)^1/2= (100/10)^1/2= 10^1/2. but the answer as given in the key of my text book is 100^1/2 for angular frequency, hows that possible? where did i go wrong?
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Old Dec 14th 2016, 05:44 AM   #2
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Originally Posted by harini View Post
A particle of mass 10g is placed in potential field given by V= (50x^2+100)erg/g. what will be the frequency of oscillation?

Note that the potential V is given in units of erg/gram. So for a mass of 10 grams the potential is $\displaystyle V = 10(50 x^2 + 100)$ erg, and $\displaystyle \frac {dV}{dx} = 10(100x) $ erg/m.
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Last edited by ChipB; Dec 14th 2016 at 05:46 AM.
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