A particle of mass $$m$$ moving along the $$x-$$ axis as a potential energy $$U(x)=a+bx^2$$ where $$a$$ and $$b$$ are positive constants. It will execute simple harmonic motion with a frequency determined by the value of :
A
$$b$$ and $$a$$ alone
B
$$b,a $$ and $$m$$ alone
C
$$b$$ alone
D
$$b$$ and $$m$$ alone
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Solution
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Correct option is C. $$b$$ and $$m$$ alone $$U=a+bx^2$$ $$F=-\dfrac {dU}{dx}=-[0+2bx]$$ $$F=ma=-2bx \ \Rightarrow \ a=-\left(\dfrac {2b}{m}\right)x$$ $$\omega =2\pi f =\left(\dfrac {2b}{m}\right)^{1/2}$$ Hence frequency depends upon $$b$$ and $$m$$
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