It may surprise you that the presence of a uniform gravitational field does not

Question

It may surprise you that the presence of a uniform gravitational field does not change the frequency of oscillation of a harmonic oscillator. Here we shall see this is the case. Consider the potential energy of a particle of mass m hanging from a spring with spring constant k in a gravitational field g directed downwards. If x is the upward direction the potential energy may he written as U(x) = 1/2 kx^2 + mgx. Find the equilibrium position x_eq for this potential energy. Does x_eq depend on the field g? Is the equilibrium position stable? Explain. Now find the frequency of oscillation omega_0 and confirm it is independent of the field g.

Explanation / Answer

part a:

equilibrium position is given by value of x for which dU/dx=0

==>k*x+m*g=0

==>x=-m*g/k

part b:

the equilibrium is stable if d^2U/dx^2 <0

here d^2U/dx^2=k

but as k is positive, the equilibrium is not stable.

part c:

force=F=-dU/dx=-k*x-m*g

==>m*(d^2x/dt^2)=-k*x-m*g....(1)

let x=A*cos(w0*t)+B

then dx/dt=-A*w0*sin(w0*t)

d^2x/dt^2=-A*w0^2*cos(w0*t)

from equation 1,

-m*A*w0^2*cos(w0*t)=-k*A*cos(w0*t)-k*B-m*g

hence k*B+m*g=0

=>B=-m*g/k

and k*A=m*A*w0^2

==>w0=sqrt(k/m)

frequency of oscillation=w0=sqrt(k/m)

hence it is independent of the field g.

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