SH oscillator behaviour when potential term is added

I came across the following question in a graduate exam it looks trivial nevertheless it is conceptual and I do not know how to connect the concept of potential and frequency to get one of the answers given. A simple harmonic oscillator has a potential energy $V=\frac{1}{2} kx^2$. An additional potential term $V = ax$ is added then,

a) It is SHM with decreased frequency around a shifted equilibrium

b) Motion is no longer SHM

c)It is SHM with decreased frequency around a shifted equilibrium

d) It is SHM with same frequency around a shifted equilibrium

e)It is SHM with increased frequency around origin

Attempt: at $x=0$ $V=0$ taking derivative $$\frac{dV}{dx}=kx+a$$

That is all I could think from what is given.

Since another has given an explicit answer, I will simply add the following.

Recall that the force is

$$F = -\frac{dV}{dx} = -(kx + a)$$

Thus there is a constant force of $-a$ in addition to the restoring force $-kx$.

Think of, for example, the case of a mass suspended by a spring from the ceiling. Gravity provides a constant downward force on the mass which shifts the equilibrium length of the spring-mass system but not the natural frequency.

• yeah that's a simpler answer actually. You can then immediately integrate to get the correct form for $V$. Apr 17 '17 at 11:50

Write the new potential $$V'=\frac{1}{2}k(x-x_0)^2+C$$ where $C$ is a constant.

Expanding $V'$: $$V'=\textstyle\frac{1}{2}kx^2-kx_0x+\frac{1}{2}kx_0^2+C$$ and choose $x_0=-a/k$ such that $-kx_0=a$. This yields $$\textstyle\frac{1}{2}kx^2+ax+\frac{1}{2}kx_0^2+C$$ and you can now find $C$ to remove the $\frac{1}{2}kx_0^2$. This shows the frequency will not change but the equilibrium position will just be shifted.

• I did not understand why the new potential was written the way you did. what about 'ax'? that is not a constant right? Apr 17 '17 at 11:38
• I just fixed some typos and added an additional line. Apr 17 '17 at 11:45
• I meant for the new potential in your answer Apr 17 '17 at 11:47
• Sorry I don't understand your comment. You basically have the new potential $\frac{1}{2}kx^2+ax$ so one simply completes the square. Apr 17 '17 at 11:49

$V(x)= \frac{1}{2}k( x-x_0)^{2}$, $x_0=-\frac{a}{k}$

$\omega=\sqrt{\frac{V''(x)}{m}}$, $V'(x)=k(x-x_0)$, $V''(x)=k$. So (d).

• Generally you want your answer to have more content than just equations without context.
– JMac
Apr 17 '17 at 12:11