Simple Harmonic Oscillator behaviour when a potential term is added

[Rahulrj]

Homework Statement

A simple harmonic oscillator has a potential energy V=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

Homework Equations

$$w= \sqrt {k/m}$$
$$x = A \sin{wt}$$

The Attempt at a Solution

$$V = 1/2 kx^2+ax$$
At x=0
$$V=0$$
Maximum potential
$$dV/dx = kx + a$$
I do not know if this is correct or if it is so then i do not know how to go further than this.
Please help.[/B]

 

[PeroK]

Perhaps think about what a "shifted equilibrium" might mean? Would a change of coordinate help?

 

[Rahulrj]

Perhaps think about what a "shifted equilibrium" might mean? Would a change of coordinate help?

I did not understand what you meant by change of coordinate.

 

[PeroK]

I did not understand what you meant by change of coordinate.

A change of coordinate is where you use another coordinate instead of $x$. That could be anything. But in this case, you could try using a new coordinate $y$ with $y = x + c$, where $c$ is a constant.

 

[Rahulrj]

A change of coordinate is where you use another coordinate instead of $x$. That could be anything. But in this case, you could try using a new coordinate $y$ with $y = x + c$, where $c$ is a constant.

Do you mean to suggest that I should write it as $$ V = \frac {1}{2}k(y-c)^2$$?

 

[PeroK]

Do you mean to suggest that I should write it as $$ V = \frac {1}{2}k(y-c)^2$$?

Yes, except that's not the potential you are supposed to be working with. It's:

$V(x) = \frac12 kx^2 + ax$

 

[Rahulrj]

Yes, except that's not the potential you are supposed to be working with. It's:

$V(x) = \frac12 kx^2 + ax$

Yes so then it becomes $$ V = \frac {1}{2}k(y-c)^2+a(y-c)$$
If i expand it, $$V = \frac{1}{2}(ky^2+kc^2)-yc+ay-ac$$

 

[PeroK]

Yes so then it becomes $$ V = \frac {1}{2}k(y-c)^2+a(y-c)$$
If i expand it, $$V = \frac{1}{2}(ky^2+kc^2)-yc+ay-ac$$

Does that give you any ideas about a suitable value for $c$? Hint: look at the terms in $y$.

Note, however, that your algebra has gone wrong.

 

[Rahulrj]

Does that give you any ideas about a suitable value for $c$? Hint: look at the terms in $y$.

Note, however, that your algebra has gone wrong.

The expansion is $$V = \frac{1}{2}(ky^2+kc^2)-kyc+ay-ac$$
Suitable value to cancel what term?
if $$ c= \frac {a}{k}$$ I would get $$V = \frac{1}{2}(ky^2+\frac {a^2}{k})-\frac {a^2}{k}$$

$$V = \frac{1}{2}(ky^2)-\frac {a^2}{k}$$
But it does not look familiar.

 

[PeroK]

The expansion is $$V = \frac{1}{2}(ky^2+kc^2)-kyc+ay-ac$$
Suitable value to cancel what term?
if $$ c= \frac {a}{k}$$ I would get $$V = \frac{1}{2}(ky^2+\frac {a^2}{k})-\frac {a^2}{k}$$

$$V = \frac{1}{2}(ky^2)-\frac {a^2}{k}$$
But it does not look familiar.

What effect does a constant term have on a potential?

 

[Rahulrj]

What effect does a constant term have on a potential?

A shift from the normal potential. So how can i infer frequency from this?

 

[PeroK]

A shift from the normal potential. So how can i infer frequency from this?

Adding a constant to a potential makes no effective difference. The force is the derivative of the potential, so the constant term has no effect.