Lagrangian Problem (Find Relation between Amplitude and Momentum)

[Wannabe Physicist]

Homework Statement

Consider the Lagrangian $$L = 1-\sqrt{1-\dot{q}^2}-\frac{q^2}{2}$$ of a particle executing oscillations whose amplitude is $A$ . If $p$ denotes the momentum of
the particle, then $4p^2$ is

a) $(A^2-q^2)(4+A^2-q^2)$
b) $(A^2+q^2)(4+A^2-q^2)$
c) $(A^2-q^2)(4+A^2+q^2)$
d) $(A^2+q^2)(4+A^2+q^2)$

Relevant Equations

1) $p = \frac{\partial L}{\partial \dot{q}}$
2) [not sure if this is relevant] $H = \left(\Sigma_{i} p_i \dot{q}_i\right) - L$

The given lagrangian doesn't seem to correspond to any of the basic systems (like simple/ coupled harmonic oscillators, etc). So I calculated the momentum $p$ which is the partial derivative of $L$ with respect to generalized velocity $\dot{q}$. Doing so I obtain
$$p = \frac{\dot{q}}{\sqrt{1-\dot{q}^2}}$$
Now, I am facing this problem: I do not know how to relate this to the amplitude of the oscillations.

In the hopes of getting a direction or line of thought to follow, I tried writing the Hamiltonian of this system, which is obtained by rewriting the above equation for $p=p(\dot{q})$ in the form $\dot{q} = \dot{q}(p) = \displaystyle{\frac{p^2}{1+p^2}}$ and substituting it in the equation (2) in the "relevant equations" listed above. I found
$$H = \sqrt{1+p^2} +\frac{q^2}{2}$$

But I am still not sure how to proceed.

After trying all of the above I looked up the answer key and the correct option is (a). But I am not sure how to get the answer.

 

[stevendaryl]

Homework Statement:: Consider the Lagrangian $$L = 1-\sqrt{1-\dot{q}^2}-\frac{q^2}{2}$$ of a particle executing oscillations whose amplitude is $A$ . If $p$ denotes the momentum of
the particle, then $4p^2$ is

a) $(A^2-q^2)(4+A^2-q^2)$
b) $(A^2+q^2)(4+A^2-q^2)$
c) $(A^2-q^2)(4+A^2+q^2)$
d) $(A^2+q^2)(4+A^2+q^2)$
Relevant Equations:: 1) $p = \frac{\partial L}{\partial \dot{q}}$
2) [not sure if this is relevant] $H = \left(\Sigma_{i} p_i \dot{q}_i\right) - L$

The given lagrangian doesn't seem to correspond to any of the basic systems (like simple/ coupled harmonic oscillators, etc). So I calculated the momentum $p$ which is the partial derivative of $L$ with respect to generalized velocity $\dot{q}$. Doing so I obtain
$$p = \frac{\dot{q}}{\sqrt{1-\dot{q}^2}}$$
Now, I am facing this problem: I do not know how to relate this to the amplitude of the oscillations.

In the hopes of getting a direction or line of thought to follow, I tried writing the Hamiltonian of this system, which is obtained by rewriting the above equation for $p=p(\dot{q})$ in the form $\dot{q} = \dot{q}(p) = \displaystyle{\frac{p^2}{1+p^2}}$ and substituting it in the equation (2) in the "relevant equations" listed above. I found
$$H = \sqrt{1+p^2} +\frac{q^2}{2}$$

But I am still not sure how to proceed.

After trying all of the above I looked up the answer key and the correct option is (a). But I am not sure how to get the answer.

I think you're almost there. You have an expression for $H$ in terms of $p$ and $q$. The definition of "amplitude" here is that $A = q_{max}$, the maximum value of $q$, which occurs when $p = 0$. So you can relate $A$ to $H$, which allows you to relate $A$, $p$ and $q$.

 

[Wannabe Physicist]

Yes! I had thought of that! But then am I not presuming that $q$ has dimensions of position?

 

[Wannabe Physicist]

Yes! I had thought of that! But then am I not presuming that $q$ has dimensions of position?

I see it now. I was assuming the amplitude itself has dimensions of position. Thanks for helping @stevendaryl !

 

[stevendaryl]

Yes! I had thought of that! But then am I not presuming that $q$ has dimensions of position?

I don't think it matters what the units of $q$ are. It only matters that $A$ means the maximum value of $q$. I actually had never heard of the word "amplitude" applied to situations other than sines and cosines, but I suppose it makes sense for more complicated functions, as well.

 

[sandeep_bhnaja]

Homework Statement: Consider the Lagrangian $$L = 1-\sqrt{1-\dot{q}^2}-\frac{q^2}{2}$$ of a particle executing oscillations whose amplitude is $A$ . If $p$ denotes the momentum of
the particle, then $4p^2$ is

a) $(A^2-q^2)(4+A^2-q^2)$
b) $(A^2+q^2)(4+A^2-q^2)$
c) $(A^2-q^2)(4+A^2+q^2)$
d) $(A^2+q^2)(4+A^2+q^2)$
Relevant Equations: 1) $p = \frac{\partial L}{\partial \dot{q}}$
2) [not sure if this is relevant] $H = \left(\Sigma_{i} p_i \dot{q}_i\right) - L$

The given lagrangian doesn't seem to correspond to any of the basic systems (like simple/ coupled harmonic oscillators, etc). So I calculated the momentum $p$ which is the partial derivative of $L$ with respect to generalized velocity $\dot{q}$. Doing so I obtain
$$p = \frac{\dot{q}}{\sqrt{1-\dot{q}^2}}$$
Now, I am facing this problem: I do not know how to relate this to the amplitude of the oscillations.

In the hopes of getting a direction or line of thought to follow, I tried writing the Hamiltonian of this system, which is obtained by rewriting the above equation for $p=p(\dot{q})$ in the form $\dot{q} = \dot{q}(p) = \displaystyle{\frac{p^2}{1+p^2}}$ and substituting it in the equation (2) in the "relevant equations" listed above. I found
$$H = \sqrt{1+p^2} +\frac{q^2}{2}$$

But I am still not sure how to proceed.

After trying all of the above I looked up the answer key and the correct option is (a). But I am not sure how to get the answer.

It is not said that the momentum is canonical therefore you can not use the canonical momentum formula

 

[Steve4Physics]

It is not said that the momentum is canonical therefore you can not use the canonical momentum formula

It may be worth noting that the thread is well over 2 years old!