How can we double the amplitude of an oscillator?

[eprparadox]

Homework Statement

The amplitude of any oscillator can be doubled by:
A. doubling only the initial displacement
B. doubling only the initial speed
C. doubling the initial displacement and halving the initial speed
D. doubling the initial speed and halving the initial displacement
E. doubling both the initial displacement and the initial speed

Homework Equations

$x(t) = A \sin(\omega t + \delta)$

The Attempt at a Solution

The answer is suppose to be E. But I have no intuition for that.

Here's what I did quantitatively. I started with a simple harmonic oscillator as x(t) = A \sin(\omega t + \delta)

Our initial conditions are $x(0) = x_0$ and $\dot{x}(0) = v(0) = v_0$.

Plugging these initial conditions in, we get:

$x_0 = A\sin(\delta)$ and

$v_0 = A \omega \cos(\delta)$

We can subtract these two equations to get

x_0 - v_0 = A \sin(\delta) - A\omega\cos(\delta)

Solving for $A$, we get

A = \frac{x_0 - v_0}{\sin(\delta) - \omega \cos(\delta)}

So if we want to double $A$, then we need to double the right side of the above equation and that amounts to doubling both the $x_0$ value and the initial speed. Two questions for the PF crew:
1. Does the above look like the right thought process to getting the answer (even though I already knew what the answer was)?
2. Is there an intuition to why it is that we need to double both the initial displacement and the initial speed to double the amplitude?

 

[mfb]

Subtracting things with different units is a bit questionable. I would subtract $x_0$ and $\frac{y_0}{\omega}$. δ depends on the initial conditions, however.

 

[Ray Vickson]

Homework Statement

The amplitude of any oscillator can be doubled by:
A. doubling only the initial displacement
B. doubling only the initial speed
C. doubling the initial displacement and halving the initial speed
D. doubling the initial speed and halving the initial displacement
E. doubling both the initial displacement and the initial speed

Homework Equations

$x(t) = A \sin(\omega t + \delta)$

The Attempt at a Solution

The answer is suppose to be E. But I have no intuition for that.

Here's what I did quantitatively. I started with a simple harmonic oscillator as x(t) = A \sin(\omega t + \delta)

Our initial conditions are $x(0) = x_0$ and $\dot{x}(0) = v(0) = v_0$.

Plugging these initial conditions in, we get:

$x_0 = A\sin(\delta)$ and

$v_0 = A \omega \cos(\delta)$

We can subtract these two equations to get

x_0 - v_0 = A \sin(\delta) - A\omega\cos(\delta)

Solving for $A$, we get

A = \frac{x_0 - v_0}{\sin(\delta) - \omega \cos(\delta)}

So if we want to double $A$, then we need to double the right side of the above equation and that amounts to doubling both the $x_0$ value and the initial speed.Two questions for the PF crew:
1. Does the above look like the right thought process to getting the answer (even though I already knew what the answer was)?
2. Is there an intuition to why it is that we need to double both the initial displacement and the initial speed to double the amplitude?

$x_0$ and $v_0$ have different units, so you cannot just blithely subtract them.

 

[eprparadox]

Hey all, thanks for the responses. I was also concerned about the idea of subtracting mismatched units.

What is the solution to this problem then? How do we get the answer? And is there an intuition to this idea that in order to double the amplitude, we need to double BOTH the displacement AND the initial velocity?

Thanks!

 

[kuruman]

What is the solution to this problem then?

Sorry, we do not provide solutions.

And is there an intuition to this idea that in order to double the amplitude, we need to double BOTH the displacement AND the initial velocity?

Intuition can often lead you astray. Have you tried using energy conservation?

 

[ehild]

The Attempt at a Solution

Our initial conditions are $x(0) = x_0$ and $\dot{x}(0) = v(0) = v_0$.

Plugging these initial conditions in, we get:

$x_0 = A\sin(\delta)$ and

$v_0 = A \omega \cos(\delta)$

delta can also change if you change the initial conditions. You have to eliminate it. Divide the second equation by ω. You get a pair of equations

$x_0 = A\sin(\delta)$ and
$v_0/ω = A \cos(\delta)$

Square both equations and add them, you get A in terms of x₀ and v₀. See at what case will it be doubled for any values of x₀ and v₀.

 

[berkeman]

Homework Statement

The amplitude of any oscillator can be doubled by:
A. doubling only the initial displacement
B. doubling only the initial speed
C. doubling the initial displacement and halving the initial speed
D. doubling the initial speed and halving the initial displacement
E. doubling both the initial displacement and the initial speed

This question does not seem well-formed to me. Is this all that was stated in the original problem statement?

In order to double the initial speed, you start with a sin() function, which has an amplitude of 0 at t=0, and has maximum "speed" at t=0. In order to double the initial amplitude, you start with a cos() function, which has its max amplitude at t=0 and no "speed" at t=0. Trying to double both the initial amplitude and the initial speed makes no sense to me.

Our initial conditions are x(0)=x0 x(0) = x_0 and ˙x(0)=v(0)=v0 \dot{x}(0) = v(0) = v_0 .

Were these initial conditions given as part of the problem statement, or did you come up with them as part of your work on the problem?

 

[kuruman]

The general solution can also be written as $x(t)=A\sin(\omega t)+B\cos(\omega t)$. By inspection this becomes $x(t)=(v_0/\omega)\sin(\omega t)+x_0\cos(\omega t)$ when the initial conditions are applied. (Calculate $x(0)$ and $\dot{x}(0)$ if you don't believe me.)
With $k = m\omega^2$, energy conservation says $\frac{1}{2}m\omega^2A^2=\frac{1}{2}m\omega^2 x^2+\frac{1}{2}m\dot{x}^2.$
Calculate the right side and see what it implies.

 

[mfb]

In order to double the initial speed, you start with a sin() function, which has an amplitude of 0 at t=0, and has maximum "speed" at t=0. In order to double the initial amplitude, you start with a cos() function, which has its max amplitude at t=0 and no "speed" at t=0. Trying to double both the initial amplitude and the initial speed makes no sense to me.

You do not know the initial phase. What if $x_0=1m$ and $\dot x_0 = 1 \frac m s$? If you double just distance or speed you don't double the amplitude. You have to double both. In the special case of $x_0=0$ doubling the distance won't matter and in the special case of $\dot x_0=0$ doubling the speed won't matter, but you cannot limit the analysis to these special cases.