# How can we double the amplitude of an oscillator?

In summary, the amplitude of a harmonic oscillator can be doubled by doubling the displacement and halving the speed.

## 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?

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.

## 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.

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!

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?

## 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.

mfb

## 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?

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.

berkeman said:
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.

## 1. How does the amplitude of an oscillator affect its performance?

The amplitude of an oscillator is a measure of its maximum displacement from its equilibrium position. It is an important factor in determining the strength and stability of an oscillating system. A higher amplitude indicates a stronger and more energetic oscillation, while a lower amplitude may result in a weaker and less stable oscillation.

## 2. What is the relationship between amplitude and frequency in an oscillator?

The amplitude and frequency of an oscillator are inversely proportional. This means that as the amplitude increases, the frequency decreases and vice versa. This relationship is governed by the natural frequency of the system and can be mathematically described by the equation A = A0sin(2πft), where A is the amplitude, A0 is the initial amplitude, f is the frequency, and t is time.

## 3. How can we measure the amplitude of an oscillator?

The amplitude of an oscillator can be measured using a variety of methods, depending on the type of oscillator. For mechanical oscillators, such as a pendulum or a spring-mass system, the amplitude can be measured directly using a ruler or a sensor. For electrical oscillators, an oscilloscope can be used to measure the voltage or current amplitude. In some cases, the amplitude can also be indirectly calculated using mathematical equations.

## 4. What factors can affect the amplitude of an oscillator?

The amplitude of an oscillator can be affected by various factors, such as the initial energy input, damping forces, and external disturbances. The initial energy input, or the amplitude of the driving force, can directly influence the amplitude of the oscillator. Damping forces, such as friction and air resistance, can cause the amplitude to decrease over time. External disturbances, such as a sudden force or a change in temperature, can also affect the amplitude of an oscillator.

## 5. What methods can we use to double the amplitude of an oscillator?

There are several methods that can be used to double the amplitude of an oscillator. One way is to increase the energy input to the system, either by increasing the amplitude of the driving force or by using a more powerful source. Another method is to reduce the damping forces, which can be achieved by minimizing friction or using a more efficient design. Additionally, adjusting the natural frequency of the system can also result in a higher amplitude oscillation.

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