Question 183: Oscillations

Question

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $$4 \mathrm{~m}, 2 \mathrm{~ms}^{-1}$$ and $$16 \mathrm{~ms}^{-2}$$ at a certain instant. The amplitude of the motion is $$\sqrt{x}, \mathrm{~m}$$ where $$x$$ is _________.

Accepted Answer

Let's begin by understanding the equations related to simple harmonic motion (SHM). For a particle executing SHM, the position $$x$$, velocity $$v$$, and acceleration $$a$$ are given by the following equations:
1. Position: $$x = A \cos(\omega t + \phi)$$
2. Velocity: $$v = -A \omega \sin(\omega t + \phi)$$
3. Acceleration: $$a = -A \omega^2 \cos(\omega t + \phi)$$
Here, $$A$$ is the amplitude of the motion, $$\omega$$ is the angular frequency, and $$\phi$$ is the phase constant.
Given the magnitudes at a certain instant:
$$x = 4 \, \mathrm{m}$$
$$v = 2 \, \mathrm{ms}^{-1}$$
$$a = 16 \, \mathrm{ms}^{-2}$$
Using the acceleration equation:
$$a = -A \omega^2 \cos(\omega t + \phi)$$
Since weÃ¢ÂÂre given the magnitude of the acceleration, we remove the negative sign:
$$16 = A \omega^2 \cos(\omega t + \phi)$$
Using the position equation:
$$x = A \cos(\omega t + \phi)$$
We already know $$x = 4 \, \mathrm{m}$$, so:
$$4 = A \cos(\omega t + \phi)$$
From these two equations, we know:
$$A \omega^2 \cos(\omega t + \phi) = 16$$
$$A \cos(\omega t + \phi) = 4$$
Therefore:
$$A \omega^2 \cdot 4/A = 16$$
$$4 \omega^2 = 16$$
$$\omega^2 = 4$$
$$\omega = 2 \, \mathrm{rad/s}$$
Next, using the velocity equation:
$$v = -A \omega \sin(\omega t + \phi)$$
Again, we consider the magnitude:
$$2 = A \cdot 2 \sin(\omega t + \phi)$$
$$2 = 2A \sin(\omega t + \phi)$$
$$\sin(\omega t + \phi) = \dfrac{1}{A}$$
We know from the position equation that:
$$\cos(\omega t + \phi) = \dfrac{4}{A}$$
Using the identity $$\sin^2(	heta) + \cos^2(	heta) = 1$$, we get:
$$\left(\dfrac{1}{A}ight)^2 + \left(\dfrac{4}{A}ight)^2 = 1$$
$$\dfrac{1}{A^2} + \dfrac{16}{A^2} = 1$$
$$\dfrac{17}{A^2} = 1$$
$$A^2 = 17$$
$$A = \sqrt{17} \, \mathrm{m}$$
Therefore, the amplitude of the motion is $$\sqrt{17} \, \mathrm{m}$$, meaning $$x$$ is 17.

Oscillations

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 \mathrm{~m}, 2 \mathrm{~ms}^{-1} and 16 \mathrm{~ms}^{-2} at a certain instant. The amplitude of the motion is \sqrt{x}, \mathrm{~m} where x$ is _________.

Choose the correct answer: