Question

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

Answer 1

Correct Answer: 17

The problem is solved by analyzing the simple harmonic motion (SHM) of a particle, given its position, velocity, and acceleration magnitudes. The general equations for SHM are:

Position: \( x(t) = A \cos(\omega t + \phi) \)

Velocity: \( v(t) = -A\omega \sin(\omega t + \phi) \)

Acceleration: \( a(t) = -A\omega^2 \cos(\omega t + \phi) \)

Where \( A \) represents the amplitude, \( \omega \) is the angular frequency, and \( \phi \) denotes the phase angle.
The given values are:

Position \( x = 4 \, \text{m} \)

Velocity \( v = 2 \, \text{ms}^{-1} \)

Acceleration \( a = 16 \, \text{ms}^{-2} \)

From the relationship \( a = -A\omega^2 \cos(\omega t + \phi) = -\omega^2 x \), we derive:

\[ 16 = \omega^2 \times 4 \] Solving for \( \omega^2 \): \[ \omega^2 = \frac{16}{4} = 4 \Rightarrow \omega = 2 \] Substituting this into the velocity equation \( v = -A\omega \sin(\omega t + \phi) \): \[ 2 = -A \cdot 2 \cdot \sin(\omega t + \phi) \Rightarrow \sin(\omega t + \phi) = -\frac{1}{2} \] The amplitude squared, \( A^2 \), is determined using the SHM identity:

\[ A^2 = x^2 + \left(\frac{v^2}{\omega^2}\right) \] Substituting the known values:

 

\[ A^2 = 4^2 + \left(\frac{2^2}{4}\right) = 16 + 1 = 17 \] Consequently, the amplitude squared is \( A^2 = 17 \). The amplitude is therefore \( A = \sqrt{17} \, \text{m} \).