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

To solve the problem, we must analyze the simple harmonic motion (SHM) of the 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 \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase angle.
Given are:

• Position \( x = 4 \, \text{m} \)
• Velocity \( v = 2 \, \text{ms}^{-1} \)
• Acceleration \( a = 16 \, \text{ms}^{-2} \)

From \( a = -A\omega^2 \cos(\omega t + \phi) = -\omega^2 x \), we have: \[ 16 = \omega^2 \times 4 \] Solving for \( \omega^2 \): \[ \omega^2 = \frac{16}{4} = 4 \Rightarrow \omega = 2 \] Substituting into the velocity equation \( v = -A\omega \sin(\omega t + \phi) \), we have: \[ 2 = -A \cdot 2 \cdot \sin(\omega t + \phi) \Rightarrow \sin(\omega t + \phi) = -\frac{1}{2} \] We now find \( A^2 \) using the identity for SHM: \[ A^2 = x^2 + \left(\frac{v^2}{\omega^2}\right) \] Substitute known values: \[ A^2 = 4^2 + \left(\frac{2^2}{4}\right) = 16 + 1 = 17 \] Thus, the amplitude squared, \( A^2 = x = 17 \), ensuring it matches the required range. Finally, \( A = \sqrt{17} \, \text{m} \).

Answer 2

The given data is:

x = 4 m, v = 2 m/s, a = 16 m/s²
For a particle in Simple Harmonic Motion (SHM), the equations for position, velocity, and acceleration are:

x = A cos ωt,
v = Aω sin ωt,
a = -Aω² cos ωt

Step 1: Using the relation between acceleration and position

The acceleration is given by:

a = -ω²x
Substitute a = 16 m/s² and x = 4 m:

16 = ω² ⋅ 4 ⟹ ω² = 4
Thus: ω = 2 rad/s

Step 2: Using the relation between velocity and amplitude

The velocity equation in SHM is:

v² = ω² (A² − x²)
Substitute v = 2 m/s, ω = 2 rad/s, x = 4 m:

2² = 2² (A² − 4²)
4 = 4 (A² − 16) ⟹ A² − 16 = 1
A² = 17

Step 3: Amplitude

The amplitude of the motion is:

A = \(\sqrt{17}\) m
Thus, x = 17.