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 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.