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 ______.
Correct Answer: 17
Approach Solution - 1
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, \( A = \sqrt{17} \, \text{m} \).
Approach Solution -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.
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