QJEEM24Phf05f: JEE Main 2024 Physics - Class 11 JEEM24Phf05f Solution: The position, velocity and acceleration of a particle executing …

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 we’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 $\sin^2(\theta) + \cos^2(\theta) = 1$ , we get:

$\left(\dfrac{1}{A}\right)^2 + \left(\dfrac{4}{A}\right)^2 = 1$

$\dfrac{1}{A^2} + \dfrac{16}{A^2} = 1$

$\dfrac{17}{A^2} = 1$

$A^2 = 17$

$A = \sqrt{17} \, \mathrm{m}$

Therefore, the amplitude of the motion is $\sqrt{17} \, \mathrm{m}$ , meaning $x$ is 17.