QJEEM24Ph9d89: JEE Main 2024 Physics - Class 11 JEEM24Ph9d89 Solution: A particle of mass executes simple harmonic motion under force…

To find the value of $x$ that represents the time period of oscillation in this simple harmonic motion (SHM) scenario, we first recall the general formula for the time period ( $T$ ) of a mass-spring system undergoing SHM, which is given by:

$T = 2\pi \sqrt{\frac{m}{k}}$

Here,

$m$ is the mass of the particle, which is $0.50 \, \mathrm{kg}$ in this case,

$k$ is the force constant of the spring or the spring constant, which is given as $50 \, \mathrm{Nm^{-1}}$ ,

and $T$ represents the time period of oscillation.

Given in the problem, $T = \frac{x}{35} \, \mathrm{s}$ and we are provided with the approximation $\pi = \frac{22}{7}$ .

Substituting the given values into the formula for $T$ :

$\frac{x}{35} = 2 \times \frac{22}{7} \times \sqrt{\frac{0.50}{50}}$

To simplify this, we first calculate the square root:

$\sqrt{\frac{0.50}{50}} = \sqrt{\frac{1}{100}} = \frac{1}{10}$

Substituting back, we get:

$\frac{x}{35} = 2 \times \frac{22}{7} \times \frac{1}{10}$

Multiplying the terms on the right side:

$\frac{x}{35} = \frac{44}{70}$

$\frac{x}{35} = \frac{22}{35}$

Multiplying both sides by $35$ to solve for $x$ :

$x = 22$

Therefore, the value of $x$ that represents the time period of oscillation is $22$ seconds.