A particle of mass $0.50 \mathrm{~kg}$ executes simple harmonic motion under force $F=-50(\mathrm{Nm}^{-1}) x$. The time period of oscillation is $\frac{x}{35} s$. The value of $x$ is _________.

(Given $\pi=\frac{22}{7}$)

<p>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:</p> <p>$T = 2\pi \sqrt{\frac{m}{k}}$</p> <p>Here,</p> <p>$m$ is the mass of the particle, which is $0.50 \, \mathrm{kg}$ in this case,</p> <p>$k$ is the force constant of the spring or the spring constant, which is given as $50 \, \mathrm{Nm^{-1}}$,</p> <p>and $T$ represents the time period of oscillation.</p> <p>Given in the problem, $T = \frac{x}{35} \, \mathrm{s}$ and we are provided with the approximation $\pi = \frac{22}{7}$.</p> <p>Substituting the given values into the formula for $T$:</p> <p>$\frac{x}{35} = 2 \times \frac{22}{7} \times \sqrt{\frac{0.50}{50}}$</p> <p>To simplify this, we first calculate the square root:</p> <p>$\sqrt{\frac{0.50}{50}} = \sqrt{\frac{1}{100}} = \frac{1}{10}$</p> <p>Substituting back, we get:</p> <p>$\frac{x}{35} = 2 \times \frac{22}{7} \times \frac{1}{10}$</p> <p>Multiplying the terms on the right side:</p> <p>$\frac{x}{35} = \frac{44}{70}$</p> <p>$\frac{x}{35} = \frac{22}{35}$</p> <p>Multiplying both sides by $35$ to solve for $x$:</p> <p>$x = 22$</p> <p>Therefore, the value of $x$ that represents the time period of oscillation is $22$ seconds.</p>