A particle of mass \( 0.50 \, \text{kg} \) executes simple harmonic motion under force \( F = -50 \, (\text{N m}^{-1}) x \). The time period of oscillation is \( \frac{x}{35} \, \text{s} \). The value of \( x \) is \( \dots \dots \dots \dots \). \[ \text{(Given } \pi = \frac{22}{7} \text{)} \]
Correct Answer: 22
Approach Solution - 1
Given a particle executing simple harmonic motion with mass \( m = 0.50 \, \text{kg} \) and force \( F = -50 \, (\text{N m}^{-1}) x \), we aim to find the value of \( x \) in the time period expression \( \frac{x}{35} \, \text{s} \). Using the formula for angular frequency \( \omega \) in simple harmonic motion, where the restoring force \( F = -kx \), with \( k = 50 \, \text{N m}^{-1} \), we write \( \omega = \sqrt{\frac{k}{m}} \).
Substitute the given values: \[ \omega = \sqrt{\frac{50}{0.50}} = \sqrt{100} = 10 \, \text{rad/s}. \]
The time period \( T \) is given by the formula: \[ T = \frac{2\pi}{\omega}. \]
Substitute \( \omega = 10 \) and \( \pi = \frac{22}{7} \): \[ T = \frac{2 \times \frac{22}{7}}{10} = \frac{44}{70} = \frac{22}{35} \, \text{s}. \]
Given \( T = \frac{x}{35} \, \text{s} \), equate and solve for \( x \): \[ \frac{x}{35} = \frac{22}{35}. \]
Clear fractions: \[ x = 22. \]
The calculated value of \( x \) is 22, which fits within the expected given range of 22,22.
Approach Solution -2
The force is given by $F = -kx$, so $k = 50 \, \mathrm{Nm}^{-1}$. The mass $m = 0.50 \, \mathrm{kg}$. The time period for simple harmonic motion is:
\[ T = 2\pi \sqrt{\frac{m}{k}}. \]
Substituting $k = 50$ and $m = 0.5$:
\[ T = 2\pi \sqrt{\frac{0.5}{50}} = 2\pi \sqrt{0.01} = 2\pi \times 0.1 = 0.2\pi \, \mathrm{s}. \]
Given $T = \frac{x}{35}$, equating:
\[ 0.2\pi = \frac{x}{35}. \]
Substituting $\pi = \frac{22}{7}$:
\[ 0.2 \times \frac{22}{7} = \frac{x}{35}. \]
Simplifying: \[ x = 0.2 \times 22 \times 5 = 22. \]
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