Question

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 \( \dots \dots \dots \dots \). \[ \text{(Given } \pi = \frac{22}{7} \text{)} \]

Answer 1

Correct Answer: 22

For a particle undergoing simple harmonic motion with mass \( m = 0.50 \, \text{kg} \) and subjected to a restoring force \( F = -50 \, (\text{N m}^{-1}) x \), we need to determine the value of \( x \) from the given time period expression \( \frac{x}{35} \, \text{s} \). The angular frequency \( \omega \) in simple harmonic motion is related to the restoring force by \( F = -kx \), where \( k = 50 \, \text{N m}^{-1} \). The formula for angular frequency is \( \omega = \sqrt{\frac{k}{m}} \).

Substituting the provided values yields: \[ \omega = \sqrt{\frac{50}{0.50}} = \sqrt{100} = 10 \, \text{rad/s}. \]

The time period \( T \) is calculated using the formula: \[ T = \frac{2\pi}{\omega}. \]

With \( \omega = 10 \) and \( \pi = \frac{22}{7} \), the time period is: \[ T = \frac{2 \times \frac{22}{7}}{10} = \frac{44}{70} = \frac{22}{35} \, \text{s}. \]

Given that \( T = \frac{x}{35} \, \text{s} \), we equate this to our calculated value: \[ \frac{x}{35} = \frac{22}{35}. \]

Solving for \( x \) by clearing fractions: \[ x = 22. \]

The determined value of \( x \) is 22, which falls within the specified range of 22,22.