Question

When a body of unknown mass is attached to an ideal spring with force constant 120 N/m, it is found to vibrate with a frequency of 6.00 Hz. Find (a) the period of the motion; (b) the angular frequency; (c) the mass of the body.

Short answer

Period: 0.1667 seconds; Angular Frequency: ~37.70 rad/s; Mass: ~0.0845 kg.

Step-by-step solution

Understanding the Relationship Between Frequency and Period

To find the period of the motion, use the relationship between frequency \( f \) and period \( T \), which is given by the formula \( T = \frac{1}{f} \). Given the frequency \( f = 6.00 \text{ Hz} \), substitute it into the formula to calculate the period.

Calculating the Period

Using the formula \( T = \frac{1}{f} \), substitute \( f = 6.00 \text{ Hz} \):\[ T = \frac{1}{6.00} \text{ } \text{seconds} = 0.1667 \text{ } \text{seconds} \] The period of the motion is 0.1667 seconds.

Understanding the Relationship for Angular Frequency

Angular frequency \( \omega \) is related to frequency \( f \) by the formula \( \omega = 2\pi f \). To find the angular frequency, substitute the given frequency into the formula.

Calculating the Angular Frequency

Using \( \omega = 2\pi f \) with \( f = 6.00 \text{ Hz} \):\[ \omega = 2 \pi \times 6.00 \text{ Hz} = 12\pi \text{ } \text{rad/s} \approx 37.70 \text{ } \text{rad/s} \] The angular frequency is approximately 37.70 rad/s.

Understanding the Mass from Angular Frequency and Spring Constant

The relationship between mass \( m \), angular frequency \( \omega \), and spring constant \( k \) is given by \( \omega = \sqrt{\frac{k}{m}} \). Rearrange this formula to solve for the mass \( m \).

Calculating the Mass of the Body

Start with the formula \( \omega = \sqrt{\frac{k}{m}} \). Square both sides to remove the square root:\[ \omega^2 = \frac{k}{m} \]Solve for \( m \): \[ m = \frac{k}{\omega^2} \]Substitute the given values \( k = 120 \text{ N/m} \) and \( \omega = 37.70 \text{ rad/s} \):\[ m = \frac{120}{(37.70)^2} \approx 0.0845 \text{ kg} \]The mass of the body is approximately 0.0845 kg.

Key concepts

Spring Constant

Understanding the concept of a spring constant is essential when dealing with simple harmonic motion. The spring constant, denoted as \( k \), measures the stiffness of a spring. It tells us how much force is needed to compress or extend a spring by a unit distance. The formula is derived from Hooke's Law, which states \( F = kx \), where \( F \) is the force applied, and \( x \) is the displacement from the equilibrium position.

In this context, a spring with a constant of 120 N/m requires 120 Newtons of force to displace it by 1 meter. This value helps us calculate how a mass attached to the spring would behave, such as determining the angular frequency and the mass itself. Springs with higher spring constants are stiffer and resist deformation more than springs with lower constants, affecting the dynamics of motion.

Angular Frequency

Angular frequency \( \omega \) is a key concept in oscillatory motion. It represents how rapidly an object moves through its cycle in radians per second. You can think of it as the rotational equivalent of linear frequency. The formula to calculate angular frequency is \( \omega = 2\pi f \), where \( f \) is the linear frequency in hertz.

For a system vibrating at 6.00 Hz, the angular frequency is 12\( \pi \) or approximately 37.70 rad/s. This measurement gives you insight into how quickly the system oscillates, which is crucial for understanding the energy and dynamics within the system. Angular frequency is particularly important because it is used in other calculations, such as finding the mass of the body involved in harmonic motion.

Mass Calculation

Calculating the mass of an object in simple harmonic motion involves understanding how tightly it is linked to its spring and how fast it oscillates. The mass \( m \) can be figured out using the rearranged formula:

• \( \omega = \sqrt{\frac{k}{m}} \)
• \( m = \frac{k}{\omega^2} \)

By substituting the spring constant \( k = 120 \text{ N/m} \) and the calculated angular frequency \( \omega = 37.70 \text{ rad/s} \), you find \( m = \frac{120}{37.70^2} \approx 0.0845 \text{ kg} \).

This calculation tells us how much mass is causing the system to oscillate at the observed frequency. It is crucial for predictions regarding system behavior and energy distribution, particularly in engineering contexts where precise motion is crucial.

Period of Motion

The period of motion \( T \) is the time it takes for a system to complete one full cycle of motion and return to its initial state. It is inversely related to frequency by the simple formula \( T = \frac{1}{f} \). If a system vibrates at 6.00 Hz, the period \( T \) would be \( \frac{1}{6.00} \approx 0.1667 \) seconds.

Knowing the period helps in understanding the timing characteristics and stability of oscillatory systems. It's like knowing how long it takes for a pendulum to swing back and forth. This value can be critical for applications that hinge on periodic timing, like clocks or even machinery that cycles through repetitive motions.