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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

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

Step by step solution

01

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.
02

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.
03

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.
04

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.
05

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 \( m \).
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

At the end of a ride at a winter-theme amusement park, a sleigh with mass 250 kg (including two passengers) slides without friction along a horizontal, snow-covered surface. The sleigh hits one end of a light horizontal spring that obeys Hooke's law and has its other end attached to a wall. The sleigh latches onto the end of the spring and subsequently moves back and forth in SHM on the end of the spring until a braking mechanism is engaged, which brings the sleigh to rest. The frequency of the SHM is 0.225 Hz, and the amplitude is 0.950 m. (a) What was the speed of the sleigh just before it hit the end of the spring? (b) What is the maximum magnitude of the sleigh's acceleration during its SHM?

A 0.0200-kg bolt moves with SHM that has an amplitude of 0.240 m and a period of 1.500 s. The displacement of the bolt is \(+\)0.240 m when \(t =\) 0. Compute (a) the displacement of the bolt when \(t =\) 0.500 s; (b) the magnitude and direction of the force acting on the bolt when \(t =\) 0.500 s; (c) the minimum time required for the bolt to move from its initial position to the point where \(x = -\)0.180 m; (d) the speed of the bolt when \(x = -\)0.180 m.

A large, 34.0-kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of mass is 0.60 m below the pivot. The bell's moment of inertia about an axis at the pivot is 18.0 kg \(\cdot\) m\(^2\). The clapper is a small, 1.8-kg mass attached to one end of a slender rod of length \(L\) and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length \(L\) of the clapper rod for the bell to ring silently\(-\)that is, for the period of oscillation for the bell to equal that of the clapper?

A certain simple pendulum has a period on the earth of 1.60 s. What is its period on the surface of Mars, where \(g =\) 3.71 m/s\(^2\)?

A 2.00-kg frictionless block is attached to an ideal spring with force constant 315 N/m. Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude of the motion, (b) the block's maximum acceleration, and (c) the maximum force the spring exerts on the block.

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