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A person bounces up and down on a trampoline, while always staying in contact with it. The motion is simple harmonic motion, and it takes \(1.90 \mathrm{s}\) to complete one cycle. The height of each bounce above the equilibrium position is \(45.0 \mathrm{cm}\). Determine (a) the amplitude and (b) the angular frequency of the motion. (c) What is the maximum speed attained by the person?

Short Answer

Expert verified
(a) Amplitude is 45.0 cm. (b) Angular frequency is 3.31 rad/s. (c) Maximum speed is 1.49 m/s.

Step by step solution

01

Understanding the problem

We need to determine the amplitude, angular frequency, and maximum speed of a person bouncing on a trampoline. The motion is described as simple harmonic with a given period and height of bounce.
02

Determine the Amplitude

In simple harmonic motion, the amplitude is the maximum displacement from the equilibrium position. Given the height of each bounce above the equilibrium position is 45.0 cm, the amplitude is simply this height.Thus, the amplitude, \(A\), is 45.0 cm.
03

Calculate the Angular Frequency

The period of the motion, \(T\), is given as 1.90 seconds. The angular frequency, \(\omega\), is related to the period by the formula:\[ \omega = \frac{2\pi}{T} \]Substitute the value of \(T\) to find \(\omega\):\[ \omega = \frac{2\pi}{1.90} \approx 3.31 \text{ rad/s} \]
04

Calculate Maximum Speed

The maximum speed, \(v_{max}\), in simple harmonic motion can be calculated using the formula:\[ v_{max} = A \cdot \omega \]Substitute the values of \(A\) (converted into meters, 0.45 m) and \(\omega\):\[ v_{max} = 0.45 \times 3.31 \approx 1.49 \text{ m/s} \]

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Key Concepts

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

Amplitude
In simple harmonic motion, the amplitude refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium. For a person bouncing up and down on a trampoline, the amplitude is simply the maximum height they reach above the equilibrium position. In other words, it's the furthest point up from where they start mid-bounce. For our specific exercise, this height or amplitude is given as 45.0 cm. This measurement tells us how far the person moves away from the resting position, each time they complete a bounce.
Understanding amplitude is pivotal because it gives us insight into the energy contained within the oscillating system. The greater the amplitude, the higher the energy involved. In many practical scenarios, such as designing a trampoline or measuring vibrations in machinery, knowing the amplitude helps us gauge the amount of work done by the system.
Keep in mind:
  • Amplitude is always a positive value.
  • It's a measure of distance, either centimeters or meters in most problems.
  • It directly affects the maximum potential energy of the system.
Angular Frequency
Angular frequency is a measure of how quickly an object is oscillating in simple harmonic motion. It is denoted by the symbol \( \omega \) (omega) and is expressed in radians per second. Angular frequency gives us the rate at which the oscillation progresses through its cycle and is closely related to the period of the motion.
For example, in our exercise, the angular frequency is calculated using the formula:\[ \omega = \frac{2\pi}{T} \]where \( T \) is the period of the motion. With a period \( T \) of 1.90 seconds, the angular frequency comes out to approximately 3.31 rad/s.
Angular frequency is important because:
  • It indicates how many cycles occur in a given time span.
  • It can affect the kinetic and potential energies within a system.
  • Higher angular frequency means faster oscillations.
By understanding angular frequency, scientists and engineers can predict how systems will behave over time, which is essential for fields ranging from acoustics to electrical engineering.
Maximum Speed
In the context of simple harmonic motion, maximum speed is a crucial parameter that indicates the highest speed an object reaches as it vibrates or oscillates. This maximum speed is achieved when the object passes through its equilibrium position, where the kinetic energy is at its peak, and potential energy is zero.
To calculate the maximum speed, we use the formula:\[ v_{max} = A \cdot \omega \]where \( A \) is the amplitude, and \( \omega \) is the angular frequency. In our example, with \( A \) as 0.45 m and \( \omega \) as 3.31 rad/s, the maximum speed \( v_{max} \) calculates to approximately 1.49 m/s.
Understanding maximum speed helps us determine:
  • How vigorous the motion is — higher speeds are usually associated with more energetic systems.
  • The safety parameters needed for structures or systems undergoing such oscillations.
  • The relationship between kinetic and potential energy during the oscillation cycle.
Knowing the maximum speed can assist engineers in designing safety measures in systems that rely on or are affected by oscillations, ensuring stability and safety.

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Most popular questions from this chapter

A 0.60-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 to 0.23 m (relative to its unstrained length), the speed of the sphere decreases from 5.70 to 4.80 m/s. What is the spring constant of the spring?

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 to \(3.0 \times 10^{-2} \mathrm{N}\). The length and radius of the collagen are, respectively, 2.5 and \(0.091 \mathrm{cm},\) and Young's modulus is \(3.1 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}\). (a) If the stretching obeys Hooke's law, what is the spring constant \(k\) for collagen? (b) How much work is done by the variable force that stretches the collagen? (See Section 6.9 for a discussion of the work done by a variable force.)

A heavy-duty stapling gun uses a \(0.140-\mathrm{kg}\) metal rod that rams against the staple to eject it. The rod is attached to and pushed by a stiff spring called a "ram spring" \((k=32000 \mathrm{N} / \mathrm{m})\). The mass of this spring may be ignored. The ram spring is compressed by \(3.0 \times 10^{-2} \mathrm{m}\) from its unstrained length and then released from rest. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{m}\) when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 59 N/m and 33 N/m. Find the angle that the rod makes with the horizontal.

A vertical spring (spring constant 5 112 N/m) is mounted on the fl oor. A 0.400-kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in Hz) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.

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