/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Estimate the stiffness of the sp... [FREE SOLUTION] | 91Ó°ÊÓ

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Estimate the stiffness of the spring in a child's pogo stick if the child has a mass of \(35 \mathrm{~kg}\) and bounces once every 2.0 seconds.

Short Answer

Expert verified
The spring constant is approximately 345.45 N/m.

Step by step solution

01

Understand the Problem

The exercise asks us to estimate the stiffness of a spring in a pogo stick. We are given the mass of the child and the time period for one bounce. The relationship we'll be using is for oscillations of a mass-spring system, where the time period (T) is related to mass (m) and spring constant (k).
02

Use the Formula for Period of Oscillation

The time period (T) of a simple spring-mass system is given by \( T = 2\pi\sqrt{\frac{m}{k}} \), where \( m \) is the mass and \( k \) is the spring constant. We will use this formula to solve for the spring constant \( k \).
03

Rearrange the Formula to Solve for k

To find \( k \), rearrange the formula: \( T = 2\pi\sqrt{\frac{m}{k}} \) becomes \( k = \frac{4\pi^2 m}{T^2} \).
04

Substitute Known Values into the Equation

Substitute the mass \( m = 35 \mathrm{~kg} \) and the period \( T = 2 \mathrm{~s} \) into the equation: \( k = \frac{4\pi^2 \times 35}{4} \).
05

Perform the Calculation

Calculate \( \pi^2 \approx 9.87 \). Then, compute \( k = \frac{4 \times 9.87 \times 35}{4} = 345.45 \mathrm{~N/m} \).
06

Conclusion

The spring constant, which is the stiffness of the pogo stick's spring, is approximately \( 345.45 \mathrm{~N/m} \).

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Key Concepts

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

Mass-Spring System
A mass-spring system is a classic example in physics to illustrate simple harmonic motion. This system involves a mass attached to a spring, which can stretch or compress. The system's behavior is governed by Hooke's law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance. This relationship is expressed in the formula:
  • Force (F) = -k * x
where:
  • k is the spring constant, a measure of stiffness
  • x is the displacement from the equilibrium position
The mass-spring system is foundational because it helps us understand how oscillatory motion occurs, not only in springs but in various real-world applications. Its simplicity yet profound implications make it an excellent starting point for exploring more complex systems.
Oscillations
Oscillations refer to any repetitive variation, typically in time, of some measure about a central value or between two or more different states. In the context of a mass-spring system, oscillations occur when the mass attached to the spring moves back and forth periodically after being displaced from its equilibrium position. This motion is typically sinusoidal and can be described using trigonometric functions. The key features of oscillations include:
  • Amplitude: The maximum extent of the oscillation measured from the equilibrium position.
  • Frequency: How many oscillations occur in a unit of time.
  • Time period: The time taken for one complete cycle of oscillation.
Understanding oscillations helps in predicting how systems behave over time, making it vital for designing systems ranging from simple toys to complex engineering mechanisms.
Period of Oscillation
The period of oscillation is the time it takes for an oscillating system to complete one full cycle of motion. For a mass-spring system, the period (T) is determined by the mass of the object and the spring constant, according to the formula:\[ T = 2\pi\sqrt{\frac{m}{k}} \]Here:
  • T is the period of oscillation.
  • m is the mass of the object attached to the spring.
  • k is the spring constant, indicating the spring's stiffness.
The period of oscillation is crucial because it characterizes the "speed" of the oscillation. In practical terms, it allows us to predict the motion of the system without external influences, providing valuable insights in various scientific and engineering fields.
Pogo Stick
A pogo stick is a fun and practical application of the mass-spring system. It is a device that allows a person to jump off the ground by utilizing the elastic potential energy stored in a spring. When you push down on the pogo stick, the spring inside compresses and stores energy. Upon release, this energy is converted to kinetic energy, propelling the user upwards. Key components of a pogo stick:
  • Spring: The main component that determines how high the user bounces.
  • Foot rests: Where the user stands while bouncing.
  • Shaft: Connects the foot rests to the handles, providing support and structure.
The spring constant of the pogo stick's spring is a crucial parameter. It dictates how stiff the spring is and, consequently, how much energy can be stored and released during each bounce. Understanding this helps in both designing pogo sticks and estimating their performance for users of different weights, ensuring safety and maximum fun.

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Most popular questions from this chapter

A physical pendulum consists of an \(85-\mathrm{cm}\) -long, 240 -g-mass, uniform wooden rod hung from a nail near one end (Fig. 14-38). The motion is damped because of friction in The motion is damped because of friction in the pivot; the damping force is approximately proportional to \(d \theta / d t\). The rod is set in oscillation by displacing it \(15^{\circ}\) from its equilibrium position and releasing it. After \(8.0 \mathrm{~s}\) the amplitude of the oscillation has been reduced to \(5.5^{\circ} .\) If the angular displacement can be written as \(\theta=A e^{-\gamma t} \cos \omega^{\prime} t,\) find (a) \(\gamma,\) (b) the approximate period of the motion, and \((c)\) how long it takes for the amplitude to be reduced to \(\frac{1}{2}\) of its original value.

(II) An 1150 kg automobile has springs with \(k=16,000 \mathrm{N} / \mathrm{m}\) . One of the tires is not properly balanced; it has a little extra mass on one side compared to the other, causing the car to shake at certain speeds. If the tire radius is \(42 \mathrm{cm},\) at what speed will the wheel shake most?

(II) A tuning fork oscillates at a frequency of 441 \(\mathrm{Hz}\) and the tip of each prong moves 1.5 \(\mathrm{mm}\) to either side of center. Calculate \((a)\) the maximum speed and \((b)\) the maximum acceleration of the tip of a prong.

(II) A 1.60 -kg object oscillates from a vertically hanging light spring once every 0.55 s. (a) Write down the equation giving its position \(y(+\) upward) as a function of time \(t,\) assuming it started by being compressed 16 \(\mathrm{cm}\) from the equilibrium position (where \(y=0 )\) , and released. \((b)\) How long will it take to get to the equilibrium position for the first time? (c) What will be its maximum speed? (d) What will be its maximum acceleration, and where will it first be attained?

(II) Estimate the stiffness of the spring in a child's pogo stick if the child has a mass of 35 \(\mathrm{kg}\) and bounces once every 2.0 seconds.

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