/*! 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 37 In a playground, a wooden horse ... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 37

In a playground, a wooden horse is attached to the ground by a stiff spring. When a 24 -kg child sits on the horse, the spring compresses by $28 \mathrm{cm} .$ With the child sitting on the horse, the spring oscillates up and down with a frequency of \(0.88 \mathrm{Hz}\). What is the oscillation frequency of the spring when no one is sitting on the horse?

Short Answer

Expert verified
Answer: 3.05 Hz

Step by step solution

01

Convert displacement to meters

As the displacement is given in centimeters, we need to convert it to meters, which is the SI unit for distance. \(x = 28 \;\mathrm{cm} = 0.28 \;\mathrm{m}\)
02

Calculate the weight of the child

We need to find the force acting on the horse due to the child. This force is equal to the weight of the child, which can be calculated using \(F = mg\) (\(m=\text{mass of the child}=24\,\mathrm{kg}\), \(g=\text{gravitational acceleration}=9.81\,\mathrm{m/s^2}\)). \(F = (24 \;\mathrm{kg})(9.81 \;\mathrm{m/s^2}) = 235.44 \;\mathrm{N}\)
03

Calculate the spring constant (k) using Hooke's Law

Hooke's Law states that the force acting on a spring is proportional to the displacement: \(F = kx\). We can rearrange the formula to solve for \(k\): \(k = \frac{F}{x}\). \(k = \frac{235.44 \;\mathrm{N}}{0.28 \;\mathrm{m}} = 841.57 \;\mathrm{N/m}\)
04

Calculate the mass when no child is sitting on the horse

Using the given frequency with the child sitting on the horse, we can calculate the mass of the wooden horse, which we will denote \(M\). The frequency formula for a mass on the spring is \(f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\), where \(m = M + m\). To find the oscillation frequency with no child on it, first we need to find the mass of the wooden horse \(M\). We are given the frequency with the child on it: \(f = 0.88 \; \mathrm{Hz}\). We can rearrange the frequency formula to solve for \(M\): \(M = \frac{k}{(\frac{2\pi}{f})^2} - m = \frac{841.57 \;\mathrm{N/m}}{(\frac{2\pi}{0.88 \;\mathrm{Hz}})^2} - 24 \;\mathrm{kg} = 5.73 \;\mathrm{kg}\)
05

Calculate the oscillation frequency with no child on the horse

Now that we have the mass of the wooden horse, we can find the oscillation frequency when no one is sitting on it by substituting the mass \(M\) into the frequency formula: \(f_{\text{no child}} = \frac{1}{2\pi} \sqrt{\frac{k}{M}} = \frac{1}{2\pi} \sqrt{\frac{841.57 \;\mathrm{N/m}}{5.73 \;\mathrm{kg}}} = 3.05 \;\mathrm{Hz}\) The oscillation frequency of the spring when no one is sitting on the horse is approximately \(3.05 \;\mathrm{Hz}\).

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