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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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Most popular questions from this chapter

Luke is trying to catch a pesky animal that keeps eating vegetables from his garden. He is building a trap and needs to use a spring to close the door to his trap. He has a spring in his garage and he wants to determine the spring constant of the spring. To do this, he hangs the spring from the ceiling and measures that it is \(20.0 \mathrm{cm}\) long. Then he hangs a \(1.10-\mathrm{kg}\) brick on the end of the spring and it stretches to $31.0 \mathrm{cm} .$ (a) What is the spring constant of the spring? (b) Luke now pulls the brick \(5.00 \mathrm{cm}\) from the equilibrium position to watch it oscillate. What is the maximum speed of the brick? (c) When the displacement is \(2.50 \mathrm{cm}\) from the equilibrium position, what is the speed of the brick? (d) How long will it take for the brick to oscillate five times?
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