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Chapter 10: Problem 60

A clock has a pendulum that performs one full swing every \(1.0 \mathrm{s}(\) back and forth). The object at the end of the pendulum weighs $10.0 \mathrm{N}$. What is the length of the pendulum?

Short Answer

Expert verified
Answer: The length of the pendulum is approximately 0.248 m.

Step by step solution

01

Identify the given values

We are given the period of the pendulum \(T = 1.0 \mathrm{s}\) and the weight of the object at the end \(W = 10.0 \mathrm{N}\).
02

Recall the formula for the period of a simple pendulum

The formula for the period of a simple pendulum is: \(T = 2\pi\sqrt{\frac{l}{g}}\), where \(T\) is the period, \(l\) is the length, and \(g\) is the acceleration due to gravity.
03

Solve the formula for the length

We need to find the length of the pendulum, so we need to rearrange the formula to solve for \(l\). By squaring both sides of the equation and isolating \(l\), we get: \(l = \frac{gT^2}{4\pi^2}\).
04

Calculate the acceleration due to gravity

Knowing that the weight of the object is \(10.0 \mathrm{N}\), we can use Newton's second law to find the mass of the object: \(W = mg\) where \(W\) is the weight, \(m\) is the mass, and \(g\) is the acceleration due to gravity. In this case, \(g\) is approximately \(9.81 \mathrm{m/s^2}\).
05

Solve for the length

Now we can plug the values of \(T\) and \(g\) into the formula for the length and find the length of the pendulum: \(l = \frac{(9.81 \mathrm{m/s^2})(1.0 \mathrm{s})^2}{4\pi^2} \approx 0.248 \mathrm{m}\).
06

Provide the answer

The length of the pendulum is approximately \(0.248 \mathrm{m}\).

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Most popular questions from this chapter

Equipment to be used in airplanes or spacecraft is often subjected to a shake test to be sure it can withstand the vibrations that may be encountered during flight. A radio receiver of mass \(5.24 \mathrm{kg}\) is set on a platform that vibrates in SHM at \(120 \mathrm{Hz}\) and with a maximum acceleration of $98 \mathrm{m} / \mathrm{s}^{2}(=10 \mathrm{g}) .$ Find the radio's (a) maximum displacement, (b) maximum speed, and (c) the maximum net force exerted on it.
The diaphragm of a speaker has a mass of \(50.0 \mathrm{g}\) and responds to a signal of frequency \(2.0 \mathrm{kHz}\) by moving back and forth with an amplitude of \(1.8 \times 10^{-4} \mathrm{m}\) at that frequency. (a) What is the maximum force acting on the diaphragm? (b) What is the mechanical energy of the diaphragm?
A marble column with a cross-sectional area of \(25 \mathrm{cm}^{2}\) supports a load of \(7.0 \times 10^{4} \mathrm{N} .\) The marble has a Young's modulus of \(6.0 \times 10^{10} \mathrm{Pa}\) and a compressive strength of $2.0 \times 10^{8} \mathrm{Pa} .$ (a) What is the stress in the column? (b) What is the strain in the column? (c) If the column is \(2.0 \mathrm{m}\) high, how much is its length changed by supporting the load? (d) What is the maximum weight the column can support?
A pendulum is made from a uniform rod of mass \(m_{1}\) and a small block of mass \(m_{2}\) attached at the lower end. (a) If the length of the pendulum is \(L\) and the oscillations are small, find the period of the oscillations in terms of \(m_{1}, m_{2}, L,\) and \(g .\) (b) Check your answer to part (a) in the two special cases \(m_{1} \gg m_{2}\) and \(m_{1}<<m_{2}\).
The gravitational potential energy of a pendulum is \(U=m g y .\) (a) Taking \(y=0\) at the lowest point, show that \(y=L(1-\cos \theta),\) where \(\theta\) is the angle the string makes with the vertical. (b) If \(\theta\) is small, \((1-\cos \theta)=\frac{1}{2} \theta^{2}\) and \(\theta=x / L\) (Appendix A.7). Show that the potential energy can be written \(U=\frac{1}{2} k x^{2}\) and find the value of \(k\) (the equivalent of the spring constant for the pendulum).
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