/*! 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 58 A bob of mass \(m\) is suspended... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 58

A bob of mass \(m\) is suspended from a string of length \(L\) forming a pendulum. The period of this pendulum is \(2.0 \mathrm{s}\) If the pendulum bob is replaced with one of mass \(\frac{1}{3} m\) and the length of the pendulum is increased to \(2 L\), what is the period of oscillation?

Short Answer

Expert verified
Answer: The period of oscillation for the pendulum with the new length and mass is approximately \(2.8314 \mathrm{s}\).

Step by step solution

01

Recall the formula for the period of a pendulum

The formula for the period of a simple pendulum is given by \(T = 2\pi\sqrt{\frac{L}{g}}\), where \(T\) is the period, \(L\) is the length of the pendulum, and \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{m/s^2}\)). Note that the mass of the pendulum bob does not affect the period.
02

Calculate the initial value of \(L\)

We are given the period of the initial pendulum (\(T_1 = 2.0 \mathrm{s}\)). We can use this to find the length of the pendulum, \(L\). Rearrange the formula to make \(L\) the subject: \(L = \frac{gT^2}{4\pi^2}\). Now, plug in the values: \(L = \frac{9.81 \cdot 2^2}{4 \cdot \pi^2} \approx 0.9934 \mathrm{m}\).
03

Calculate the new length and mass

We are told that the length of the pendulum is increased to \(2L\), and the mass of the bob is changed to \(\frac{1}{3} m\). However, since the mass does not affect the period, we don't need to consider it in our calculation. The new length is \(L_2 = 2L = 2(0.9934) = 1.9868 \mathrm{m}\).
04

Calculate the new period

Now, we can find the new period \(T_2\) using the new length \(L_2\). Substitute the new length into the formula: \(T_2 = 2\pi\sqrt{\frac{L_2}{g}} = 2\pi\sqrt{\frac{1.9868}{9.81}} \approx 2.8314 \mathrm{s}\).
05

Answer

The period of oscillation for the pendulum with the new length and mass is approximately \(2.8314 \mathrm{s}\).

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Most popular questions from this chapter

A pendulum of length \(L_{1}\) has a period \(T_{1}=0.950 \mathrm{s}\). The length of the pendulum is adjusted to a new value \(L_{2}\) such that $T_{2}=1.00 \mathrm{s} .\( What is the ratio \)L_{2} / L_{1} ?$
A \(230.0-\mathrm{g}\) object on a spring oscillates left to right on a frictionless surface with a frequency of \(2.00 \mathrm{Hz}\). Its position as a function of time is given by \(x=(8.00 \mathrm{cm})\) sin \(\omega t\) (a) Sketch a graph of the elastic potential energy as a function of time. (b) The object's velocity is given by \(v_{x}=\omega(8.00 \mathrm{cm}) \cos \omega t .\) Graph the system's kinetic energy as a function of time. (c) Graph the sum of the kinetic energy and the potential energy as a function of time. (d) Describe qualitatively how your answers would change if the surface weren't frictionless.
(a) Given that the position of an object is \(x(t)=A\) cos \(\omega t\) show that \(v_{x}(t)=-\omega A\) sin \(\omega t .\) [Hint: Draw the velocity vector for point \(P\) in Fig. \(10.17 \mathrm{b}\) and then find its \(x\) component.] (b) Verify that the expressions for \(x(t)\) and \(v_{x}(t)\) are consistent with energy conservation. [Hint: Use the trigonometric identity $\sin ^{2} \omega t+\cos ^{2} \omega t=1.1$.]
A body is suspended vertically from an ideal spring of spring constant $2.5 \mathrm{N} / \mathrm{m}$. The spring is initially in its relaxed position. The body is then released and oscillates about its equilibrium position. The motion is described by $$y=(4.0 \mathrm{cm}) \sin [(0.70 \mathrm{rad} / \mathrm{s}) t]$$ What is the maximum kinetic energy of the body?
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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