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Chapter 14: Problem 46

A Pendulum on Mars. A certain simple pendulum has a period on the earth of 1.60 s. What is its period on the surface of Mars, where \(g=3.71 \mathrm{~m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
The period of the pendulum on Mars is 2.57 seconds.

Step by step solution

01

Write out the known values

The given values are: The period of the pendulum on Earth \(T_{earth} = 1.60 s\) and the acceleration due to gravity on Mars \(g_{mars} = 3.71 m/s^2\). The acceleration due to gravity on Earth is \(g_{earth} = 9.81 m/s^2\).
02

Express the period equation for Earth

The formula for the period of a pendulum on Earth is \(T_{earth} = 2\pi \sqrt{\frac{l}{g_{earth}}}\). It can be rearranged as \(l= \frac{T_{earth}^2g_{earth}}{4\pi^2}\).\
03

Express the period equation for Mars

The formula for the period of a pendulum on Mars is \(T_{mars} = 2\pi \sqrt{\frac{l}{g_{mars}}}\). However, since \(l\) is constant, one can substitute \(l\) from step 2, Hence, the formula becomes \(T_{mars}=2\pi \sqrt{\frac{ \frac{T_{earth}^2g_{earth}}{4\pi^{2}}}{g_{mars}}}\).\
04

Simplify the equation for Mars and solve

Simplify the equation to find \(T_{mars}\). This results in \(T_{mars} = T_{earth} \sqrt{\frac{g_{earth}}{g_{mars}}}\). Substituting the known values \(T_{mars} = 1.60s \sqrt{\frac{9.81 m/s^2}{3.71 m/s^2}} = 2.57 s\).

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Key Concepts

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

Simple Pendulum
A simple pendulum is a classic example of a harmonic oscillator, used to demonstrate basic principles of physics such as periodic motion. It consists of a weight, known as a pendulum bob, hanging from a fixed point by a string or rod, which has negligible mass. When the bob is moved to one side and released, it swings back and forth around the equilibrium position in a regular, repeating pattern known as an oscillation.

This type of pendulum demonstrates two important behavior aspects: it is a resonant system with a specific resonance frequency, and it has a period of swing that is independent of the bob's mass. This makes it a powerful tool in various scientific applications, such as timekeeping and measuring gravitational acceleration.
Acceleration Due to Gravity
The acceleration due to gravity, symbolized as 'g', is the rate at which an object accelerates when falling under the sole influence of gravity. This value varies slightly depending on the location and altitude on Earth, averaging about 9.81 m/s虏. When considering other celestial bodies like Mars, 'g' differs due to variations in size, density, and composition. Mars, for example, has a gravitational acceleration of approximately 3.71 m/s虏, which is substantially less than Earth's gravity.

Understanding the acceleration due to gravity is crucial in physics as it affects the motion of objects, the weight force, and plays an integral role in calculations involving gravitational forces.
Period of a Pendulum
The period of a pendulum refers to the time it takes for the pendulum to complete one full swing, from one side to the other and back again. It is one of the most fascinating aspects of a pendulum's motion because, for small angles of swing, the period is actually independent of the amplitude鈥攎eaning, the length of the swing does not affect the time it takes to swing. This property is known as isochronous behavior.

For simple pendulums, the only factors that affect the period are the length of the pendulum and the local acceleration due to gravity. Heavier or lighter pendulum bobs with the same length and under the same gravitational influence swing with the same period. This characteristic time of swing is a fundamental concept that paves the way for understanding more complex oscillatory systems.
Pendulum Period Equation
The period of a simple pendulum can be quantified using the pendulum period equation: \[ T = 2\pi \sqrt{\frac{l}{g}} \] where \( T \) is the period, \( l \) represents the length of the pendulum, and \( g \) is the acceleration due to gravity at that location. This equation shows that the period is proportional to the square root of the length of the pendulum and inversely proportional to the square root of the gravitational acceleration.

If we know the period on Earth and want to find the period on Mars, we take into account that the length of the pendulum remains constant while the gravitational acceleration will change to that of Mars. With some algebraic manipulation, as demonstrated in the step-by-step solution, we can derive the period on Mars using Earth's period and the known gravitational accelerations of both planets. This provides students with a real-world application of how pendulum motion can vary across different planetary environments, deepening their understanding of gravitational effects on periodic motion.

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Most popular questions from this chapter

DATA Experimenting with pendulums, you attach a light string to the ceiling and attach a small metal sphere to the lower end of the string. When you displace the sphere \(2.00 \mathrm{~m}\) to the left, it nearly touches a vertical wall; with the string taut, you release the sphere from rest. The sphere swings back and forth as a simple pendulum, and you measure its period \(T\). You repeat this act for strings of various lengths \(L\), each time starting the motion with the sphere displaced \(2.00 \mathrm{~m}\) to the left of the vertical position of the string. In each case the sphere's radius is very small compared with \(L\). Your results are given in the table: $$ \begin{array}{l|rrrrrrrr} \boldsymbol{L}(\mathbf{m}) & 12.00 & 10.00 & 8.00 & 6.00 & 5.00 & 4.00 & 3.00 & 2.50 & 2.30 \\ \hline \boldsymbol{T}(\mathbf{s}) & 6.96 & 6.36 & 5.70 & 4.95 & 4.54 & 4.08 & 3.60 & 3.35 & 3.27 \end{array} $$ (a) For the five largest values of \(L,\) graph \(T^{2}\) versus \(L\). Explain why the data points fall close to a straight line. Does the slope of this line have the value you expected? (b) Add the remaining data to your graph. Explain why the data start to deviate from the straight-line fit as \(L\) decreases. To see this effect more clearly, plot \(T / T_{0}\) versus \(L,\) where \(T_{0}=2 \pi \sqrt{L / g}\) and \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} .\) (c) Use your graph of \(T / T_{0}\) versus \(L\) to estimate the angular amplitude of the pendulum (in degrees) for which the equation \(T=2 \pi \sqrt{L / g}\) is in error by \(5 \%\).

A sinusoidally varying driving force is applied to a damped harmonic oscillator of force constant \(k\) and mass \(m .\) If the damping constant has a value \(b_{1},\) the amplitude is \(A_{1}\) when the driving angular frequency equals \(\sqrt{k / m}\). In terms of \(A_{1}\), what is the amplitude for the same driving frequency and the same driving force amplitude \(F_{\max },\) if the damping constant is (a) \(3 b_{1}\) and (b) \(b_{1} / 2 ?\)

Quantum mechanics is used to describe the vibrational motion of molecules, but analysis using classical physics gives some useful insight. In a classical model the vibrational motion can be treated as SHM of the atoms connected by a spring. The two atoms in a diatomic molecule vibrate about their center of mass, but in the molecule HI, where one atom is much more massive than the other, we can treat the hydrogen atom as oscillating in SHM while the iodine atom remains at rest. (a) A classical estimate of the vibrational frequency is \(f=7 \times 10^{13} \mathrm{~Hz}\). The mass of a hydrogen atom differs little from the mass of a proton. If the HI molecule is modeled as two atoms connected by a spring, what is the force constant of the spring? (b) The vibrational energy of the molecule is measured to be about \(5 \times 10^{-20} \mathrm{~J}\). In the classical model, what is the maximum speed of the H atom during its SHM? (c) What is the amplitude of the vibrational motion? How does your result compare to the equilibrium distance between the two atoms in the HI molecule, which is about \(1.6 \times 10^{-10} \mathrm{~m} ?\)

Object \(A\) has mass \(m_{A}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{A} .\) Object \(B\) has mass \(m_{B}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{B}\). The amplitude \(A_{A}\) for object \(A\) is twice the amplitude \(A_{B}\) for the motion of object \(B\). Also, \(m_{B}=4 m_{A}\) and \(k_{A}=9 k_{B}\). (a) What is the ratio of the maximum speeds of the two objects, \(v_{\max , A} / v_{\max , B} ?\) (b) What is the ratio of their maximum accelerations, \(a_{\max , A} / a_{\max , B} ?\)

A small sphere with mass \(m\) is attached to a massless rod of length \(L\) that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle \(\theta\) from the vertical, and released from rest. (a) In a diagram, show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere? (b) Repeat part (a) for the instant when the pendulum rod is at an angle \(\theta / 2\) from the vertical. (c) Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?
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