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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

A uniform, solid metal disk of mass \(6.50 \mathrm{~kg}\) and diameter \(24.0 \mathrm{~cm}\) hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of \(4.23 \mathrm{~N}\) tangent to the rim of the disk to turn it by \(3.34^{\circ},\) thus twisting the wire. You now remove this force and release the disk from rest. (a) What is the torsion constant for the metal wire? (b) What are the frequency and period of the torsional oscillations of the disk? (c) Write the equation of motion for \(\theta(t)\) for the disk.

SHM of a Floating Object. An object with height \(h\)mass \(M\), and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\). (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M,\) and the cross-sectional area \(A\) of the object. You can ignore the damping due to fluid friction (see Section 14.7).

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is \(0.165 \mathrm{~m}\). The maximum speed of the block is \(3.90 \mathrm{~m} / \mathrm{s}\). What is the maximum magnitude of the acceleration of the block?

BIO Weighing a Virus. In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just \(30 \mathrm{nm}\) long) with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring. (a) Show that the ratio of the frequency with the virus attached \(\left(f_{\mathrm{S}+\mathrm{V}}\right)\) to the frequency without the virus \(\left(f_{\mathrm{S}}\right)\) is given by \(f_{\mathrm{S}+\mathrm{V}} / f_{\mathrm{S}}=1 / \sqrt{1+\left(m_{\mathrm{V}} / m_{\mathrm{S}}\right)},\) where \(m_{\mathrm{V}}\) is the mass of the virus and \(m_{\mathrm{S}}\) is the mass of the silicon sliver. Notice that it is not necessary to know or measure the force constant of the spring. (b) In some data, the silicon sliver has a mass of \(2.10 \times 10^{-16} \mathrm{~g}\) and a frequency of \(2.00 \times 10^{15} \mathrm{~Hz}\) without the virus and \(2.87 \times 10^{14} \mathrm{~Hz}\) with the virus. What is the mass of the virus, in grams and in femtograms?

A harmonic oscillator has angular frequency \(\omega\) and amplitude \(A\). (a) What are the magnitudes of the displacement and velocity when the elastic potential energy is equal to the kinetic energy? (Assume that \(U=0\) at equilibrium.) (b) How often does this occur in each cycle? What is the time between occurrences? (c) At an instant when the displacement is equal to \(A / 2,\) what fraction of the total energy of the system is kinetic and what fraction is potential?
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