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A simple pendulum is \(5.00 \mathrm{m}\) long. (a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at \(5.00 \mathrm{m} / \mathrm{s}^{2} ?\) (b) What is its period if the elevator is accelerating downward at \(5.00 \mathrm{m} / \mathrm{s}^{2} ?\) (c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at \(5.00 \mathrm{m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
The periods for the pendulum are as follows: in an elevator accelerating upward, \(2\pi\sqrt{\frac{5.00m }{14.80 m/s^2}}\); in an elevator accelerating downward, \(2\pi\sqrt{\frac{5.00m}{4.80 m/s^2}}\); and in a truck accelerating horizontally, \(2\pi\sqrt{\frac{5.00m}{9.80 m/s^2}}\). Please remember to calculate these equations to get the actual values.

Step by step solution

01

Understand the question and relevant equations

The period of small oscillations for a simple pendulum can be calculated by \(T = 2\pi\sqrt{\frac{L}{g}}\), where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity. However, in a non-inertial frame of reference such as the moving elevator or truck, the effective gravity is altered and the revised equation becomes \(T = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}}\). The value of \(g_{\text{eff}}\) will vary based on the scenario - when the elevator is moving upwards, the effective gravity \(g_{\text{eff}} = g + a\); when it's moving downwards, \(g_{\text{eff}} = g - a\); and when the truck is moving horizontally, the effective gravity remains \(g_{\text{eff}} = g\). Here, \(a\) is the acceleration of the elevator/truck.
02

Calculate the period for the upward moving elevator

Let's first calculate the period of small oscillations for the pendulum in an elevator accelerating upward at \(5.00 \mathrm{m} / \mathrm{s}^{2}\). Using the formula mentioned before, and considering \(g = 9.80 \mathrm{m/s}^{2}\), we have \(g_{\text{eff}} = g + a = 9.80 \mathrm{m/s}^{2} + 5.00 \mathrm{m/s}^{2} = 14.80 \mathrm{m/s}^{2}\). Thus, the period \(T\) of small oscillations is \(T = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi\sqrt{\frac{5.00 \mathrm{m}}{14.80 \mathrm{m/s}^{2}}}\).
03

Calculate the period for the downward moving elevator

Now, let's calculate the period of small oscillations for the pendulum in an elevator accelerating downward at \(5.00 \mathrm{m} / \mathrm{s}^{2}\). Again using the formula, we get \(g_{\text{eff}} = g - a = 9.80 \mathrm{m/s}^{2} - 5.00 \mathrm{m/s}^{2} = 4.80 \mathrm{m/s}^{2}\). Thus, the period \(T\) of small oscillations is \(T = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi\sqrt{\frac{5.00 \mathrm{m}}{4.80 \mathrm{m/s}^{2}}}\).
04

Calculate the period for the horizontally moving truck

Finally, let's calculate the period of small oscillations for the pendulum in a truck accelerating horizontally at \(5.00 \mathrm{m} / \mathrm{s}^{2}\). In this case, the acceleration doesn't affect the effective gravity, so \(g_{\text{eff}} = g = 9.80 \mathrm{m/s}^{2}\). Thus, the period \(T\) of small oscillations is \(T = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi\sqrt{\frac{5.00 \mathrm{m}}{9.80 \mathrm{m/s}^{2}}}\).

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Key Concepts

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

Oscillation Period
The oscillation period of a pendulum describes how long it takes for the pendulum to complete one full swing back and forth. This is a crucial concept for understanding pendulum motion. For a simple pendulum, the formula to calculate the period is:- \(T = 2\pi\sqrt{\frac{L}{g}}\)- where \(T\) is the period,- \(L\) is the length of the pendulum,- and \(g\) is the acceleration due to gravity.The period is independent of the pendulum mass, which means a heavier pendulum would swing with the same timing as a lighter one. However, this formula applies in ideal conditions where the pendulum is not affected by additional forces, such as air resistance or friction. Understanding this basic formula enables one to calculate the pendulum's period in various settings by adjusting for any changes in the effective gravity, discussed in the next section.
Effective Gravity
Effective gravity is a key concept in determining the behavior of a pendulum under different conditions. Normal gravity, designated as \(g\), is approximately \(9.8 \text{ m/s}^2\) on Earth's surface. However, "effective gravity" reflects the net force acting on the pendulum when additional accelerations are considered, such as when an elevator or vehicle is accelerating.- When the elevator moves **upward**, effective gravity increases: \(g_{\text{eff}} = g + a\)- When it moves **downward**, effective gravity decreases: \(g_{\text{eff}} = g - a\)- For horizontal truck motion, effective gravity is unchanged: \(g_{\text{eff}} = g\)These adjustments in effective gravity directly impact the pendulum's period of oscillation. For example, if a pendulum is in an upward accelerating elevator, the effective gravity is greater, thus reducing the period. Conversely, downward acceleration results in a lower effective gravity, increasing the period. Recognizing these scenarios and their influence on effective gravity is crucial for solving problems involving pendulums in motion.
Non-Inertial Reference Frame
A non-inertial reference frame is one where the observer or point of reference is accelerating rather than staying still or moving at a constant velocity. This is an important concept when analyzing pendulum motion in contexts like elevators or trucks. - Within a non-inertial frame, forces like acceleration must be accounted for in calculations. - This is achieved by adjusting gravity to "effective gravity" to reflect these extra accelerations. - Resultant forces in a non-inertial frame appear to modify physical laws, like inertia or momentum. When solving pendulum problems, non-inertial frames require the observer to perceive changes in gravity. Thus, the period of oscillation is recalculated using the effective gravity as explained earlier. This adjustment is essential because, in non-inertial frames, conventional equations do not directly apply until compensated by such calculations. Understanding this concept helps conquer trickier pendulum setups where straightforward physics require deeper analysis.

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Most popular questions from this chapter

A simple pendulum has a mass of \(0.250 \mathrm{kg}\) and a length of \(1.00 \mathrm{m} .\) It is displaced through an angle of \(15.0^{\circ}\) and then released. What are (a) the maximum speed, (b) the maximum angular acceleration, and (c) the maximum restoring force? What If? Solve this problem by using the simple harmonic motion model for the motion of the pendulum, and then solve the problem more precisely by using more general principles.

A \(1.00-\mathrm{kg}\) glider attached to a spring with a force constant of \(25.0 \mathrm{N} / \mathrm{m}\) oscillates on a horizontal, frictionless air track. At \(t=0\) the glider is released from rest at \(x=-3.00 \mathrm{cm} .\) (That is, the spring is compressed by \(3.00 \mathrm{cm} .)\) Find (a) the period of its motion, (b) the maximum values of its speed and acceleration, and (c) the position, velocity, and acceleration as functions of time.

The angular position of a pendulum is represented by the equation \(\theta=(0.320 \mathrm{rad}) \cos \omega t,\) where \(\theta\) is in radians and \(\omega=4.43 \mathrm{rad} / \mathrm{s} .\) Determine the period and length of the pendulum.

A simple pendulum with a length of \(2.23 \mathrm{m}\) and a mass of \(6.74 \mathrm{kg}\) is given an initial speed of \(2.06 \mathrm{m} / \mathrm{s}\) at its equilibrium position. Assume it undergoes simple harmonic motion, and determine its (a) period, (b) total energy, and (c) maximum angular displacement.

A physical pendulum in the form of a planar body moves in simple harmonic motion with a frequency of \(0.450 \mathrm{Hz}\) If the pendulum has a mass of \(2.20 \mathrm{kg}\) and the pivot is located \(0.350 \mathrm{m}\) from the center of mass, determine the moment of inertia of the pendulum about the pivot point.

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