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Chapter 10: Problem 44

Astronauts on a distant planet set up a simple pendulum of length \(1.2 \mathrm{m} .\) The pendulum executes simple harmonic motion and makes 100 complete vibrations in 280 s. What is the magnitude of the acceleration due to gravity on this planet?

Short Answer

Expert verified
The acceleration due to gravity on the planet is approximately 6.05 m/s虏.

Step by step solution

01

Understand the Principle of a Simple Pendulum

A simple pendulum's period, \( T \), is the time it takes to complete one full oscillation. For a pendulum of length \( L \), the period can be expressed using the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \( g \) is the acceleration due to gravity.
02

Calculate the Period of the Pendulum

We are given that the pendulum makes 100 complete vibrations in 280 seconds. The period \( T \) of the pendulum is the total time divided by the number of vibrations: \[ T = \frac{280 \, \text{s}}{100} = 2.8 \, \text{s} \]
03

Rearrange the Formula to Solve for \( g \)

Rearrange the formula for the period of a pendulum to solve for \( g \): \[ T = 2\pi \sqrt{\frac{L}{g}} \] which implies \[ g = \frac{4\pi^2 L}{T^2} \]
04

Substitute Known Values

Substitute the known values into the rearranged formula: \( L = 1.2 \, \text{m} \) and \( T = 2.8 \, \text{s} \): \[ g = \frac{4 \pi^2 \times 1.2}{(2.8)^2} \]
05

Calculate the Magnitude of \( g \)

Perform the calculation to find \( g \): \[ g = \frac{4 \times (3.14159)^2 \times 1.2}{7.84} \approx 6.05 \, \text{m/s}^2 \]

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Key Concepts

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

harmonic motion
Simple harmonic motion is a type of periodic motion where an object moves back and forth along a path in such a way that its acceleration is directly proportional to its displacement from a specific point, known as the equilibrium position. This motion occurs in both directions, creating symmetry as it continuously swings through its path.
For pendulums, this harmonic motion is driven by the force of gravity, which tries to return the pendulum to its equilibrium position. Importantly, in an ideal simple pendulum, without any form of friction or air resistance, this motion would continue perpetually. However, in real-world scenarios, factors like air resistance will eventually dampen the pendulum's oscillations.
In the context of simple harmonic motion, the displacement, velocity, and acceleration of an object can be described using trigonometric functions such as sine and cosine. For example, the pendulum's position with respect to time is often modeled as a sine wave, which helps in understanding how it behaves over time.
acceleration due to gravity
The acceleration due to gravity ( "g" ) is a critical factor that influences the motion of a pendulum. It is the force that pulls objects towards the center of a planet. On Earth, this value is typically about 9.81 m/s虏. However, on other celestial bodies like the moon or distant planets, this value can differ significantly.
In our exercise, the pendulum is used to measure the acceleration due to gravity on a distant planet. The greater the value of "g" , the stronger the gravitational pull and the quicker the pendulum will swing back to its starting point.
This gravitational pull is crucial because it determines how quickly a pendulum accelerates as it moves away from and back towards its equilibrium position. Understanding this concept is essential for calculating properties like the pendulum's period, which in turn helps in determining distances and forces in physics.
period of pendulum
The period of a pendulum is the duration required for it to complete one full cycle of swing. This cycle includes a motion from one side to another and back again. The period is a critical parameter in characterizing harmonic motion.
The formula for calculating the period (T) of a simple pendulum is given by: \[ T = 2\pi \sqrt{\frac{L}{g}} \]Here "L" denotes the length of the pendulum, and "g" is the acceleration due to gravity. This formula shows how the period does not depend on the mass of the pendulum bob and is only affected by the length of the pendulum and the gravitational acceleration acting upon it.
  • If the pendulum is longer, it will take more time to complete one oscillation, thereby increasing the period.
  • If the gravitational pull is stronger, each oscillation will happen faster, decreasing the period.
In our problem, knowing the period helps in solving for the unknown gravity on the distant planet, indicating how intertwined these concepts are.

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Most popular questions from this chapter

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of \(7.0 \mathrm{rad} / \mathrm{s} .\) The drawing indicates the position of the block when the spring is unstrained. This position is labeled " \(x=\) \(0 \mathrm{m} .\) "The drawing also shows a small bottle located \(0.080 \mathrm{m}\) to the right of this position. The block is pulled to the right, stretching the spring by \(0.050 \mathrm{m},\) and is then thrown to the left. In order for the block to knock over the bottle, it must be thrown with a speed exceeding \(v_{0} .\) Ignoring the width of the block, find \(v_{0}\).

A spiral staircase winds up to the top of a tower in an old castle. To measure the height of the tower, a rope is attached to the top of the tower and hung down the center of the staircase. However, nothing is available with which to measure the length of the rope. Therefore, at the bottom of the rope a small object is attached so as to form a simple pendulum that just clears the floor. The period of the pendulum is measured to be 9.2 s. What is the height of the tower?

A heavy-duty stapling gun uses a 0.140 -kg metal rod that rams against the staple to eject it. The rod is attached to and pushed by a stiff spring called a "ram spring" \((k=32000 \mathrm{N} / \mathrm{m})\). The mass of this spring may be ignored. The ram spring is compressed by \(3.0 \times 10^{-2} \mathrm{m}\) from its unstrained length and then released from rest. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{m}\) when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

A vertical spring (spring constant \(=112 \mathrm{N} / \mathrm{m}\) ) is mounted on the floor. A 0.400-kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in Hz) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.

A 70.0 -kg circus performer is fired from a cannon that is elevated at an angle of \(40.0^{\circ}\) above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by \(3.00 \mathrm{m}\) from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as the height of the net into which he is shot. He takes 2.14 s to travel the horizontal distance of \(26.8 \mathrm{m}\) between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.
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