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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 is approximately 6.04 m/s².

Step by step solution

01

Identify Known Quantities

The length of the pendulum is given as \( L = 1.2 \text{ m} \). The pendulum completes 100 vibrations in 280 seconds, which means the period \( T \) can be calculated by dividing the total time by the number of vibrations: \[ T = \frac{280 \text{ s}}{100} = 2.8 \text{ s}.\]
02

Recall the Formula for the Period of a Pendulum

The period \( T \) of a simple pendulum is related to the length \( L \) and the acceleration due to gravity \( g \) by the formula: \[ T = 2\pi\sqrt{\frac{L}{g}}. \] We need to solve for \( g \).
03

Solve for Acceleration Due to Gravity

Rearrange the formula to find \( g \): \[ g = \frac{4\pi^2L}{T^2}. \] Substitute \( L = 1.2 \text{ m} \) and \( T = 2.8 \text{ s} \) into the formula: \[ g = \frac{4\pi^2 \times 1.2}{(2.8)^2}. \]
04

Perform the Calculations

Calculate \( 4\pi^2 \times 1.2 \): \[ 4\pi^2 \approx 39.4784, \] so \[ 39.4784 \times 1.2 \approx 47.37408. \] Then, calculate \( (2.8)^2 = 7.84 \). Finally, compute \( \frac{47.37408}{7.84} \approx 6.04 \).
05

Conclusion

The magnitude of the acceleration due to gravity on the distant planet is approximately \( 6.04 \text{ m/s}^2 \).

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Key Concepts

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

Acceleration Due to Gravity
The concept of acceleration due to gravity, often represented as \( g \), is a fundamental principle in physics. It refers to the acceleration that a mass experiences as a result of the gravitational pull of a planet or celestial body. On Earth, the standard value of \( g \) is approximately \( 9.81 \, \text{m/s}^2 \), but this value can differ significantly on other planets.

The difference in the acceleration due to gravity is due to the varying mass and size of each celestial body. These differences affect the gravitational pull that influences the motion of objects. In the pendulum case, calculating \( g \) involved using the pendulum's period and length. This method is effective in determining \( g \) on planets where direct measurement is not feasible.

When evaluating planets like the one in the exercise, understanding their gravitational pull is crucial, as it affects various aspects such as atmosphere retention and weight of objects. These insights are not only valuable for physics problems but also for practical applications in space exploration.
Simple Harmonic Motion
Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that displacement. This is a common form of motion exhibited by systems like pendulums and springs.

  • Key Characteristics: In SHM, the object moves back and forth in a regular, repeating cycle.
  • Restoring Force: The force that brings the object back to its equilibrium position is proportional to the displacement.
  • Sinusoidal Waveform: The oscillations can be described by sine or cosine functions, indicating predictable patterns.


a simple pendulum, like the one in our scenario, undergoes SHM, making it a practical example for studying such motion. As it swings, the pendulum's potential and kinetic energy persistently exchange, maintaining its motion unless damped by external forces.
Pendulum Period Calculation
Calculating the period of a pendulum involves understanding its relationship with the pendulum's length and the gravitational acceleration. The period \( T \) is the time taken for one complete cycle of the pendulum's swing and is a key component in studying pendulum dynamics.

The formula \[ T = 2\pi\sqrt{\frac{L}{g}} \] indicates a direct relationship between the length \( L \) of the pendulum, the gravitational acceleration \( g \), and the period \( T \).

  • Longer Length: Increasing the length results in a longer period, meaning the pendulum swings more slowly.
  • Higher Gravity: A greater gravitational force tends to decrease the period, causing faster swings.


In the exercise, the calculations demonstrated this relationship by rearranging the formula to find \( g \), given the known values of \( L \) and \( T \). This method effectively determines the acceleration due to gravity by exploiting the predictable nature of pendulum motion.

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Most popular questions from this chapter

A die is designed to punch holes with a radius of \(1.00 \times 10^{-2} \mathrm{m}\) in a metal sheet that is \(3.0 \times 10^{-3} \mathrm{m}\) thick, as the drawing illustrates. To punch through the sheet, the die must exert a shearing stress of \(3.5 \times 10^{8} \mathrm{Pa} .\) What force \(\overrightarrow{\mathbf{F}}\) must be applied to the die?

ao \(A\) spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is \(0.80 \mathrm{kg},\) and the spring has a spring constant of \(59 \mathrm{N} / \mathrm{m} .\) The coefficient of static friction between the box and the table on which it rests is \(\mu_{\mathrm{s}}=0.74\) How far can the spring be stretched from its unstrained position without the box moving when it is released?

A \(1.00 \times 10^{-2}\) -kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is \(124 \mathrm{N} / \mathrm{m} .\) The block is shoved parallel to the spring axis and is given an initial speed of \(8.00 \mathrm{m} / \mathrm{s},\) while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 59 N/m and 33 N/m. Find the angle that the rod makes with the horizontal.

A small object oscillates back and forth at the bottom of a frictionless hemispherical bowl, as the drawing illustrates. The radius of the bowl is \(R\), and the angle \(\theta\) is small enough that the object oscillates in simple harmonic motion. Derive an for the angular frequency \(\omega\) of the motion. Express your answer in terms of \(R\) and \(g,\) the magnitude of the acceleration due to gravity.

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