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Chapter 14: Problem 49

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 cm. She finds that the pendulum makes 100 complete swings in 136 s. What is the value of \(g\) on this planet?

Short Answer

Expert verified
The acceleration due to gravity, \( g \), is approximately 10.67 m/s虏 on that planet.

Step by step solution

01

Understand the Pendulum Formula

The period of a simple pendulum, which is the time for one complete swing back and forth, is given by the formula \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \( T \) is the period in seconds, \( L \) is the length of the pendulum in meters, and \( g \) is the acceleration due to gravity.
02

Find the Period of One Swing

The period \( T \) is the total time for the pendulum to make one complete swing. The total time for 100 swings is 136 seconds. Therefore, the period for one swing is: \[ T = \frac{136 \text{ s}}{100} = 1.36 \text{ s} \]
03

Convert Length to Meters

The length of the pendulum is 50.0 cm. Convert this into meters by dividing by 100:\[ L = \frac{50.0}{100} = 0.50 \text{ m} \]
04

Rearrange Formula to Solve for g

We need to solve for \( g \) in the pendulum formula. Rearrange the formula to:\[ g = \frac{4\pi^2L}{T^2} \]
05

Substitute Values and Calculate g

Now, substitute \( L = 0.50 \text{ m} \) and \( T = 1.36 \text{ s} \) into the rearranged formula to calculate \( g \):\[ g = \frac{4\pi^2 (0.50)}{(1.36)^2} \]Calculate the result:1. \( 1.36^2 = 1.8496 \)2. \( 4\pi^2 = 39.4784 \) (since \( \pi^2 \approx 9.8696 \))3. Therefore, \( \frac{39.4784 \times 0.50}{1.8496} \equiv \frac{19.7392}{1.8496} \approx 10.67 \text{ m/s}^2 \)

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Key Concepts

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

Acceleration Due to Gravity
Acceleration due to gravity, often symbolized as \( g \), is a key factor in understanding how objects fall or oscillate. It represents the rate at which an object accelerates when it is in free fall. On Earth, this value is approximately 9.81 m/s虏, but it can differ on other planets depending on their masses and radii.
In pendulum physics, \( g \) can be calculated using the pendulum's period and length. The acceleration due to gravity affects how swiftly a pendulum swings back and forth. This is essential in determining time periods in various planetary environments, as seen in our exercise where all other variables aside from the pendulum's length and time period are a mystery.
  • Gravity differs from planet to planet.
  • It determines the speed of oscillation for a pendulum.
  • Knowing \( g \) is crucial for astronomical calculations.
Pendulum Period
The pendulum period is the time it takes for a pendulum to make one full oscillation, swinging forward and then back to the starting point. The period (\( T \)) is linked with the pendulum's length (\( L \)) and the acceleration due to gravity (\( g \)).
In our exercise, the space explorer determines the period of the pendulum's swing by timing how long it takes for the pendulum to complete 100 swings 鈥� a total of 136 seconds. This measurement is then averaged to find a single swing time: 1.36 seconds.
  • The period helps in determining gravitational forces on planets.
  • It's derived by dividing total swing time by the number of swings.
  • A longer period generally suggests weaker gravity if \( L \) remains constant.
Simple Pendulum Formula
The simple pendulum formula is crucial in analyzing the motion of pendulums. It ties together the pendulum's period, length, and the gravity it experiences.\[T = 2\pi \sqrt{\frac{L}{g}}\]This formula describes how the three elements interact: \( T \) (period) is the time for a complete swing, \( L \) (length) is measured in meters, and \( g \) (acceleration due to gravity) affects everything.
To isolate \( g \), we rearrange the formula:\[g = \frac{4\pi^2L}{T^2}\]By plugging in the known values of period and length, we can accurately find \( g \) on an unknown planet, as we did in the exercise.
  • The formula showcases the balance between length, period, and acceleration due to gravity.
  • Rearranging the formula lets us solve for unknown gravitational forces.
  • It is an essential tool for physics, especially in environments with unknown gravitational conditions.

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Most popular questions from this chapter

A 175-g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 155 N/m. At the instant you make measurements on the glider, it is moving at 0.815 m/s and is 3.00 cm from its equilibrium point. Use \({energy\ conservation}\) to find (a) the amplitude of the motion and (b) the maximum speed of the glider. (c) What is the angular frequency of the oscillations?

A 5.00-kg partridge is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of 4.20 s. (a) What is its speed as it passes through the equilibrium position? (b) What is its acceleration when it is 0.050 m above the equilibrium position? (c) When it is moving upward, how much time is required for it to move from a point 0.050 m below its equilibrium position to a point 0.050 m above it? (d) The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?

You hang various masses \(m\) from the end of a vertical, 0.250-kg spring that obeys Hooke's law and is tapered, which means the diameter changes along the length of the spring. Since the mass of the spring is not negligible, you must replace \(m\) in the equation \(T =\) 2\(\pi\sqrt{ m/k }\) with \(m + m_\mathrm{eff}\), where \(m_\mathrm{eff}\) is the effective mass of the oscillating spring. (See Challenge Problem 14.93.) You vary the mass m and measure the time for 10 complete oscillations, obtaining these data: (a) Graph the square of the period \(T\) versus the mass suspended from the spring, and find the straight line of best fit. (b) From the slope of that line, determine the force constant of the spring. (c) From the vertical intercept of the line, determine the spring's effective mass. (d) What fraction is \(m_\mathrm{eff}\) of the spring's mass? (e) If a 0.450-kg mass oscillates on the end of the spring, find its period, frequency, and angular frequency.

A building in San Francisco has light fixtures consisting of small \(2.35-\mathrm{kg}\) bulbs with shades hanging from the ceiling at the end of light, thin cords \(1.50 \mathrm{~m}\) long. If a minor earthquake occurs, how many swings per second will these fixtures make?

When a 0.750-kg mass oscillates on an ideal spring, the frequency is 1.75 Hz. What will the frequency be if 0.220 kg are (a) added to the original mass and (b) subtracted from the original mass? Try to solve this problem \(without\) finding the force constant of the spring.
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