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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/sĀ² 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/sĀ², 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 ā€“ 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 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At \(t\) = 0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.

You pull a simple pendulum 0.240 m long to the side through an angle of 3.50\(^\circ\) and release it. (a) How much time does it take the pendulum bob to reach its highest speed? (b) How much time does it take if the pendulum is released at an angle of 1.75\(^\circ\) instead of 3.50\(^\circ\)?

A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of the beam. The beam has mass 225 kg, and a 175-kg sack of gravel sits on the middle of it. The beam is oscillating in SHM with an amplitude of 40.0 cm and a frequency of 0.600 cycle/s. (a) The sack falls off the beam when the beam has its maximum upward displacement. What are the frequency and amplitude of the subsequent SHM of the beam? (b) If the sack instead falls off when the beam has its maximum speed, repeat part (a).

Two uniform solid spheres, each with mass \(M =\) 0.800 kg and radius \(R =\) 0.0800 m, are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant \(k =\) 160 N/m has one end attached to the wall and the other end attached to a frictionless ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring. Show that the motion of the center of mass of the spheres is simple harmonic and calculate the period.

The wings of the blue-throated hummingbird \((Lampornis\) \(clemenciae)\), which inhabits Mexico and the southwestern United States, beat at a rate of up to 900 times per minute. Calculate (a) the period of vibration of this bird's wings, (b) the frequency of the wings' vibration, and (c) the angular frequency of the bird's wing beats.

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