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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

An object is undergoing SHM with period 0.300 s and amplitude 6.00 cm. At \(t =\) 0 the object is instantaneously at rest at \(x =\) 6.00 cm. Calculate the time it takes the object to go from \(x =\) 6.00 cm to \(x = -\)1.50 cm.

In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just 30 nm long) with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring. (a) Show that the ratio of the frequency with the virus attached \((f_S+\,_V)\) to the frequency without the virus \((f_S)\) is given by \(f_S+\,_V/f_S = 1\sqrt{ 1 + (m_V/m_S) }\), where \(m_V\) is the mass of the virus and \(m_S\) is the mass of the silicon sliver. Notice that it is \(not\) necessary to know or measure the force constant of the spring. (b) In some data, the silicon sliver has a mass of 2.10 \(\times\) 10\(^{-16}\) g and a frequency of 2.00 \(\times\) 10\(^{15}\) Hz without the virus and 2.87 \(\times\) 10\(^{14}\) Hz with the virus. What is the mass of the virus, in grams and in femtograms?

A 40.0-N force stretches a vertical spring 0.250 m. (a) What mass must be suspended from the spring so that the system will oscillate with a period of 1.00 s? (b) If the amplitude of the motion is 0.050 m and the period is that specified in part (a), where is the object and in what direction is it moving 0.35 s after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is 0.030 m below the equilibrium position, moving upward?

A small sphere with mass \(m\) is attached to a massless rod of length \(L\) that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle \(\theta\) from the vertical, and released from rest. (a) In a diagram, show the pendulum just after it is released. Draw vectors representing the \(forces\) acting on the small sphere and the \(acceleration\) of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere? (b) Repeat part (a) for the instant when the pendulum rod is at an angle \(\theta\)/2 from the vertical. (c) Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

When a body of unknown mass is attached to an ideal spring with force constant 120 N/m, it is found to vibrate with a frequency of 6.00 Hz. Find (a) the period of the motion; (b) the angular frequency; (c) the mass of the body.

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