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

Short Answer

Expert verified
The gravitational acceleration \( g \) on this planet is approximately 10.67 m/s².

Step by step solution

01

Recall the formula for pendulum period

A simple pendulum's period (T) is related to the length (L) and gravitational acceleration (g) by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \] This formula relates the time taken for one complete oscillation (period) with the gravitational field of the planet.
02

Calculate one period of the pendulum

Since the pendulum makes 100 complete swings in 136 seconds, each swing (which has a back-and-forth period) lasts:\[ T = \frac{136\, \text{s}}{100} = 1.36\, \text{s} \]This is the period (T) of the pendulum on this planet.
03

Rearrange formula to solve for g

Rearrange the period formula to express g:\[ g = \frac{4\pi^2 L}{T^2} \]This formula now allows us to determine the gravitational acceleration by substituting known quantities.
04

Substitute known values into the formula

Insert the values into the equation: - Length (L) = 0.50 m (convert 50.0 cm to meters),- Period (T) = 1.36 s,\[ g = \frac{4\pi^2 \times 0.50\, \text{m}}{(1.36\, \text{s})^2} \]
05

Calculate the value of g

Perform the calculations:- Calculate the squared period: \( (1.36)^2 = 1.8496 \)- Multiply: \( 4\pi^2 \approx 39.4784 \)- Calculate g: \[ g = \frac{39.4784 \times 0.50}{1.8496} \approx 10.67\, \text{m/s}^2 \]

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Key Concepts

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

Gravitational Acceleration
Gravitational acceleration, often represented by the symbol \( g \), is the acceleration due to gravity that an object experiences in a specific location, such as on the surface of a planet. It's a measure of how fast objects will accelerate towards the planet when they are in free fall.
The standard value of \( g \) on Earth is approximately 9.81 \( \text{m/s}^2 \), but this can vary based on a planet's mass and radius. In the context of the pendulum experiment on an unfamiliar planet, finding \( g \) helps us understand the planet's gravitational pull as compared to Earth's.
Using the pendulum formula \( T = 2\pi \sqrt{\frac{L}{g}} \), we can rearrange it to find \( g \) when the period \( T \) and the length \( L \) of the pendulum are known. This way, the cycle of the pendulum swing gives direct insight into possible characteristics of the planet's gravity.
Simple Harmonic Motion
Simple Harmonic Motion (SHM) refers to a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.
  • A simple pendulum is one classic example of SHM, where a mass (the bob) oscillates back and forth on a string or rod fixed at a point.
  • The pendulum swings in an arc under the influence of gravity, exhibiting a predictable manner described by laws of SHM.
When the pendulum is displaced from its equilibrium position, gravity causes it to accelerate back towards that position. This continuous cycle of swinging back and forth embodies SHM.
The period \( T \) of a simple pendulum, or the time it takes for one complete cycle of motion, depends on the length of the pendulum \( L \) and the gravitational acceleration \( g \), as given by the formula \( T = 2\pi \sqrt{\frac{L}{g}} \). This relationship illustrates the key characteristics of SHM: regularity and predictability in motion.
Physics Problem Solving
Physics problem solving is an essential skill often practiced through experiments and theoretical calculations. These problems, like the pendulum experiment, help build a deep understanding of physical concepts and enhance analytical skills.
To solve physics problems effectively:
  • First, carefully identify the known variables and the formula needed to connect these variables to the unknowns.
  • Rearrange the necessary equations to isolate the unknown variable, often requiring algebraic manipulation.
  • Substitute the known values into the equation and solve for the unknown.
In our pendulum problem, systematic problem solving involved identifying the pendulum as operating under SHM, using the pendulum formula to rearrange for \( g \), and then carefully substituting the length of the pendulum and observed period.
The accuracy of calculations is crucial. Each step should ensure no errors in substituting numbers, units, or performing calculations as errors compound over problem-solving steps. Through practice, these problem-solving techniques reveal the interconnections within physics concepts like time, distance, and force on varied scales across the universe.

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Most popular questions from this chapter

(a) What is the change \(\Delta T\) in the period of a simple penduIum when the acceleration of gravity \(g\) changes by \(\Delta g ?\) (Hint: The new period \(T+\Delta T\) is obtained by substituting \(g+\Delta g\) for \(g :\) $$\boldsymbol{T}+\Delta \boldsymbol{T}=2 \pi \sqrt{\frac{\boldsymbol{L}}{\boldsymbol{g}+\Delta \boldsymbol{g}}}$$ To obtain an approximate expression, expand the factor \((g+\Delta g)^{-1 / 2}\) using the binomial theorem (Appendix B) and keep only the first two terms: $$(g+\Delta g)^{-1 / 2}=g^{-1 / 2}-\frac{1}{2} g^{-3 / 2} \Delta g+\cdots$$ The other terms contain higher powers of \(\Delta g\) and are very small if \(\Delta g\) is small.) Express your result as the fractional change in period \(\Delta T / T\) in terms of the fractional change \(\Delta g / g .\) (b) A pendulum clock keeps correct time at a point where \(g=9.8000 \mathrm{m} / \mathrm{s}^{2},\) but is found to lose 4.0 \(\mathrm{s}\) each day at a higher elevation. Use the result of part (a) to find the approximate value of \(g\) at this new location.

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

A \(1.80-\mathrm{kg}\) monkey wrench is pivoted 0.250 \(\mathrm{m}\) from its center of mass and allowed to swing as a physical pendulum. The period for small-angle oscillations is 0.940 s. (a) What is the moment of inertia of the wrench about an axis through the pivot? (b) If the wrench is initially displaced 0.400 rad from its equilibrium position, what is the angular speed of the wrench as it passes through the equilibrium position?

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 \(\mathrm{m}\) below its equilibrium position and released, it vibrates with a period of 4.20 \(\mathrm{s}\) . (a) What is its speed as it passes through the equilibrium position? (b) What is its acceleration when it is 0.050 \(\mathrm{m}\) above the equilibrium position? (c) When it is moving upward, how much time is required for it to move from a point 0.050 \(\mathrm{m}\) below its equilibrium position to a point 0.050 \(\mathrm{m}\) above it? (d) The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?

A thrill-seeking cat with mass 4.00 \(\mathrm{kg}\) is attached by a harness to an ideal spring of negligible mass and oscillates verticallyin SHM. The amplitude is \(0.050 \mathrm{m},\) and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies when the cat is (a) at its highest point; ( \((b)\) at its lowest point; (c) at its equilibrium position.

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